tag:blogger.com,1999:blog-37414104180967168272020-09-18T02:46:52.754-07:00Unizor - Creative Mind through Art of MathematicsUnizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.comBlogger437125tag:blogger.com,1999:blog-3741410418096716827.post-31010889866327480662020-09-08T09:07:00.001-07:002020-09-08T09:07:47.172-07:003-Phase AC Problem: UNIZOR.COM - Physics4Teens - Electromagnetism - Alte...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Problems on AC Induction</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />Three-phase generator has four wires coming out from it connected to a "star" with three <b>phase</b> wires carrying sinusoidal EMF shifted by 120°=2π/3 from each other and one <b>neutral</b> wire.<br /><br />The angular speed of a generator's rotor is <i><b>ω</b></i>.<br /><br />Assume that the peak difference in electric potential between each phase wire and a neutral one is <i><b>E</b></i>.<br /><br />Describe the difference in electric potential between each pair of phase wires as a function of time.<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />The difference in electric potential between phase wires and a neutral one can be described as<br /><br />Phase 1: <i><b>E<sub>1</sub>(t)=E·sin(ωt)</b></i><br /><br />Phase 2: <i><b>E<sub>2</sub>(t)=E·sin(ωt−2π/3)</b></i><br /><br />Phase 3: <i><b>E<sub>3</sub>(t)=E·sin(ωt+2π/3)</b></i><br /><br />The difference in electric potential between phase 1 wire and phase 2 wire can be represented as<br /><br /><i><b>E<sub>1,2</sub>(t) = E<sub>1</sub>(t) − E<sub>2</sub>(t)</b></i><br /><br />Similarly, the difference in electric potential between two other pairs of phase wires is<br /><br /><i><b>E<sub>2,3</sub>(t) = E<sub>2</sub>(t) − E<sub>3</sub>(t)</b></i><br /><br /><i><b>E<sub>3,1</sub>(t) = E<sub>3</sub>(t) − E<sub>1</sub>(t)</b></i><br /><br />Let's calculate all these voltages.<br /><br /><i><b>E<sub>1,2</sub>(t) = E·sin(ωt) − E·sin(ωt−2π/3) =<br /><br />= E·sin((ωt−π/3)+π/3) − E·sin((ωt−π/3)−π/3) =</b></i><br /><br />[substitute <i>φ=ωt−π/3</i>]<br /><br /><i><b>= E·sin(φ+π/3)−E·sin(φ−π/3) =<br /><br />= E·</b></i>[<i><b>sin(φ+π/3)−sin(φ−π/3)</b></i>]<br /><br />Let's simplify the trigonometric expression.<br /><br /><i><b>sin(φ+π/3) − sin(φ−π/3) =<br /><br />= sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) −<br /><br />− sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) =<br /><br />= 2cos(φ)·sin(π/3) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(φ) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(ωt−π/3)</b></i><br /><br />Therefore,<br /><br /><i><b>E<sub>1,2</sub>(t) = E√<span style="text-decoration: overline;">3</span>cos(ωt−π/3)</b></i><br /><br />As we see, the electric potential between phase 1 and phase 2 wires is also sinusoidal (since <i>cos(x)=sin(x+π/2), cos(ωt−π/3)</i> equals to <i>sin(ωt+π/6)</i>),<br /> but shifted in time, and its peak voltage is greater than the peak <br />voltage between a phase wire and a neutral one by a factor of √<span style="text-decoration: overline;">3</span>.<br /><br />Similar factor difference of √<span style="text-decoration: overline;">3</span> is between <b>effective voltages</b> of these pairs of wires.<br /><br />Analogous calculations for the other pairs of phase wires produce the following.<br /><br /><i><b>E<sub>2,3</sub>(t) = E·sin(ωt−2π/3) − E·sin(ωt+2π/3) =</b></i><br /><br />[substitute <i>φ=ωt</i>]<br /><br /><i><b>= sin(φ)·cos(2π/3) −<br /><br />− cos(φ)·sin(2π/3) −<br /><br />− sin(φ)·cos(2π/3) −<br /><br />− cos(φ)·sin(2π/3) =<br /><br />= −2cos(φ)·sin(2π/3) =<br /><br />= −√<span style="text-decoration: overline;">3</span>cos(φ) =<br /><br />= −√<span style="text-decoration: overline;">3</span>cos(ωt)</b></i><br /><br />Therefore,<br /><br /><i><b>E<sub>2,3</sub>(t) = −E√<span style="text-decoration: overline;">3</span>cos(ωt)</b></i><br /><br />Also the same factor difference of √<span style="text-decoration: overline;">3</span> relative to phase/neutral voltage.<br /><br />Finally, the third phase/phase voltage calculations produce the following.<br /><br /><i><b>E<sub>3,1</sub>(t) = E·sin(ωt+2π/3) − E·sin(ωt) =<br /><br />= E·sin((ωt+π/3)+π/3) − E·sin((ωt+π/3)−π/3) =</b></i><br /><br />[substitute <i>φ=ωt+π/3</i>]<br /><br /><i><b>= E·sin(φ+π/3)−E·sin(φ−π/3) =<br /><br />= E·</b></i>[<i><b>sin(φ+π/3)−sin(φ−π/3)</b></i>]<br /><br />Let's simplify the trigonometric expression.<br /><br /><i><b>sin(φ+π/3) − sin(φ−π/3) =<br /><br />= sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) −<br /><br />− sin(φ)·cos(π/3) +<br /><br />+ cos(φ)·sin(π/3) =<br /><br />= 2cos(φ)·sin(π/3) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(φ) =<br /><br />= √<span style="text-decoration: overline;">3</span>cos(ωt+π/3)</b></i><br /><br />Therefore,<br /><br /><i><b>E<sub>3,1</sub>(t) = E√<span style="text-decoration: overline;">3</span>cos(ωt+π/3)</b></i><br /><br />Again, the same factor difference of √<span style="text-decoration: overline;">3</span> relative to phase/neutral voltage.<br /><br />CONCLUSION<br /><br />The voltage between any two phase wires is in magnitude greater than phase to neutral voltage by √<span style="text-decoration: overline;">3</span>. Both are sinusoidal, but one is shifted in time relatively to another.<br /><br />The phase to phase <b>effective voltage</b> (which is by √<span style="text-decoration: overline;">2</span> less then peak voltage) also is greater than phase to neutral by the same √<span style="text-decoration: overline;">3</span>.<br /><br />EXAMPLES<br /><br />If phase to neutral effective voltage is 127V, the phase to phase effective voltage is 220V.<br /><br />If phase to neutral effective voltage is 220V, the phase to phase effective voltage is 380V.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-74959013741698563602020-08-27T08:59:00.001-07:002020-08-27T08:59:06.657-07:00Three-phase AC: UNIZOR.COM - Physics4Teens - Electromagnetism - Alternat...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Three Phases AC</u><br /><br /><br /><br /><i>The Basic Principle<br /><br /> of 3-Phase AC Generation</i><br /><br /><br /><br />Recall the process of generating alternating current (AC) using a pair <br />of permanent magnets and a wire frame (a coil) rotating around the axis <br />perpendicular to magnetic field lines.<br /><br />The pictures below represent the schematic design of such a system and a<br /> graph of an electromotive force (EMF) generated between the ends <i>a<sub>1</sub></i> and <i>a<sub>2</sub></i> of the wire frame<br /><br /><i><b>E<sub>a<sub>1</sub>a<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i>E<sub>max</sub></i> is the maximum absolute value of EMF,<br /><br /><i>ω</i> is the angular speed of rotation of the wire frame,<br /><br /><i>t</i> is time.<br /><br /><img src="http://www.unizor.com/Pictures/3phase1.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/3phase1sin.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br />An obvious improvement to this design is to use the power of rotation <br />more efficiently by having two wire frames on the same axis positioned <br />perpendicularly to each other. In this case we can have two independent <br />sources of EMF with the only difference of one of them to be shifted in <br />time relatively to another by 1/2 of the time of rotation.<br /><br />This shift is related to a simple fact that at the moment one wire <br />frame, aligned along the magnetic line, crosses these magnetic lines <br />with the highest rate, while another wire frame, that is perpendicular <br />to magnetic field lines, moves along these lines without actual <br />crossing. Then the roles are changed, as the coils rotate.<br /><br />The EMF between the ends <i>b<sub>1</sub></i> and <i>b<sub>2</sub></i> of the second wire frame will then be<br /><br /><i><b>E<sub>b<sub>1</sub>b<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt−π/2)</b></i><br /><br /><img src="http://www.unizor.com/Pictures/3phase2.png" style="height: 130px; width: 200px;" /><br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/3phase2sin.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br />Why stop at two wire frames? Let's have three coils positioned at 120° <br />relative to each other. Now we will have three independent sources of <br />EMF shifted in time from each other by 1/3 of the time of rotation (<b>phase</b><br /> shift) - the time needed by one wire frame to take the position between<br /> the magnet poles, previously taken by another wire frame.<br /><br />Three different EMF, therefore, will be equal to<br /><br /><i><b>E<sub>a<sub>1</sub>a<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt)</b></i><br /><br /><i><b>E<sub>b<sub>1</sub>b<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt−2π/3)</b></i><br /><br /><i><b>E<sub>c<sub>1</sub>c<sub>2</sub></sub> = E<sub>max</sub>·sin(ωt−4π/3) = E<sub>max</sub>·sin(ωt+2π/3)</b></i><br /><br /><img src="http://www.unizor.com/Pictures/3phase3.png" style="height: 140px; width: 200px;" /><br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/3phase3sin.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br /><br /><br /><i>Practical Implementation</i><br /><br /><br /><br />The design of a three phase generator, as depicted above, is just the <br />first try of an idea. If the magnet is fixed and three wire frames <br />(coils) are rotating between its poles, it presents a problem to connect<br /> these coils to transmit the generated electricity to consumers, we need<br /> sliding contacts, brushes and other impractical devices.<br /><br /><br /><br />In real life generators the three coils make up a stator - a fixed part <br />of a generator, while the magnet is rotating inside a circle of coils by<br /> external power (like steam, water, wind etc.), generating the <br />alternating current in the coils, which allows to make an electric <br />connection to coils fixed.<br /><br /><br /><br />At the first glance, three coils have three pairs of connections with <br />sinusoidal EMF generated in each pair and, to transfer AC electricity <br />from all coils to consumers, we seem to need three pairs of wires, two <br />from each coil - six wires altogether. This, however, can be improved by<br /> using the following technique.<br /><br />Let's connect ends <i>a<sub>2</sub></i>, <i>b<sub>2</sub></i> and <i>c<sub>2</sub></i><br /> of three coils together (see a picture below, it's a black wire at the <br />bottom connected to a black circle going around all coils) and see what <br />kind of resulting voltage will be observed on each end of the coils. <br />This is called a <b>star</b> connection of the generator's coils.<br /><br /><img src="http://www.unizor.com/Pictures/3phaseLoad.jpg" style="height: 160px; width: 200px;" /><br /><br />We know that the electric potential (EMF) on each of the above contacts <br />has a sinusoidal magnitude with a time shift by 1/3 of a period relative<br /> to each other. When we connect these three contacts, the potential at <br />the joint will be<br /><br /><i><b>E<sub>0</sub> = E<sub>max</sub>·sin(ωt) + E<sub>max</sub>·sin(ωt−2π/3) + E<sub>max</sub>·sin(ωt+2π/3) = E<sub>max</sub>·X</b></i><br /><br />where<br /><br /><i><b>X = sin(ωt) + sin(ωt−2π/3) + sin(ωt+2π/3) =<br /><br />= sin(ωt) + sin(ωt)·cos(2π/3) − cos(ωt)·sin(2π/3) + sin(ωt)·cos(2π/3) + cos(ωt)·sin(2π/3) =<br /><br />= sin(ωt) + sin(ωt)·(−1/2) − cos(ωt)·(√<span style="text-decoration: overline;">3</span>/2) + sin(ωt)·(−1/2) +<br />cos(ωt)·(√<span style="text-decoration: overline;">3</span>/2) = 0</b></i><br /><br />Therefore, <i><b>E<sub>0</sub> = 0</b></i><br /><br />There will be no difference in electrical potential between the joint and the ground.<br /><br /><br /><br />This fact enables to transmit all three phases of generated electricity along four wires - one from <i>a<sub>1</sub></i> (<b>phase 1</b>) contact, one from <i>b<sub>1</sub></i> (<b>phase 2</b>) contact, one from <i>c<sub>1</sub></i> (<b>phase 3</b>) contact and one <b>neutral</b> from a joint connection to <i>a<sub>2</sub></i>, <i>b<sub>2</sub></i> and <i>c<sub>2</sub></i>.<br /><br />The <b>neutral</b> wire is usually grounded since its electric potential is equal to zero.<br /><br /><br /><br />With this arrangement we still have an advantage of having three <br />independent phases of alternating current, but we need only four wires <br />to transmit it - three <b>phase</b> wires and one <b>neutral</b>.<br /><br />Connecting any device to any phase and a neutral wires, we will get a closed circuit with AC running in it.<br /><br /><br /><br /><br /><br /><i>Energy Consideration</i><br /><br /><br /><br />Obviously, putting two or three coils in a stator of a generator doubles<br /> or triples the energy output carried by outgoing wires. The Law of <br />Energy Conservation must work, so where is the energy is coming from?<br /><br /><br /><br />Recall the electromagnetic induction experiment described in the lecture<br /> "Faraday's Law" in the "Electromagnetic Induction" chapter of this <br />course with a wire moving in the uniform magnetic field.<br /><br /><img src="http://www.unizor.com/Pictures/Faraday.png" style="height: 111px; width: 200px;" /><br /><br />Since we physically move wire's electrons in one direction <br />perpendicularly to magnetic field lines, the Lorentz force pushes them <br />perpendicularly to both, the direction of the movement of a wire and the<br /> direction of the magnetic field lines, that is, along the wire, thereby<br /> creating an electric current between wire ends.<br /><br /><br /><br />Now electrons are moving with a wire in one direction and along the wire in another.<br /><br />The first movement maintains the electric current in the wire, but the <br />second, again, is a subject of the Lorentz force that pushes the <br />electrons perpendicularly to their direction, that is opposite to the <br />original direction of a wire movement.<br /><br />This force resists the movement of a wire in its original direction. We <br />have to perform work against this force to move the wire.<br /><br /><br /><br />Similar considerations are true in a case of a circular movement of a <br />wire frame in a magnetic field or, if wire coils are in a stator, the <br />force is needed to rotate the magnet in a rotor. That is, we have to <br />spend energy to generated the electricity, the rotor's rotation is <br />possible only if we apply the force against the Lorentz forces resisting<br /> this rotation. The magnetic field generated by the electric current in a<br /> wire coil of a stator resists the rotation of a magnet in a rotor.<br /><br /><br /><br />If we have more than one coil in a stator, each one resists the rotation<br /> of a rotor, so we have to spend proportionally more effort to rotate <br />the rotor.<br /><br />The Law of Energy Conservation works. The more coils we have in a stator<br /> - the more electricity is generated, but the more resistance to a <br />rotor's rotation needs to be overcome.<br /><br /><br /><br /><br /><br /><i>Three Phase AC Motor</i><br /><br /><br /><br />The lecture "AC Motors" of this chapter described the necessity of having a rotating magnetic field to make an AC motor.<br /><br />To achieve such a rotating field we had to resort to artificially create<br /> a second AC current with a phase shift by 90° using a capacitor or a <br />transformer.<br /><br />Most of household AC motors (like in a fan) work on this principle, they need only two wires, which are, as we can say now, a <b>phase</b> wire and a <b>neutral</b> one.<br /><br /><br /><br />Powerful industrial level AC motors (like in a water pump that works in a<br /> tall building to pump water to the roof tank) needs more power, and we <br />can use all three phases to create a rotating magnetic field.<br /><br /><br /><br />So, all four wires coming from the AC generator, three <b>phase</b> wires and one <b>neutral</b><br /> one go into an AC motor, whose principal construction very much <br />resembles the one described in the previous lecture. The only difference<br /> is, we already have three wires with AC phase shifted by 120° <br />relatively to each other. So, we have to position three wire coils in a <br />stator at 120° angles to each other, connect one end of each coil to a <br />corresponding <b>phase</b> wire and another end - to a common <b>neutral</b><br /> wire, and the rotating field is ready. Then it will work pretty much as<br /> it was described in the "AC Motors" lecture, but smoother because three<br /> phases make a smoother rotation of a magnetic field than two phases.<br /><br /><br /><br />At the end I would like to say again, that it was Nicola Tesla's genius <br />that created all the basic principles, based on which all the current AC<br /> motors are working now. His contributions to our industrial development<br /> are grossly underappreciated. Calling an electric car model "Tesla" is a<br /> late but well deserved tribute to his creativity.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-3118095052187790112020-08-23T13:14:00.001-07:002020-08-23T13:14:08.114-07:00AC MotorsUNIZOR.COM - Physics4Teens - Electromagnetism - Alternating Cur...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Alternating Current Motors</u><br /><br /><br /><br /><i>Alternating current (AC)</i> motors are used where the sources of electricity produce alternating current - in our homes, at industrial facilities etc.<br /><br />For example, AC motors work in refrigerators, water pumps, air conditioners, large fans and many other devices.<br /><br /><br /><br />The idea of AC motors belongs to Nicola Tesla, who invented it at the end of 19th century.<br /><br />And it was a brilliant idea!<br /><br /><br /><br />Before talking about AC motors, let's recall how DC motors are working. <br />Their basic design was presented in a lecture Physics 4 Teens - <br />Electromagnetism - Magnetism of Electric Current - DC Motors.<br /><br /><br /><br />The important feature of DC motor, which was the starting idea behind <br />its design, was the rotational momentum exerted by a magnetic field of a<br /> permanent magnet onto a wire frame with the direct electric current <br />going through it due to Lorentz force. Without a <i>commutator</i> or <br />any electronic switches the rotating frame would rotate until its plane <br />will reach a perpendicular position relative to magnetic lines, which <br />will be its equilibrium position.<br /><br /><br /><br />Let's start with this idea of an AC motor and try to make it work.<br /><br /><br /><br />Firstly, as in the case of DC motors, let's replace the permanent magnet<br /> in a stator with two coils around an iron cores - electromagnets that <br />create magnetic field. Notice, however, that the current in these <br />electromagnets is alternating, which means that the magnetic field <br />between these electromagnets is alternating as well in magnitude and <br />direction.<br /><br /><br /><br />Secondly, we will place a permanent magnet on an axis, so it freely rotates between the poles of the electromagnets<br /><br /><br /><br />What will happen if we switch on the alternating current in the electromagnets?<br /><br />Well, nothing noticeable. The problem is, the polarity of electromagnets<br /> will start switching with frequency of the AC - 50 or 60 oscillations <br />per second in usual commercial wiring, and a permanent magnet in-between<br /> will be forced in two opposite directions with the same frequency and <br />it will not start rotating.<br /><br /><br /><br />But let's manually start rotating the magnet in any direction strong <br />enough to force it to rotate with significant angular speed.<br /><br />At different positions the variable external magnetic field of <br />electromagnetic stator interferes with rotating permanent magnet rotor, <br />sometimes slowing it, sometimes speeding its rotation.<br /><br />In a short while the rotation of the permanent magnet rotor will <br />synchronize with variable external field of electromagnetic stator and <br />rotation will be maintained with the same angular speed as the frequency<br /> of AC in the coils of electromagnets.<br /><br /><br /><br />Let's analyze this rotation in steps after the synchronization is achieved.<br /><br />Let a pole facing the rotor of one electromagnet be X and an opposite pole of another electromagnet be Y.<br /><br />As AC changes its direction and magnitude, pole X is gradually changing <br />from North (X=N) to zero (X=0), to South (X=S), again to zero (X=0), <br />again to North (X=N) etc.<br /><br />At the same time pole Y of an opposite electromagnet is gradually <br />changing from South (Y=S) to zero (Y=0), to North (Y=N), again to zero <br />(Y=0), again to South (X=N) etc.<br /><br /><br /><br />Synchronously rotating permanent magnet of a rotor should with its North<br /> pole approach X=S and, simultaneously, with its South pole approach <br />Y=N.<br /><br /><img src="http://www.unizor.com/Pictures/ACelectromagnetSN.jpg" style="height: 150px; width: 200px;" /><br /><br />As a rotor approaches with its poles the poles of a stator, X and Y <br />should gradually weaken and, when the permanent magnet is fully aligned <br />along XY line, the magnitudes of the magnetic fields of electromagnets <br />should diminish to zero (X=0, Y=0).<br /><br /><br /><br />Permanent magnet rotor will pass this point by inertia and the polarity of the electromagnets switches, so now X=N and Y=S.<br /><br /><img src="http://www.unizor.com/Pictures/ACelectromagnetNS.jpg" style="height: 150px; width: 200px;" /><br /><br />That causes rotation to continue until the permanent magnet of a rotor again takes a position along XY line.<br /><br />By that time the electromagnets will be at X=0 and Y=0, rotor will <br />continue rotation by inertia, then AC switches the polarity of <br />electromagnets again and rotation continues in a similar manner <br />indefinitely.<br /><br /><br /><br />Our first version of an AC motor is functional, but it has two obvious disadvantages.<br /><br />One disadvantage is the usage of permanent magnet, they are very <br />expensive. We could not avoid it in a DC motor, but in an AC case we <br />might think that variable magnetic field might help to use the induction<br /> effect to avoid it.<br /><br />Another disadvantage is that switching the AC on does not really start <br />the rotation, we manually started it, and only then, if we gave a strong<br /> push, it started to rotate and maintained this rotation.<br /><br /><br /><br />Let's use a wire loop instead of a permanent magnet rotor. But, to act <br />as a permanent magnet, it needs an electric current running through the <br />wire, and we don't want any extra sources of electricity to connect to <br />it, it's complicated, needs a commutator, like in the original DC <br />motors.<br /><br />Instead, we will count on the induction effect created by an alternating<br /> current in a stator, which creates an alternating magnetic field, <br />which, in turn, induces the electric current in the wire loop of a <br />rotor.<br /><br /><br /><br />The induced electric current in a wire loop of our rotor appears when a <br />wire crosses the magnetic field lines. Magnetic field produced by a <br />stator is directed always along a center line XY and is changing in <br />magnitude from a maximum in one direction (from X to Y) to zero, to a <br />maximum in another direction (from Y to X), again to zero etc.<br /><br /><br /><br />The problem is, if the rotor is standing still, its wire does not cross <br />magnetic field lines. Therefore, no electric current will be produced in<br /> it, and there will be no rotation. Rotor needs an initial push <br />sufficient to synchronize its rotation with the frequency of alternating<br /> magnetic field to continue the rotation. Our second disadvantage is <br />still unresolved.<br /><br /><br /><br />Let's imagine that, besides two main electromagnets with poles X and Y, <br />we have another pair of auxiliary electromagnets in a stator, that are <br />positioned perpendicularly to line XY. Let their poles be A and B and <br />(very important!) the current in them, also sinusoidal, is shifted in <br />time relatively to a current in main electromagnets by 90°.<br /><br />So, when the magnetic field in one direction is maximum along line XY, <br />it's zero along line AB; then it gradually decreases along XY and <br />increases along AB until it's zero along XY and maximum along AB. This <br />cycle repeats itself indefinitely.<br /><br /><br /><br />Graphically, it can be represented as follows.<br /><br /><img src="http://www.unizor.com/Pictures/ACrotation.png" style="height: 300px; width: 200px;" /><br /><br />Graphs on the left and on the right of a picture represent the electric <br />current in each pair of electromagnets - XY and AB. Notice, they are <br />shifted by π/2=90°. The middle part of a picture represents the <br />direction of the magnetic field created by both pairs of electromagnets <br />of a stator.<br /><br />As the time goes, the magnetic field direction is rotating. The rotating<br /> magnetic field around a closed wire loop of a rotor causes the rotor's <br />wire to cross the magnetic field lines of a stator, which, in turn, <br />causes induction of electric current in a rotor's closed wire loop, <br />which will act now as a permanent magnet, which is supposed to follow <br />the rotation of the magnetic field around it.<br /><br />This causes the rotation of the rotor. The rotor's actual physical <br />rotation will follow the stator's magnetic field rotation, created <br />without any physical movement.<br /><br /><br /><br />The rotor cannot go after the stator's magnetic field rotation exactly <br />synchronously because then there will be no crossing of magnetic field <br />lines. So, the rotor, after it reached the same speed as the stator's <br />magnetic field rotation, will lose its rotational momentum and slow <br />down. Then the rotor's wire, rotating slower than the magnetic field <br />around it, will cross the magnetic field lines, there will be induction <br />current in it, it's magnetic properties will be restored and it will try<br /> to catch up with the stator's magnetic field rotation. This process <br />will continue as long as AC is supplied.<br /><br /><br /><br />An obvious improvement is to use multiple wire loops as a rotor with <br />common axis of rotation. That will cause more uniform rotation.<br /><br /><br /><br />What's remaining in our project of designing the AC motor is to create <br />another alternating current for the second pair of electromagnets in a <br />stator with a time delay of π/2=90° relatively to the main AC.<br /><br />This problem is resolved and described in the previous lecture about AC <br />capacitors. If we introduce another circuit fed from the same AC source,<br /> but with a capacitor in it, this circuit will have the alternating <br />current shifted in time exactly as we want. This current will go through<br /> the second pair of electromagnets and both pairs will create a <br />revolving magnetic field.<br /><br /><br /><br />The interaction between magnetic fields of a stator and a rotor is quite<br /> complex, when the rotor rotates. The magnetic flux going through a <br />rotor wire loop depends on a variable magnetic field of a stator and <br />variable area of a rotor wire loop in a direction of a magnetic field <br />lines of a stator. The exact calculations of this process are beyond the<br /> scope of this course.<br /><br />The main idea, however, is clear - <b>to create an AC motor we have to create a revolving magnetic field</b> - the great idea of Nicola Tesla. That can be accomplished by using known methods described above.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-75921051130765982892020-08-19T08:47:00.001-07:002020-08-19T08:47:43.493-07:00DC Motors: UNIZOR.COM - Physics4Teens - Electromagnetism - Magnetism of ...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Direct Current Motors</u><br /><br /><br /><br /><i>Direct current (DC)</i> motors are used where the sources of electricity produce direct current, like batteries.<br /><br />For example, DC motors work in car engine starter, computers, toys, drones and many other devices.<br /><br /><br /><br />We will concentrate on principles of their work without going into many <br />details. Basically, we will describe these motors as they were first <br />thought of by their inventors. Obviously, initial ideas were further <br />improved by many people and improvements are still introduced after more<br /> than 200 years after their invention, but the basic ideas are still <br />there.<br /><br /><br /><br />Let's start with the idea of the DC motor.<br /><br />The beginning of a development of a DC motor is associated with a simple<br /> experiment we described before - the one that demonstrated the Lorentz <br />force on a current in a magnetic field.<br /><br />If the direction of a current <i><b>I</b></i> is perpendicular to a direction of magnetic field lines <i><b>B</b></i>,<br /> the magnetic field "pushes" the conductor that carries this current in a<br /> direction perpendicular to both magnetic field lines and a current by <br />force <i><b>F</b></i>.<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce.jpg" style="height: 180px; width: 200px;" /><br /><br /><br /><br />The first modification of this experiment was to use a rectangular wire <br />frame instead of a straight line conductor and let it spin around an <br />axis. The force exerted by a magnetic field will act on both sides of a <br />wire frame that are perpendicular to magnetic lines in opposite <br />directions, creating a rotational momentum, so the frame with rotate.<br /><br /><img src="http://www.unizor.com/Pictures/WireLoopRotation.jpg" style="height: 120px; width: 200px;" /><br /><br />The force <i><b>F</b></i> of a magnetic field "pushes" segment <i><b>AB</b></i> up and segment <i><b>CD</b></i>, where the electric current goes in an opposite to segment <i><b>AB</b></i> direction, is "pushed" down by this force. This creates a rotational momentum.<br /><br /><br /><br />The problem is, the rotation will stop when the wire frame plane will be<br /> perpendicular to a direction of a magnetic field, because the forces <br />exerted by a magnetic field on both sides of a wire frame <i><b>AB</b></i> and <i><b>CD</b></i><br /> will no longer create a rotating momentum, they will act against each <br />other within the wire frame plane and after a short oscillation caused <br />by a rotational inertia the rotation stops.<br /><br /><br /><br />On the picture above, when the plane of a wire loop reaches the position perpendicular to the magnetic field force, the segment <i><b>AB</b></i> will be on the top and the magnetic field force <i><b>F</b></i> will "push" this segment up. Segment <i><b>CD</b></i> will be on the bottom and the magnetic field force <i><b>F</b></i><br /> will "push" this segment down. Both forces are acting within the same <br />plane with the axis of rotation and nullify each other. No rotational <br />momentum is created.<br /><br /><br /><br />Our next improvement is related to overcoming this problem.<br /><br />What happens if exactly at the moment when our wire frame plane is <br />perpendicular to magnetic field, when the forces exerted by a magnetic <br />field no longer create a rotational moment, we switch off the electric <br />current in a wire and a very short moment later we switch it on, but in <br />the opposite direction of the electric current?<br /><br />First of all, the rotation will continue by inertia during the time <br />electricity is off. Then, we turn the electric current on but in <br />opposite direction. That means, the direction of the magnetic field <br />force will change to an opposite. Segment <i><b>AB</b></i> will be "pushed" down, segment <i><b>CD</b></i><br /> will be "pushed" up. Since the wire frame has passed the point of <br />perpendicularity to magnetic field by rotational inertia during electric<br /> current off time, the newly formed magnetic field forces will create a <br />rotational moment and the direction of rotation will be the same as <br />before.<br /><br /><br /><br />Our frame will continue the rotation in the same direction until segment <i><b>AB</b></i> will be on the bottom and segment <i><b>CD</b></i> - on the top.<br /><br />At that moment we will do the same switching off the electric current to<br /> let our frame pass the perpendicular position towards the magnetic <br />field force, and a very short moment later we switch electric current on<br /> in an opposite direction. Now segment <i><b>AB</b></i> will be "pushed" up again, segment <i><b>CD</b></i> will be "pushed" down, which will allow to complete the cycle of rotation and the rotation will continue in the same direction.<br /><br /><br /><br />All we need is to explain how to switch the direction of an electric current.<br /><br />Here is a drawing of this simple device called <i>commutator</i>.<br /><br /><img src="http://www.unizor.com/Pictures/DCcommutator.jpg" style="height: 111px; width: 200px;" /><br /><br />Direct current comes through brushes to a commutator and to a revolving <br />frame connected to it. As this construction revolves, the brushes lose <br />the contact with a commutator for a very short period, then again touch <br />the commutator, but with opposite poles.<br /><br />That's how we change the direction of the current that forces the frame to constantly rotate in the same direction.<br /><br /><br /><br />The inner rotating part of this type of a DC motor is called <i>rotor</i>, the outer stationary part with a magnet is called <i>stator</i>.<br /><br />The detail implementation of this design of a DC motor is outside the scope of this course. Our purpose is to convey the idea.<br /><br /><br /><br />The weak part of a DC motor design with brushes is that these brushes <br />wear out with time and are the source of sparks, which might be <br />prohibitive in some environments. Plus, they are noisy.<br /><br />With development of electronics inventors came up with a better design <br />that does not involve brushes at all, rotation is accomplished without <br />any mechanical switches.<br /><br /><br /><br />First step on the design of this brushless DC motor is to realize that <br />we have to invert the functionality of permanent magnet and a wire <br />frame. The need for a <i>commutator</i> with mechanical brushes was related to the fact that electric circuit was rotating.<br /><br />Let's invert roles and put two parallel wire frames on opposite ends of a circle as a <i>stator</i> and a permanent magnet on an axis in the middle as a <i>rotor</i>.<br /><br />Recall that a wire loop with direct current running through it acts like a magnet<br /><br /><img src="http://www.unizor.com/Pictures/LoopMagnet.jpg" style="height: 200px; width: 200px;" /><br /><br />Putting two such loops on opposite ends of a stator creates a magnetic <br />field and a freely rotating permanent magnet in between will have to <br />turn to align itself with the field of these two wire loops<br /><br /><img src="http://www.unizor.com/Pictures/DCmotorBrushless.png" style="height: 200px; width: 200px;" /><br /><br />If at the moment our permanent magnet aligns with magnetic field of both<br /> wire loops we change the direction of electric current in the loops to <br />an opposite, thus changing the polarity of the magnetic field they <br />generate, the permanent magnet in between will have to continue the <br />rotation until it will align again. Repeating this cycle creates the <br />rotation of a permanent magnet.<br /><br /><br /><br />This rotation will not be smooth. The force of magnetic field attraction<br /> will create a stronger moment of rotation when the direction of the <br />permanent magnet is perpendicular to a direction of the magnetic field <br />of the loops. This can be improved by certain additional details <br />described below.<br /><br /><br /><br />The changing of the direction of the current in the wire loops now is <br />much easier than in the previous design because the wires are not <br />moving. Simple electronic switch working off some kind of a marker on a <br />rotating permanent magnet can signal its position and trigger the switch<br /> of direction of the current.<br /><br /><br /><br />Basically, the idea is finished here. But some very important improvements should be mentioned.<br /><br /><br /><br />1. Instead of a single wire loop we can use a copper coil around an iron<br /> core to make the magnetic field of this electromagnet stronger.<br /><br /><br /><br />2. Instead of a pair of such electromagnets with switching the direction<br /> after the permanent magnet in the middle turns by 180°, we can use two <br />pairs and properly engage another pair after 90° turn, switching off the<br /> previous pair, thus creating a rotating magnetic field that results in a<br /> smoother movement of a permanent magnet rotor. Even better, we can use <br />three pairs of electromagnets and engage another pair after each 60°, <br />switching off the previous two pairs. This will allow even smoother <br />rotation. In some DC motors they use even six pairs of electromagnets, <br />which results in a very uniform rotation with practically constant <br />angular speed.<br /><br />The electronic switches that engage and disengage the coils can be designed for any type of coil arrangement.<br /><br /><br /><br />3. Instead of a bar permanent magnet inside the circle of electromagnets<br /> we can use a ring permanent magnet outside it. It creates a better <br />response to a revolving magnetic field of a stator, the rotation will be<br /> smoother because of inertia of a rotating ring.<br /><br /><br /><br />With all the above improvements the DC motor used in hard disk of computers and other devices looks like the one below<br /><br /><img src="http://www.unizor.com/Pictures/DCmotorReal.jpg" style="height: 480px; width: 210px;" /><br /><br />This DC motor has 9 electromagnets sequentially engaged after each 40° turn of a rotor.<br /><br />Rotor is a ring-shaped permanent magnet that rotates around the electromagnets.<br /><br />A marker on the rotor sends a signal to an electronic switch to indicate<br /> the position of a rotor, which is used by electronics on the attached <br />board to properly engage the electromagnets of a stator.<br /><br /><br /><br />The most important detail of this design of a DC motor is the rotating <br />magnetic field achieved through purely electronic means without any <br />moving parts. This results in a smooth rotation of a rotor made of a <br />ring permanent magnet around a stator.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-82782206556858281582020-08-15T13:26:00.001-07:002020-08-15T13:26:02.359-07:00AC Capacitors: UNIZOR.COM - Physics4Teens - Electromagnetism - AC Induction<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/-q4WLYIEdqw" width="480"></iframe><br /><br /><br /><br /><u><i>Notes to a video lecture on http://www.unizor.com</i></u><br /><br /><br /><br /><u>Alternating Current and Capacitors</u><br /><br /><br /><br />Let's start with a clear illustration of what we will discuss in this lecture.<br /><br /><img src="http://www.unizor.com/Pictures/ACcapacitor.png" style="height: 300px; width: 200px;" /><br /><br />The main idea is that alternating current (AC) goes through a capacitor, while direct current (DC) does not.<br /><br /><br /><br />To explain this fact, consider how electricity is produced in both cases DC and AC.<br /><br /><br /><br />For DC case we will use a battery that uses some chemical reaction to <br />separate some outer layer ("free") electrons from their atoms and moves <br />them toward a battery's negative pole, where excess of electrons will be<br /> present, leaving deficiency of electrons on the positive pole.<br /><br /><br /><br />If we connect the poles of this battery to two plates of a capacitor, <br />the excess of electrons from the negative pole of a battery will travel <br />to one plate, while the deficiency of electrons will be on the other <br />plates. It will continue until we reach a saturation of electrons on one<br /> plate and their absence on another. The electromotive power of the <br />battery could not push electrons anymore to one plate, because their <br />repulsive force will be too strong. And that's it. Electrons will not <br />jump from one plate of a capacitor to another (unless the EMF of a <br />battery is so strong that it breaks the isolation between the plates of a<br /> capacitor, which defeats the purpose), the circuit is not closed, the <br />current will not go through.<br /><br /><br /><br />In AC case the generated electromotive force (EMF) is variable in magnitude and direction.<br /><br /><br /><br />It starts with separating "free" electrons from their atoms, gradually <br />increasing the concentration of electrons on one pole and on the <br />capacitor's plate connected to it.<br /><br />At the same time deficiency of electrons is observed on the other pole and a capacitor's plate connected to it.<br /><br /><br /><br />That lasts for some short period of time (half a period), when the <br />generated EMF gets weaker and goes down to zero. This causes the excess <br />electrons from one capacitor plate to go back to a generator and <br />compensate the deficiency of electrons on the other plate. This ends <br />half a period of EMF oscillation.<br /><br /><br /><br />On the next half a period the EMF changes the sign, electrons will be <br />accumulated on the plate, where previously we observed their deficiency.<br /> On the plate where we had excess of electrons during the previous half a<br /> period we will have their deficiency.<br /><br /><br /><br />So, during the full cycle of two half-periods electrons move to one <br />plate of a capacitor, then back to a generator and to another plate of a<br /> capacitor. Then the cycle repeats itself, electrons move within a <br />circuit with a capacitor back and forth, thus facilitating the <br />alternating current in this circuit.<br /><br /><br /><br />Let's compare the alternating current in a case of a closed circuit without a capacitor with a circuit that has a capacitor.<br /><br />We start in both cases with a device that generates alternating sinusoidal EMF<br /><br /><i><b>U(t) = U<sub>max</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>U(t)</b></i> is the generated EMF as a function of time <i><b>t</b></i>,<br /><br /><i><b>U<sub>max</sub></b></i> is the peak value of EMF,<br /><br /><i><b>ω</b></i> is the angular speed of rotation of a device generating the alternating EMF.<br /><br /><br /><br />In a closed circuit with an alternating EMF and no capacitors, according to the Ohm's Law, the electric current <i><b>I(t)</b></i>, as a function of time <i><b>t</b></i>, will be proportional to an EMF and will alternate synchronously with it<br /><br /><i><b>I(t) = I<sub>max</sub>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>I<sub>max</sub></b></i> is the peak value of electric current in a circuit.<br /><br /><br /><br />In case of a capacitor being a part of a circuit the situation is more complex.<br /><br />As we know, the amount of electricity <i><b>Q(t)</b></i> accumulated in the capacitor is proportional to voltage <i><b>U(t)</b></i> applied to its plates, and <i>capacity</i> of a capacitor <i><b>C</b></i><br /> (see lecture "Electric Fields" - "Capacitors" in this course) is the <br />constant proportionality factor that depends on a type of a capacitor<br /><br /><i><b>C = Q(t)/U(t)</b></i><br /><br /><br /><br />Therefore,<br /><br /><i><b>Q(t) = C·U(t) = C·U<sub>max</sub>·sin(ωt)</b></i><br /><br />Knowing the amount of electricity <i><b>Q(t)</b></i> accumulated in a capacitor as a function of time <i><b>t</b></i>, we can determine the electric current <i><b>I(t)</b></i> in a circuit, which is a rate of change (that is, derivative by time) of the amount of electricity<br /><br /><i><b>I(t) = </b>d<b>Q(t)/</b>d<b>t =<br /><br />= C·U<sub>max</sub>·ω·cos(ωt) =<br /><br />= I<sub>max</sub>·cos(ωt) =<br /><br />= I<sub>max</sub>·sin(ωt+π/2)</b></i><br /><br />where<br /><br /><i><b>I<sub>max</sub> = C·U<sub>max</sub>·ω</b></i><br /><br /><br /><br />Incidentally, expression which is called <i>reactance</i>, plays for a capacitor similar role to that of a resistance because, using this <i>reactance</i>, the formula for a peak value of the electric current in a circuit with a capacitor resembles the Ohm's Law<br /><br /><i><b>I<sub>max</sub> = U<sub>max</sub>/X<sub>C</sub></b></i><br /><br /><br /><br />What's most important in the formula <i><b>I(t)=I<sub>max</sub>·sin(ωt+π/2)</b></i><br /> and the most important property of a capacitor in an AC circuit is <br />that, while an EMF in a circuit oscillates with angular speed <i><b>ω</b></i>, the <b>electric current oscillates with the same angular speed <i><b>ω</b></i> as EMF, but its period is shifted in time by <i>π/2</i> relative to EMF</b>.<br /><br /><br /><br />This is a very important capacitor's property used in AC motors, which will be a subject of the next lecture.<br /><br /><nobr><i><b>X<sub>C</sub> = 1/(ω·C)</b></i>,</nobr>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-40804915874119424402020-08-13T09:44:00.001-07:002020-08-13T09:44:03.855-07:00AC Induction: UNIZOR.COM - Physics4Teens - Electromagnetism<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Alternating Current Induction</u><br /><br /><br /><br />The Faraday's Law of electromagnetic induction states that a <b>variable <i>magnetic flux Φ(t)</i> going through a wire loop generates in this wire loop an <i>electromotive force EMF=U(t)</i> equal in magnitude to a rate of change of the magnetic flux</b>:<br /><br /><i><b>U(t) = −</b>d<b>Φ(t)/</b>d<b>t</b></i><br /><br />where the meaning of the minus sign was explained in the lecture about <i>self-induction</i> previously in this course.<br /><br /><br /><br />There are different ways to generate a variable magnetic flux. For <br />example, we can physically move a permanent magnet through or around a <br />wire loop, which will generate EMF in the wire.<br /><br />But, having <i>alternating current (AC)</i> at our disposal and knowing <br />that any electric current creates a magnetic field around it, we can <br />create a variable magnetic field just by running AC through a wire.<br /><br /><br /><br />Let's start from a simple experiment just to demonstrate the principle of AC induction.<br /><br /><img src="http://www.unizor.com/Pictures/ACinduction.png" style="height: 225px; width: 200px;" /><br /><br />Here the bottom wire loop (we will call it <i>primary</i>) is connected to a source of an <i>alternating current</i>, while the top wire loop (<i>secondary</i>)<br /> has no source of electricity, but is connected to a device to measure <br />the effective electric current in it - an AC ammeter (or amp-meter, or <br />ampermeter).<br /><br /><br /><br />Since the AC in the primary loop is variable (more precisely, sinusoidal), the <i>intensity</i> of the magnetic field generated by it is also variable.<br /><br /><br /><br />Therefore, the <i>magnetic flux</i> going through the secondary loop is <br />variable (also sinusoidal) with the generated EMF (first derivative of <br />magnetic flux) not equal to zero and also sinusoidal since the <br />derivative of <i>sin(x)</i> is <i>cos(x)</i> and derivative of <i>cos(x)</i> is <i>−sin(x)</i>, as we know.<br /><br /><br /><br />Variable (sinusoidal) EMF in the secondary loop causes the variable <br />(sinusoidal) electric current in it, which we can observe on the AC <br />ammeter.<br /><br />We have just demonstrated the principle of <i>AC induction</i> - <br />transformation of the alternating electric current from one circuit <br />(primary loop) to another (secondary loop) without physical connection <br />between them.<br /><br /><br />Let's switch to more practical aspects of AC induction.<br /><br /><br /><br />Recall from the earlier presented lecture "Magnetism of Electric Current in a Loop" that the intensity <i><b>B</b></i> of a magnetic field at the center of a wire loop in a vacuum equals to<br /><br /><i><b>B = μ<sub>0</sub>·I/(2R)</b></i><br /><br />where<br /><br /><i><b>μ<sub>0</sub></b></i> is the <i>permeability</i> of free space,<br /><br /><i><b>I</b></i> is the electric current going through a wire loop (maybe variable as in the case of AC),<br /><br /><i><b>R</b></i> is the radius of a wire loop.<br /><br /><br /><br />If, instead of a vacuum, we have some other material inside a loop, the <br />intensity of a magnetic field will be different. The formula will be <br />almost the same, just the <i>permeability</i> of free space <i><b>μ<sub>0</sub></b></i> should be replaced with <i>permeability</i> of the corresponding material <i><b>μ</b></i>.<br /><br /><br /><br />Experimentally obtained data tell us that one of the greatest <i>permeability</i><br /> is that of so-called ferromagnetic materials - those that contain iron.<br /> Actually, pure iron has the permeability of about 200,000 greater than <br />vacuum. So, if we have a ferromagnetic material inside the primary wire <br />loop in our experiment, the magnetic energy will have a choice - to go <br />through highly resistant space around this ferromagnetic material or to <br />go through it with much less resistance.<br /><br /><br /><br />The general principle of field propagation is that the most energy goes <br />through least resistance. So, we can expect that ferromagnetic material <br />inside the wire loop will concentrate the magnetic field generated by a <br />primary wire loop and much less of the field energy will be dissipated <br />around it.<br /><br /><br /><br />Our first improvement to the experiment above is to use a cylinder made of <i>ferromagnetic</i> material going through both our loops.<br /><br /><img src="http://www.unizor.com/Pictures/ACinductionFerro.png" style="height: 225px; width: 200px;" /><br /><br />Ferromagnetic material has temporary magnetic properties and quickly <br />reacts to external magnetic field by rearranging the orientation of its <br />atoms. The variable magnetic field created by AC inside the primary loop<br /> will force the atoms inside this cylinder to react by orienting along <br />the vector of intensity of the magnetic field, thus increasing this <br />intensity.<br /><br />So, the purpose of this ferromagnetic cylinder going through both loops <br />is to increase the intensity of the magnetic field going through the <br />secondary wire loop without spending any more energy. The ferromagnetic <br />cylinder just concentrates and directs more of the energy of the <br />magnetic field to go through the secondary wire loop and less will be <br />dissipated outside the secondary loop.<br /><br /><br /><br />Still, the energy of magnetic field generated by the primary wire loop <br />and directed through the ferromagnetic cylinder through both loops will <br />be dissipated from the ends of this cylinder.<br /><br />To make our induction even more efficient, we can close the magnetic <br />field on the ends of a ferromagnetic cylinder as on this picture:<br /><br /><img src="http://www.unizor.com/Pictures/ACinductionClosed.png" style="height: 150px; width: 200px;" /><br /><br />Now the magnetic field is not dissipated outside the ferromagnetic <br />material, and almost all energy generated by the primary wire loop with <br />AC running through it is used to create the magnetic flux inside the <br />secondary wire loop.<br /><br /><br /><br />Our next practical improvement is related to the fact that the vector of<br /> intensity of a magnetic field is an additive function. That is, the <br />intensity of a combined magnetic field of two or more sources equals to a<br /> vector sum of intensities of all component fields.<br /><br /><br /><br />We can use multiple loops in the primary coil, which is a source of <br />magnetic field, instead of a single primary wire loop in our experiment.<br /><br />Let's wind <i>N<sub>P</sub></i> primary loops of wire in a coil around <br />our ferromagnetic core. Connected to the same source of AC and having <br />the same sinusoidal current going through them, each loop will create a <br />magnetic field and all these <i>N<sub>P</sub></i> fields will have to be added together to get the resulting magnetic field intensity.<br /><br /><br /><br />So, the generated magnetic flux of such a coil will be <i>N<sub>P</sub></i> times greater than the one by a single loop. Its inductive effect on the secondary wire loop will be <i>N<sub>P</sub></i> times stronger.<br /><br /><br /><br />As the next practical improvement of our experiment, let's use a coil of <i>N<sub>S</sub></i> wire turns instead of a single secondary loop.<br /><br />Each turn of this secondary winding will have the same magnetic flux <br />flowing through it and will induce an electromotive force (EMF) <br />proportional to a rate of change of magnetic flux. All these EMF in each<br /> turn of a secondary wire are sequential to each other and should be <br />added together to get the resulting EMF in the secondary circuit.<br /><br />So, the EMF induced in a secondary coil of <i>N<sub>S</sub></i> wire turns will be <i>N<sub>S</sub></i> stronger then the EMF induced in a single secondary loop.<br /><br /><br /><br />So, changing the number of turns in the primary wire coil <i>N<sub>P</sub></i>,<br /> we proportionally change the variable magnetic flux, which, in turn, <br />proportionally change the induced EMF in the secondary coil.<br /><br />Analogously, changing the number of turns in the secondary wire coil <i>N<sub>S</sub></i>, we proportionally change the induced EMF in this wire coil.<br /><br />These two facts lead us to use the device we have constructed to <br />transform the EMF of alternating current for whatever purpose we need.<br /><br /><br /><br />Indeed, if <i>N<sub>P</sub></i>=<i>N<sub>S</sub></i> and our device is <br />ideal in a sense that the energy does not dissipate and the coils are <br />not heated up because of internal resistance, the same magnetic flux <br />goes through the same coils, which means the EMF in both primary and <br />secondary coils is the same.<br /><br /><br /><br />If we want to induce the secondary EMF twice as stronger as the primary <br />one, we can double the number of turns in the secondary coil.<br /><br />If we want to induce the secondary EMF twice as weaker as the primary <br />one, we can reduce the number of turns in the secondary coil by half.<br /><br /><br /><br />This can be expressed as<br /><br /><i><b>U<sub>P</sub>(t)/U<sub>S</sub>(t) = N<sub>P</sub>/N<sub>S</sub></b></i><br /><br />where<br /><br /><i>U<sub>P</sub>(t)</i> is the EMF in the primary coil,<br /><br /><i>U<sub>S</sub>(t)</i> is the EMF in the secondary coil,<br /><br /><i>N<sub>P</sub></i> is the number of wire turns in the primary coil,<br /><br /><i>N<sub>S</sub></i> is the number of wire turns in the secondary coil.<br /><br /><br /><br />But changing the EMF has its consequences.<br /><br />The Conservation of Energy Law is still supposed to be held. The power <br />produced by AC in the primary coil in an ideal transformer is completely<br /> transferred to the secondary coil, which means<br /><br /><i><b>P(t) = U<sub>P</sub>(t)·I<sub>P</sub>(t) = U<sub>S</sub>(t)·I<sub>S</sub>(t)</b></i><br /><br />where<br /><br /><i>P(t)</i> is the power produced by primary and transferred to secondary coil, variable for AC,<br /><br /><i>U<sub>P</sub>(t)</i> is the EMF in the primary coil,<br /><br /><i>I<sub>P</sub>(t)</i> is the electric current in the primary coil,<br /><br /><i>U<sub>S</sub>(t)</i> is the EMF in the secondary coil,<br /><br /><i>I<sub>S</sub>(t)</i> is the electric current in the secondary coil.<br /><br /><br /><br />From the equation above follows that increasing the EMF in the secondary<br /> coil causes proportional decreasing in the electric current in it and <br />vice versa:<br /><br /><i><b>U<sub>P</sub>(t)/U<sub>S</sub>(t) = I<sub>S</sub>(t)/I<sub>P</sub>(t)</b></i><br /><br /><br /><br />The formulas above are the basic principle of working of transformers - <br />devices used to increase the AC voltage, while proportionally decrease <br />the electric current and, therefore, decrease losses related to <br />transmission of electricity along long wires and also used to decrease <br />this voltage, while proportionally increase the electric current at the <br />location of homes and electric devices that need the lower voltage in <br />the outlets with higher amperage.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-26067326828103625712020-08-07T09:42:00.001-07:002020-08-07T09:42:46.858-07:00AC - Effective Voltage: UNIZOR.COM - Physics4Teens - Electromagnetism<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Effective Voltage<br />and Current (RMS)</u><br /><br /><br /><br />We all know the voltage in our outlets at home. In some countries it's <br />220V, in others 110V or some other. But we also know that electric <br />current distributed to consumers from power plants is <i>alternating (AC)</i>.<br /><br />The actual value of <i>voltage</i> and <i>amperage</i> are sinusoidal, changing with time as:<br /><br /><i><b>U(t) = U</b><sub>max</sub><b>·sin(ωt)</b></i><br /><br /><i><b>I(t) = U(t)/R = I</b><sub>max</sub><b>·sin(ωt)</b></i><br /><br />where<br /><br /><i><b>ω</b></i> is some parameter related to the generation of electricity at the power plant,<br /><br /><i><b>t</b></i> is time,<br /><br /><i><b>U</b><sub>max</sub></i> is the voltage <i>amplitude</i>,<br /><br /><i><b>I</b><sub>max</sub><b> = U</b><sub>max</sub><b>/R</b></i> is the amperage <i>amplitude</i>,<br /><br /><i><b>R</b></i> is resistance of the circuit where <i>alternating current</i> is running through.<br /><br /><br /><br />If <i>voltage</i> is variable, what does it mean that in the outlet it is, for example, 220V?<br /><br /><br /><br />The answer to this question is in comparing energy the electric current is carrying in case of a variable <i>voltage</i> with energy of a <i>direct current</i>.<br /><br /><br /><br />The AC <i>voltage</i> is sinusoidal, so we can calculate the energy it carries through a circuit during the time of its <i>period</i> <i><b>T=2π/ω</b></i> and calculate the corresponding DC <i>voltage</i> that carries the same amount of energy during the same time through the same circuit.<br /><br />That corresponding value of the DC <i>voltage</i> is, <u>by definition</u>, the <i>effective voltage</i> in the AC circuit, which is usually called just <i>voltage</i> for alternating current.<br /><br /><br /><br />First, let's evaluate how much energy the AC with voltage <i><b>U(t)=U</b><sub>max</sub><b>·sin(ωt)</b></i> carries through some circuit of resistance <i><b>R</b></i> during the time of its period <i><b>T=2π/ω</b></i>.<br /><br /><br /><br />Consider an infinitesimal time interval from <i><b>t</b></i> to <i><b>t+</b>d<b>t</b></i>.<br /> During this interval we can consider the voltage and amperage to be <br />constant and, therefore, use the expression of the energy flowing <br />through a circuit of resistance <i><b>R</b></i> for direct current:<br /><br /><i>d<b>W(t) = U(t)·I(t)·</b>d<b>t = U²(t)·</b>d<b>t<span style="font-size: medium;">/</span>R</b></i><br /><br />(see "Electric Heat" topic of this course and "Ohm's Law" for direct current)<br /><br /><br /><br />To calculate the energy going through a circuit during any period <i><b>T=2π/ω</b></i>, we have to integrate this expression for <i>d<b>W</b></i> on an interval from <i><b>0</b></i> to <i><b>T</b></i>.<br /><br /><i><b>W</b></i><sub>[0,T]</sub> = <b><span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i>d<b>W(t) = </b></i><br /><br /><b>= <span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i><b>U²(t)·</b>d<b>t<span style="font-size: medium;">/</span>R</b></i> =<br /><br /><b>= <span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i><b>U²</b><sub>max</sub><b>·sin²(ωt)·</b>d<b>t<span style="font-size: medium;">/</span>R</b></i> =<br /><br />= (<i><b>U²</b><sub>max</sub><b><span style="font-size: medium;">/</span>R</b></i>)·<b><span style="font-size: large;">∫</span></b><sub>[0,T]</sub><i><b>sin²(ωt)·</b>d<b>t</b></i><br /><br /><br /><br />To simplify the integration we will use the known trigonometric identity<br /><br /><i><b>cos(2x) = cos²(x) − sin²(x) = 1 − 2sin²(x)</b></i><br /><br />from which follows<br /><br /><i><b>sin²(x) = </b></i>[<i><b>1 − cos(2x)</b></i>]<i><b> <span style="font-size: medium;">/</span>2</b></i><br /><br /><br /><br />Next we will do the substitution<br /><br /><i><b>x = ω·t</b></i><br /><br /><i><b>t = x/ω</b></i><br /><br /><i>d<b>t = </b>d<b>x/ω</b></i><br /><br /><i><b>t</b></i> ∈ [0,T] <span style="font-size: medium;">⇒</span> <i><b>x</b></i> ∈ [0,ωT]=[0,2π]<br /><br /><br /><br />The integral above, expressed in terms of <i><b>x</b></i>, is<br /><br /><b><span style="font-size: large;">∫</span></b><sub>[0,2π]</sub><i><b>sin²(x)·</b>d<b>x/ω =<br /><br />= </b></i><b><span style="font-size: large;">∫</span></b><sub>[0,2π]</sub>[<i><b>1−cos(2x)</b></i>]<i><b>·</b>d<b>x/(2ω) =<br /><br />= </b></i>[<i><b>2π−</b></i><span style="font-size: large;">∫</span><sub>[0,2π]</sub><i><b>cos(2x)·</b>d<b>x</b></i>]<i><b>/(2ω) =<br /><br />= π/ω</b></i><br /><br />Therefore,<br /><br /><i><b>W</b></i><sub>[0,T]</sub><i><b> = π·U²</b><sub>max</sub><b><span style="font-size: medium;">/</span>(ω·R)</b></i><br /><br /><br /><br />At the same time we have defined the <i>effective</i> voltage as the one that delivers the same amount of energy as <i><b>W</b></i><sub>[0,T]</sub> to the same circuit of resistance <i><b>R</b></i> during the same time <i><b>T=2π/ω</b></i> if the current is constant and direct.<br /><br />That is<br /><br /><i><b>U²</b><sub>eff</sub><b>·2π/(R·ω) = W</b></i><sub>[0,T]</sub><i><b> = π·U²</b><sub>max</sub><b><span style="font-size: medium;">/</span>(ω·R)</b></i><br /><br /><br /><br />Cancelling and simplifying the above equality, we conclude<br /><br /><i><b>2·U²</b><sub>eff</sub><b> = U²</b><sub>max</sub></i><br /><br />or<br /><br /><i><b>√<span style="text-decoration: overline;">2</span>·U</b><sub>eff</sub><b> = U</b><sub>max</sub></i><br /><br />For <i>effective</i> voltage 110V the peak voltage (amplitude) is 156V,<br /><br />for <i>effective</i> voltage 120V the peak voltage (amplitude) is 170V,<br /><br />for <i>effective</i> voltage 220V the peak voltage (amplitude) is 311V.<br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-36665845220884295822020-08-04T09:17:00.001-07:002020-08-04T09:17:01.989-07:00Properties of Alternating Current: UNIZOR.COM - Physics4Teens - Electrom...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Properties of<br /><br />Alternating Current (AC)</u><br /><br /><br /><br />Let's talk about practical aspects of electricity. We use it to light up<br /> our homes, to move the motors that pump the water, to ignite the car <br />engine etc.<br /><br />These are practical issues that must be resolved in the most efficient way.<br /><br />A short statement about <b>alternating electric current</b> is that it allows to do things easier and more efficient than <b>direct electric current</b>.<br /><br /><br /><br />The primary reason for using alternating electric current that allows to<br /> perform all these functions more efficiently than using direct electric<br /> current is that <b>alternating electric current</b> creates <b>alternating magnetic field</b>, that, in turn, can generate <b>alternating electric current</b> in another circuit through a mechanism of <i>induction</i>,<br /> and at many points on the way of these transformations we can use some <br />device to extract some useful functionality, like rotate the motor.<br /><br /><br /><br />Let's start from the beginning.<br /><br />In order to use electricity, we have to generate it. In most cases we <br />convert mechanical movement into electricity at power plants. There are <br />other ways, like chemical reaction in the battery or solar panels, but <br />they produce limited amount of electricity currently used.<br /><br /><br /><br />The most common way to transform mechanical motion into electricity is through <i>induction</i>. Recall, when we rotate a wire frame in the uniform magnetic field, the electric current is induced in the wire frame.<br /><br /><br /><br />This process of generating electric current by transforming mechanical energy into electricity using the <i>induction</i> is at the heart of all power plants.<br /><br />The simplest mechanical motion that can be used on industrial scale is <br />rotation. So, some mechanical source of energy can rotate the wire frame<br /> in a permanent magnetic field, which generates the electric current, <br />that we can distribute through different circuits to different devices.<br /><br /><br /><br />As we know, <i>induced electromotive force (EMF)</i> in a wire frame, rotating in a magnetic field, depends on the rate of change of <i>magnetic flux</i> through this frame. In a simple case of a uniform magnetic field of intensity <i><b>B</b></i> and a wire frame rotating with constant angular speed <i><b>ω</b></i> around an axis perpendicular to the direction of magnetic field lines the <i>magnetic flux</i><br /> going through a frame is changing with time, depending on the angle of <br />rotation of a wire frame and the direction of the vector of magnetic <br />field intensity.<br /><br /><br /><br />The picture below represents the top view of a flat rectangular wire frame of dimensions <i><b>a</b></i> by <i><b>b</b></i> at some moment of time <i><b>t</b></i>, when the frame deviated by angle <i><b>φ</b></i> from its original position, which was parallel to the magnetic field lines.<br /><img src="http://www.unizor.com/Pictures/InductionFrameTop.png" style="height: 200px; width: 200px;" /><br /><br />As was calculated in a lecture about a rotation of a wire frame, the <i>magnetic flux</i> going through a frame is<br /><br /><i><b>Φ(t) = B·S·sin(ωt)</b></i><br /><br />where <i><b>S=a·b</b></i> is the area of a wire frame.<br /><br /><br /><br />The <i>induced EMF</i> <i><b>U(t)</b></i> in a wire frame equals to the rate of change of this flux:<br /><br /><i><b>U(t) = −</b>d<b>Φ(t)/</b>d<b>t = −B·S·ω·cos(ωt)</b></i><br /><br />What's most important for us in this formula is that it shows the change of a magnitude and a direction of the <i>induced EMF</i>.<br /><br />The dependence of <i><b>U(t)</b></i> on <i><b>cos(ωt)</b></i> shows that<br /> the induced electric current in the wire loop and, consequently, in all<br /> devices that receive this current from this wire loop and have combined<br /> resistance <i><b>R</b></i> is changing from some maximum value <i><b>I<sub>max</sub>=B·S·ω/R</b></i> down to <i><b>0</b></i>, then it changes the direction to an opposite and grows in absolute value to <i><b>I<sub>max</sub></b></i> in that opposite direction, then again diminishes in absolute value to <i><b>0</b></i>, changes the direction etc.<br /><br /><br /><br />This electric current that increases and decreases in absolute values and changes the direction in <i>sinusoidal</i> way is called <b>alternating electric current</b>.<br /><br /><br /><br /><i>So, the most direct and effective way to produce an electric current causes this current to be <b>alternating</b></i>.<br /><br /><br /><br />As a side note, which we might discuss later on in the course, it's <br />easier to rotate the magnetic field, keeping the wire frame steady than <br />described above for purely educational purposes. In this type of <br />arrangement the contacts between the wire frame and a circuit going to <br />consumers of electricity is permanent and, therefore, more reliable.<br /><br /><br /><br />It is possible to produce <b>direct electric current</b>, transforming <br />rotation into induced electric current, but it involves more complex and<br /> less efficient engineering. Even then the electric current will not be <br />perfectly constant, but will change between maximum and minimum values, <br />while flowing in the same direction.<br /><br /><br /><br />The next step after generating the electricity is delivering it to <br />consumers. This requires an extensive distribution network with very <br />long wires and many branches.<br /><br /><br /><br />The amount of energy produced by a power plant per unit of time is<br /><br /><i><b>P<sub>gen</sub> = U·I</b></i><br /><br />where<br /><br /><i><b>U</b></i> is the output voltage<br /><br /><i><b>I</b></i> is the output electric current<br /><br />and all the above values are changing in time in case of alternating electric current.<br /><br /><br /><br />The amount of energy that is delivered to a user of electricity is less <br />that this by the amount lost to overcome the resistance of wires that <br />deliver the electricity, which is per unit of time equals to<br /><br /><i><b>P<sub>lost</sub> = I²·R</b></i><br /><br />where<br /><br /><i><b>I</b></i> is the electric current<br /><br /><i><b>R</b></i> is the resistance of wires.<br /><br /><br /><br />Using <i>transformers</i>, we can increase the <i>voltage</i> and, simultaneously, decrease the <i>amperage</i><br /> of an electric current. We will do it right after the electricity is <br />generated, which reduces the loss of energy in a wire that transmits <br />electricity on a long distance.<br /><br /><br /><br />One of the important steps in the distribution of electricity is <br />lowering the voltage, because we cannot connect any normal electric <br />device, like a vacuum cleaner, to an outlet with hundreds of thousands <br />of volts.<br /><br />This process of lowering the voltage is also easily accomplished using <i>transformers</i>.<br /><br /><br /><br />As you see, <i>transformers</i> play an extremely important role in the <br />distribution of electric power. But the transformers work only when the <br />electric current is changing. So, the consideration of distribution of <br />electricity are also a very important reason why we use alternating <br />current.<br /><br /><br /><br />Consider two <i>solenoids</i>, one, primary, connected to a source of an<br /> alternating electric current, another, secondary, - to a circuit <br />without a source of electricity. Since the electric current in the <br />primary solenoid is alternating, the magnetic field inside and around it<br /> will also be alternating.<br /><br />We have derived a formula for generated magnetic field for infinitely <br />long solenoid, but, approximately, it is correct for finite ones too. In<br /> this case the density of wire loops is proportional to a total number <br />of loops, since the length of a solenoid is fixed.<br /><br />The formula for an intensity of a magnetic field generated by a solenoid is<br /><br /><i><b>B = μ<sub>0</sub>·I·N</b></i><br /><br />where<br /><br /><i><b>B</b></i> is magnetic field intensity<br /><br /><i><b>I</b></i> is electric current in a solenoid (alternating in our case)<br /><br /><i><b>N</b></i> is density of wire loops in the solenoid (proportional to a number of wire loops for finite in length solenoid)<br /><br /><i><b>μ<sub>0</sub></b></i> is a constant representing the permeability of space.<br /><br /><br /><br />According to this formula, magnetic field characterized by <i><b>B</b></i> will be <i>sinusoidal</i> because the electric current <i><b>I</b></i> is <i>sinusoidal</i>.<br /><br />If our secondary solenoid is located inside the first one or close <br />nearby, the alternating magnetic field generated by the primary solenoid<br /> will induce <i>alternating EMF</i> in the secondary one proportional to a number of loops in it, since each loop will be a source of EMF on its own.<br /><br />If the number of loops in the secondary solenoid is smaller than in the <br />primary one, the induced EMF in the secondary solenoid will have smaller<br /> voltage than in the primary one. This is the idea behind the <i>transformers</i> that are used to lower the voltage coming from the power plant.<br /><br /><br /><br />Real <i>transformers</i> are built on the above principle with some <br />practical modifications that we will discuss in a separate lecture <br />dedicated to this topic.<br /><br />What's important is that alternating electric current allows to change the voltage in the circuit for distribution purposes.<br /><br /><br /><br />As we see, generation and distribution of <i>alternating electric current</i> is more efficient from practical standpoint than generation and distribution of <i>direct electric current</i>. That's why <i>alternating electric current</i> is so widely used.<br /><br /><br /><br />Most important property of the <i>alternating electric current</i> is its <i>sinusoidal</i> dependency on time.<br /><br /><i>Voltage</i> oscillates between <i><b>−U<sub>max</sub></b></i> and <i><b>+U<sub>max</sub></b></i> with certain frequency <i><b>ω</b></i>:<br /><br /><i><b>U(t) = U<sub>max</sub>·cos(ωt)</b></i> or<br /><br /><i><b>U(t) = U<sub>max</sub>·sin(ωt)</b></i><br /><br />(both formulas, based on <i>sin(ωt)</i> and <i>cos(ωt)</i> describe the same behavior and differ only in when is the start time <i>t=0</i>).<br /><br /><br /><br />The maximum value <i><b>U<sub>max</sub></b></i> is called <i>amplitude</i> or <i>peak</i> voltage of an alternating current.<br /><br /><br /><br />The time during which the value of <i><b>U(t)</b></i> changes from its positive peak to another positive peak (or from negative peak to another negative peak) is called a <i>period</i> of an alternating electric current.<br /><br />If <i><b>ω</b></i> is the angular speed of rotation that caused the generation of alternating electric current, the period is, obviously, <i><b>T=2π/ω</b></i>.<br /><br /><br /><br />Another characteristic of an alternating electric current is <i>frequency</i>, which is the number of <i>periods</i> per unit of time, which is, obviously, an inverse of a <i>period</i>, that is <i><b>f=1/T=ω/2π</b></i> and the unit of <i>frequency</i> is <b>hertz (Hz)</b>, so the frequency of 50Hz means that the voltage changes from one positive peak to another 50 times per second..<br /><br /><br /><br />The <i>amperage</i> behaves, according to the Ohm's Law, similarly, oscillating between <i><b>−I<sub>max</sub></b></i> and <i><b>+I<sub>max</sub></b></i> with the same as voltage frequency <i><b>ω</b></i>:<br /><br /><i><b>I(t) = I<sub>max</sub>·cos(ωt)</b></i> or<br /><br /><i><b>I(t) = I<sub>max</sub>·sin(ωt)</b></i><br /><br />(depending on when is the start time <i>t=0</i>)<br /><br />where <i><b>I<sub>max</sub> = U<sub>max</sub> <span style="font-size: medium;">/</span> R</b></i> and <i><b>R</b></i> is the resistance of the circuit.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-74725120239845908002020-07-31T16:38:00.001-07:002020-07-31T16:38:54.991-07:00Induced Variable EMF: UNIZOR.COM - Physics4Teens - Electromagnetism - La...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Induced Variable EMF</u><br /><br /><br /><br />The most important fact about <i>electromagnetic induction</i> is that variable <i>magnetic flux</i> <i><b>Φ(t)</b></i> going through some wire frame generates in it an <i>electromotive force (EMF)</i> <i><b>U(t)</b></i> proportional to a rate of change of the <i>magnetic flux</i> and oppositely directed.<br /><br /><i><b>U(t) = −</b>d<b>Φ(t)/</b>d<b>t</b></i><br /><br /><br /><br />Let's apply it to the following experiment.<br /><br /><img src="http://www.unizor.com/Pictures/InductionLoop.png" style="height: 135px; width: 200px;" /><br /><br /><br /><br />The experiment consists of a permanent magnet moving into a wire loop, <br />as shown on the above picture, with different positions of a wire loop <br />relatively to a magnet at different times marked as <i><b>t=1</b></i>, <i><b>t=2</b></i> etc.<br /><br />At <i><b>t=1</b></i> a magnet is still outside of the area of a wire loop, facing a loop with its North pole.<br /><br />At <i><b>t=2</b></i> a North pole of a magnet enters the area of a wire loop.<br /><br />At <i><b>t=3</b></i> a magnet covered half of its way through a wire loop, so the loop is positioned around a midpoint of a magnet.<br /><br />At <i><b>t=4</b></i> a South pole of a magnet leaves the area of a wire loop.<br /><br />At <i><b>t=5</b></i> a magnet is completely out of the area of a wire loop.<br /><br /><br /><br />While the magnet is far from a wire loop (even before time <i><b>t=1</b></i>), practically no electricity can be observed in a wire loop. There are two reasons for it.<br /><br />Firstly, the magnetic field is weak on a large distance.<br /><br />Secondly, the direction of <i>magnetic field lines</i> in an area where a<br /> wire loop is located at a large distance from a magnet is practically <br />parallel to the movement of a magnet (or movement of a wire loop <br />relative to a magnet). In this position a wire crosses hardly any <br />magnetic field lines and, consequently, the <i>magnetic flux</i> going through a wire loop is almost constant. Hence, the rate of the <i>magnetic flux</i> change is almost zero and no EMF is induced.<br /><br /><br /><br />As we move the magnet closer (time <i><b>t=1</b></i>), its magnetic field in the area of a loop is increasing in magnitude and the angle between the <i>magnetic field lines</i> in the area of a wire loop and the direction of a wire loop relative to a magnet becomes more and more perpendicular. The <i>magnetic flux</i> going through a wire loop is increasing and, consequently, the EMF induced in a wire becomes more noticeable.<br /><br /><br /><br />At time <i><b>t=2</b></i> the magnitude of the magnetic field reaches <br />its maximum, the direction of magnetic field lines crossed by a wire <br />loop is close to perpendicular to a relative movement of a wire. The <br />magnetic flux changes rapidly and induced EMF in the wire is strong.<br /><br /><br /><br />At time <i><b>t=3</b></i> the wire loop is at the midpoint of a magnet. <br />The strength of a magnetic field at this position is very weak, magnetic<br /> field lines are practically parallel to the relative movement of a <br />wire, and no noticeable EMF is generated.<br /><br /><br /><br />At time <i><b>t=4</b></i> the situation is analogous to that of <i><b>t=2</b></i><br /> with only difference in the direction of the magnetic field lines. Now <br />the wire is crossing them in the opposite direction, which causes the <br />induced EMF to have a different sign than at time <i><b>t=2</b></i>.<br /><br /><br /><br />After that, at time <i><b>t=5</b></i> the induced EMF is diminishing in magnitude and becomes almost not noticeable.<br /><br /><br /><br />In the next experiment we replace the permanent magnet with a wire loop <i><b>A</b></i><br /> with direct current running through it. As we stated many times, the <br />properties of a wire loop with direct electric current going through it <br />are similar to those of a permanent magnet.<br /><br /><img src="http://www.unizor.com/Pictures/InductionTwoLoops.jpg" style="height: 300px; width: 200px;" /><br /><br />So, if we move the wire loop <i><b>A</b></i> with the direct current closer and closer to a wire loop <i><b>B</b></i> not connected to any source of electricity, the magnetic flux going through a wire loop <i><b>B</b></i> will be changing and, similarly to the first experiment, the induced EMF will be observed in this wire loop <i><b>B</b></i>.<br /><br /><br /><br />Two positions of the loop <i><b>B</b></i> relative to a wire loop <i><b>A</b></i> are pictured above at times <i><b>t=1</b></i>, when wire loop <i><b>A</b></i> is very close to wire loop <i><b>B</b></i>, and <i><b>t=2</b></i>, when wire loop <i><b>A</b></i> has passed the location of wire loop <i><b>B</b></i>.<br /><br /><br /><br />Similar considerations lead us to conclude that at times <i><b>t=1</b></i> and <i><b>t=2</b></i> the rate of change of magnetic flux through wire loop <i><b>B</b></i> will be the greatest and, consequently, the induced EMF in it will be the greatest.<br /><br /><br /><br />However, these two EMF's will be opposite in direction because the <br />direction of crossing the magnetic lines will be opposite - at time <i><b>t=1</b></i> magnetic lines are directed from inside wire loop <i><b>B</b></i>, but at <i><b>t=2</b></i> the lines are directed into inside loop <i><b>B</b></i>.<br /><br /><br /><br />In both experiments above the most important is that variable magnetic <br />flux going through a wire loop, not connected to any source of <br />electricity, induces the EMF in it.<br /><br /><br /><br />Our third and final experiment is set as the second, except, instead of using constant direct current in the wire loop <i><b>A</b></i> and moving it through wire loop <i><b>B</b></i> to cause changing magnetic flux, we will keep the wire loop <i><b>A</b></i><br /> stationary, but will change the intensity of its magnetic field by <br />changing the magnitude of the electric current going through it.<br /><br /><br /><br />The result of changing electric current <i><b>I(t)</b></i> going through wire loop <i><b>A</b></i> of radius <i><b>R</b></i> (as a function of time <i><b>t</b></i>) is changing magnetic field vector <i><b><span style="text-decoration: overline;">B(t)</span></b></i> it produces.<br /><br />We did calculate the value of the intensity of this field in the center of a loop as<br /><br /><i><b>B(t) = μ<sub>0</sub>·I(t)<span style="font-size: medium;">/</span>(2R)</b></i><br /><br /><br /><br />If electric current <i><b>I(t)</b></i> is variable, magnetic field intensity vector <i><b><span style="text-decoration: overline;">B(t)</span></b></i> is variable at all points around wire loop <i><b>A</b></i>, including points around wire loop <i><b>B</b></i>, if it's relatively close.<br /><br />Therefore, the magnetic flux <i><b>Φ(t)</b></i> through wire loop <i><b>B</b></i> will be variable.<br /><br />Consequently, the induced EMF<br /><br /><i><b>U(t) = −</b>d<b>Φ(t)/</b>d<b>t</b></i><br /><br />will not be zero.<br /><br /><br /><br />At this point it's appropriate to mention that, if wire loop <i><b>A</b></i> is a tight wire spiral of <i>N</i> loops, the variable magnetic field and, therefore, variable magnetic flux through wire loop <i><b>B</b></i> will be <i>N</i> times stronger.<br /><br />That results in <i>N</i> times stronger EMF induced in wire loop <i><b>B</b></i>.<br /><br /><br /><br />Analogously, if wire loop <i><b>B</b></i> is a tight wire spiral of <i>M</i> loops, the same EMF will be induced in each loop and, therefore, combined EMF will be <i>M</i> times stronger than if <i><b>B</b></i> consists of only a single loop.<br /><br /><br /><br />The above considerations bring us to an idea of a <i>transformer</i> - a device allowing to change <i>voltage</i> in a circuit. This will be discussed in a separate lecture.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-20063956104149660302020-07-26T09:16:00.001-07:002020-07-26T09:16:04.559-07:00Solenoid: UNIZOR.COM - Physics4Teens - Electromagnetism - Magnetism of E...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Solenoid</u><br /><br /><br /><br />Solenoid is a wire spiral like on this picture<br /><br /><img src="http://www.unizor.com/Pictures/Solenoid.jpg" style="height: 100px; width: 200px;" /><br /><br />It can be considered as a combination of individual circular wire loops.<br /><br />We will consider an ideal case when the wire is infinitesimally thin, <br />the length of a solenoid along its axis is infinite and there is a <br />direct electric current running through it, that is running through each<br /> of its loops.<br /><br />Our purpose, as in a case of a single loop, is to determine the magnetic field inside such a solenoid.<br /><br /><br /><br />Position our infinite in length solenoid along the X-axis such that the <br />X-axis is the axis of a solenoid. Let the radius of a solenoid's loops <br />be <i><b>R</b></i> and the density of the loops per unit of length along X-axis be <i><b>N</b></i>, that is, each, however small, segment <i>d<b>X</b></i> on the X-axis contains <i><b>N·</b>d<b>X</b></i> loops.<br /><br />Assume a direct electric current <i><b>I</b></i> is running through all loops.<br /><br /><br /><br />Since a solenoid is infinite, magnetic field created by an electric <br />current running through it should be the same at any point on its axis.<br /><br />So, let's calculate the intensity of a magnetic field <i><b>B</b></i> at the origin of coordinates. At any other point on the X-axis it will be the same.<br /><br /><br /><br />The method of calculation the magnetic field is as follows.<br /><br /><b>Firstly</b>, we will calculate the intensity of the magnetic field at<br /> the origin of coordinates, generated by a single loop at the distance <i><b>X</b></i> from this point.<br /><br /><b>Secondary</b>, we will take into account the density of the loops and<br /> increase the calculated intensity of a single loop by a number of loops<br /> on a segment from <i><b>X</b></i> to <i><b>X+</b>d<b>X</b></i>, assuming <i>d<b>X</b></i> is infinitesimal increment.<br /><br />The <b>final step</b> is to integrate the obtained intensity for all <i><b>X</b></i> from <i><b>−∞</b></i> to <i><b>+∞</b></i>.<br /><br /><br /><br />1. <i>Magnetic field from a single loop at distance <i><b>X</b></i> from the origin of coordinates</i><br /><br /><br /><br />Calculations of the magnetic field at any point on the axis going <br />through a wire loop is similar to but more complex than the calculation <br />of the magnetic field at the center of a loop we studied in the previous<br /> lecture.<br /><br />The picture below represents the general view of the wire loop with a point <i><b>O</b></i>,<br /> where we want to evaluate its magnetic field and, below this general <br />view, the same loop as viewed from a plane this loop belongs to <br />perpendicularly to the X-axis.<br /><br /><img src="http://www.unizor.com/Pictures/Solenoid_Loop.jpg" style="height: 400px; width: 200px;" /><br /><br />As we know, the intensity <i><b>OB</b></i> of the magnetic field at point <i><b>O</b></i>, generated by an infinitesimal segment <i>d<b>S</b></i> of a wire loop located at point on a loop <i><b>A</b></i> is perpendicular to both electric current <i><b>I</b></i> at point <i><b>A</b></i> and vector <i><b>AO</b></i> from a point on a loop <i><b>A</b></i> to a point of interest - the origin of coordinates <i><b>O</b></i>.<br /><br /><br /><br />The direction of the electric current at point <i><b>A</b></i> is perpendicular to the plane of a picture. Therefore, vector of intensity <i><b>OB</b></i> is within a plane of a picture, as well as line <i><b>AO</b></i>. Now all lines we are interested in are on the plane of a picture and vector of magnetic field intensity <i><b>OB</b></i> at point of interest - the origin of coordinates <i><b>O</b></i> is perpendicular to a line from point <i><b>A</b></i> on a loop to a point of interest <i><b>O</b></i>:<br /><br /><i><b>AO</b></i><b>⊥<i>OB</i></b>.<br /><br /><br /><br />Let <i><b>B'</b></i> be a projection of the endpoint of vector <i><b>OB</b></i> onto X-axis. From perpendicularity of <i><b>AO</b></i> to <i><b>OB</b></i> follows that <i><b>∠AOP=∠OBB'</b></i> and triangles Δ<i><b>AOP</b></i> and Δ<i><b>OBB'</b></i> are similar.<br /><br /><br /><br />As we know, the magnitude of magnetic field intensity at point <i><b>O</b></i><br /> depends on the distance from the loop segment with electric current, <br />the length of this segment, the magnitude of the electric current in it <br />and a sine of the angle between the direction of an electric current in a<br /> segment and the line from a segment to a point of interest:<br /><br /><br /><br />Since the direction of the electric current is perpendicular to a plane <br />of a picture, the angle between the direction of an electric current in a<br /> segment and the line from a segment to a point of interest is 90°, so <br />its sine is equal to 1, and the formula for intensity of a magnetic <br />field at point <i><b>O</b></i> produced by a segment <i>d<b>S</b></i> at point <i><b>A</b></i> on the loop is<br /><br /><i>d<b>B = μ<sub>0</sub>·I·</b>d<b>S <span style="font-size: medium;">/</span> </b></i>[<i><b>4π(R²+X²</b></i>]<br /><br /><br /><br />We should take into consideration only the X-component of the vector of <br />magnetic field intensity because the Y-component will be canceled by the<br /> corresponding Y-component of a vector of intensity produced by a <br />segment of a loop at diametrically opposite to point <i><b>A</b></i> location.<br /><br />Let <i><b>k = sin(∠OBB').</b></i><br /><br />Hence,<br /><br /><i><b>k = sin(∠AOP) = R<span style="font-size: medium;">/</span>√<span style="text-decoration: overline;">X²+R²</span></b></i>.<br /><br />Then the X-component of vector <i><b>OB</b></i> is<br /><br /><i>d<b>B<sub>x</sub> = μ<sub>0</sub>·I·</b>d<b>S·k <span style="font-size: medium;">/</span> </b></i>[<i><b>4π(R²+X²)</b></i>]<i><b> =<br /><br />= μ<sub>0</sub>·I·</b>d<b>S·R <span style="font-size: medium;">/</span> </b></i>[<i><b>4π(R²+X²)<sup>3/2</sup></b></i>]<br /><br /><br /><br />To get a magnitude of the magnetic field intensity at point <i><b>O</b></i> from an entire loop we have to integrate the above expression by <i><b>S</b></i> from <i><b>0</b></i> to <i><b>2πR</b></i>, and the result will be<br /><br /><i><b>B<sub>x</sub>= μ<sub>0</sub>·I·2πR² <span style="font-size: medium;">/</span> </b></i>[<i><b>4π(R²+X²)<sup>3/2</sup></b></i>]<br /><br />Simplifying this gives<br /><br /><i><b>B(X) = μ<sub>0</sub>·I·R² <span style="font-size: medium;">/</span> </b></i>[<i><b>2(R²+X²)<sup>3/2</sup></b></i>]<br /><br />Incidentally, if our loop is located at <i><b>X=0</b></i>, the formula above gives<br /><br /><i><b>B<sub>x</sub> = μ<sub>0</sub>·I <span style="font-size: medium;">/</span> (2R)</b></i>,<br /><br />that corresponds to the results presented in the previous lecture for intensity of a magnetic field at the center of a loop.<br /><br /><br /><br />2. <i>Magnetic field from all loops located at distance from <i><b>X</b></i> to <i><b>X+</b>d<b>X</b></i> from the origin of coordinates</i><br /><br /><br /><br />This is a simple multiplication of the above expression for <i><b>B<sub>x</sub></b></i> by the number of loops located at distance from <i><b>X</b></i> to <i><b>X+</b>d<b>X</b></i> from the origin of coordinates, which is <i><b>N·</b>d<b>X</b></i>, where <i><b>N</b></i> is a density of the loops in a solenoid.<br /><br />The result is the intensity of a magnetic field at point <i><b>O</b></i> produced by an infinitesimal segment of our infinitely long solenoid from <i><b>X</b></i> to <i><b>X+</b>d<b>X</b></i><br /><i><b>μ<sub>0</sub>·I·R²·N·</b>d<b>X <span style="font-size: medium;">/</span> </b></i>[<i><b>2(R²+X²)<sup>3/2</sup></b></i>]<br /><br /><br /><br />3. <i>Finding the magnetic field at any point inside a solenoid by integrating the intensity produced by its segment from <b>X</b> to <b>X+</b>d<b>X</b> by <b>X</b> for an entire length of a solenoid from </i>−∞<i> to </i>+∞<br /><br /><br /><br />The above expression should be integrated by <i><b>X</b></i> from <i><b>−∞</b></i> to <i><b>+∞</b></i> to get a total value of the intensity of the magnetic field produced by an entire infinitely long solenoid at point <i><b>O</b></i> (or any other point on its axis).<br /><br /><i><b>B = γ·</b></i><span style="font-size: large;">∫</span><sub>[−∞;+∞]</sub><i><b> </b>d<b>X <span style="font-size: medium;">/</span> (R²+X²)<sup>3/2</sup></b></i><br /><br />where<br /><br /><i><b>γ = μ<sub>0</sub>·I·N·R²<span style="font-size: medium;">/</span>2</b></i><br /><br /><br /><br />Substitute <i><b>X=x·R</b></i> in the integral with <i>d<b>X=R·</b>d<b>x</b></i>.<br /><br />Infinite limits for <i><b>X</b></i> will be infinite limits for <i><b>x</b></i>.<br /><br />Then integral will be<br /><br /><span style="font-size: large;">∫</span><sub>[−∞;+∞]</sub><i><b> R·</b>d<b>x <span style="font-size: medium;">/</span> (R²+R²·x²)<sup>3/2</sup> =<br /><br />= (1/R²)·</b></i><span style="font-size: large;">∫</span><sub>[−∞;+∞]</sub><i><b> </b>d<b>x <span style="font-size: medium;">/</span> (1+x²)<sup>3/2</sup></b></i><br /><br />The value of the integral can be easily calculated.<br /><br />The indefinite integral is<br /><br /><i><b>x/√<span style="text-decoration: overline;">1+x²</span> + C</b></i><br /><br />So, the value of the definite integral between infinite limits is equal to<br /><br /><i><b>1 − (−1) = 2</b></i>.<br /><br /><br /><br />Therefore,<br /><br /><i><b>B = γ·2/R² = μ<sub>0</sub>·I·N</b></i><br /><br /><br /><br />IMPORTANT: The intensity of a magnetic field at any point on an axis of <br />an infinitely long solenoid does not depend on its radius.<br /><br />Textbooks usually state that the intensity calculated above is the same <br />at all points inside an infinitely long solenoid, not only at the points<br /> along its axis. The justification and the proof of this are beyond the <br />scope of this course.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-63807857207172051952020-07-19T10:21:00.001-07:002020-07-19T10:21:06.283-07:00Magnetic Flux: UNIZOR.COM - Physics4Teens - Electromagnetism - Magnetism...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetic Flux</u><br /><br /><br /><br />Consider a flat lying completely within an XY-coordinate plane closed <br />wire loop that includes a battery with direct electric current <i><b>I</b></i> running through it.<br /><br /><img src="http://www.unizor.com/Pictures/MagField_Flux.png" style="height: 200px; width: 200px;" /><br /><br />Since electric current generates a magnetic field around each however small segment of a wire <i>d<b>S</b></i><br /> with circular magnetic field lines around it in the plane perpendicular<br /> to this segment of a wire, the general direction of all magnetic lines <br />inside a wire loop will be parallel to Z-axis.<br /><br />More precisely, vectors of magnetic field intensity <i><b>B</b></i> at <br />all points of XY-plane inside a wire loop are similarly directed <br />perpendicularly to this plane and parallel to Z-axis. They are not <br />necessarily equal in magnitude, though.<br /><br /><br /><br />Consider an infinitesimal two-dimensional area of XY-plane inside a wire loop <i>d<b>A</b></i> around some point <i><b>P</b></i>.<br /> Since it's infinitesimal, we can assume that the intensity of the <br />magnetic field inside it is uniform and equal to the intensity at point <i><b>P</b></i>, which we will denote as vector <i><b><span style="text-decoration: overline;">B(P)</span></b></i>.<br /><br />This intensity is a result of combined magnetic field generated at that point by all infinitesimal segments <i>d<b>S</b></i> of a wire.<br /><br />Each one of these components of a resulting intensity vector has a <br />direction perpendicular to a direction of electric current in segment <i>d<b>S</b></i> (its source) and to a vector <i><b><span style="text-decoration: overline;">R</span></b></i> from a chosen infinitesimal segment <i>d<b>S</b></i> of a wire loop to point <i><b>P</b></i>, from which follows the perpendicularity of all vectors of magnetic field intensity to XY-plane and parallel to Z-axis.<br /><br /><br /><br />Magnetic field, as other force fields, is <i>additive</i>, that is, the <br />combined intensity vector generated by an entire wire loop is a vector <br />sum of intensity vectors generated by each segment of a wire.<br /><br /><br /><br />All these individual magnetic fields generated by different segments of a<br /> wire are directed along the Z-axis, so we will concentrate only on <br />their magnitude to find the result of their addition.<br /><br /><br /><br />Knowing the geometry of a wire and the electric current <i><b>I</b></i> in it, we can use the law we described in the previous lecture that determines the magnetic field intensity at any point <i><b>P</b></i> inside a wire loop generated by its any infinitesimal segment <i>d<b>S</b></i>:<br /><br /><i>d<b>B = μ<sub>0</sub>·I·</b>d<b>S·sin(α) <span style="font-size: medium;">/</span> (4πR²)</b></i><br /><br />where<br /><br /><i><b>R</b></i> is a magnitude of a vector <i><b><span style="text-decoration: overline;">R</span></b></i> from a chosen infinitesimal segment <i>d<b>S</b></i> of a wire loop to point <i><b>P</b></i>,<br /><br /><i>d<b>S</b></i> is the infinitesimal segment of a wire,<br /><br /><i><b>α</b></i> is an angle between the direction of electric current in the segment <i>d<b>S</b></i> and a vector <i><b><span style="text-decoration: overline;">R</span></b></i> from this segment to a point <i><b>P</b></i>,<br /><br /><i><b>μ<sub>0</sub></b></i> is the <i>permeability</i> of free space.<br /><br /><br /><br />Integrating this along the entire wire (this is integration along a curve) gives the value of magnetic field intensity <i><b>B(P)</b></i> at point <i><b>P</b></i>.<br /><br /><br /><br />Now we assume that this process is done and function <i><b>B(P)</b></i> is determined, that is we know the magnitude <i><b>B(P)</b></i> of the intensity vector of magnetic field generated by our wire loop at any point <i><b>P</b></i> inside a loop.<br /><br />Recall that the direction of this vector is always along Z-axis, that is perpendicular to a plane of a wire loop.<br /><br /><br /><br />We define <i>magnetic field flux</i> <i><b>Φ</b></i> as a two-dimensional integral of a product of intensity <i><b>B(P)</b></i> by area <i>d<b>A</b></i>.<br /><br />In a simple case of a constant magnetic field intensity <i><b>B(P)=B</b></i> at any point <i><b>P</b></i> inside a loop this is a simple product of a constant magnetic field intensity <i><b>B</b></i> by the area of a wire loop <i><b>A</b></i><br /><br /><i><b>Φ = B · A</b></i><br /><br />If <i><b>B(P)</b></i> is variable inside a loop, the two-dimensional integration by area of the loop produces the <i>magnetic field flux</i><br /><br /><i><b>Φ = </b></i><span style="font-size: large;">∫∫</span><sub>A</sub><i><b>B(P)·</b>d<b>A</b></i><br /><br /><br /><br />The complexity of exact calculation of magnetic field intensity <i><b>B(P)</b></i> at any point <i><b>P</b></i> inside a loop and subsequent calculation of the <i>magnetic field flux</i><br /> flowing through the wire loop are beyond the scope of this course, so <br />in our practice problems we will usually assume the uniformity of the <br />magnetic field with constant <i><b>B(P)=B</b></i>.<br /><br /><br /><br />A concept of <i>magnetic field flux</i> is applicable not only for a <br />magnetic field generated by a wire loop with electric current running <br />through it. It is a general concept used to characterize the combined <br />effect of a magnetic field onto the area where we observe it.<br /><br /><br /><br />Consider an analogy:<br /><br /><i>magnetic field intensity</i> and amount of grass growing in the backyard in a season per unit of area,<br /><br /><i>wire loop area</i> and area of a backyard,<br /><br /><i>magnetic field flux</i> and how much grass growth in an entire backyard in a season.<br /><br />Another analogy:<br /><br /><i>magnetic field intensity</i> and productivity of each individual doing some work,<br /><br /><i>wire loop area</i> and a team of people doing this work,<br /><br /><i>magnetic field flux</i> and productivity of a team doing this work.<br /><br /><br /><br />A general <i>magnetic field flux</i> is defined for a given magnetic field with known intensity vector <i><b><span style="text-decoration: overline;">B(P)</span></b></i> at each point <i><b>P</b></i> of space and given two-dimensional finite surface <i><b>S</b></i> in space, through which magnetic field lines are going.<br /><br /><img src="http://www.unizor.com/Pictures/MagFlux.png" style="height: 200px; width: 200px;" /><br /><br />The purpose of our definition of <i>magnetic field flux</i> in this general case is to quantify the total amount of "magnetic field energy" flowing through a surface <i><b>S</b></i>.<br /><br /><br /><br />Notice a small area around point <i><b>P</b></i> on a picture above.<br /><br />If we consider it to be infinitesimally small, we can assume that <br />magnetic field intensity at any point of this small area is the same as <br />at point <i><b>P</b></i>.<br /><br /><br /><br />If this small area is perpendicular to vector <i><b><span style="text-decoration: overline;">B(P)</span></b></i>, we would just multiply the magnitude of intensity vector <i><b>B(P)</b></i> in <i>tesla</i> by the area Δ<i><b>A</b></i> of a small area around point <i><b>P</b></i> in <i>square meters</i> and obtain the magnetic field flux Δ<i><b>Φ</b></i> going through this small area in units of magnetic flux -<br /><i>weber = tesla · meter²</i>.<br /><br />In a more general case, when this small area around point <i><b>P</b></i> is not perpendicular to magnetic field intensity vector <i><b><span style="text-decoration: overline;">B(P)</span></b></i>, to obtain the <i>magnetic field flux</i> going through it, we have to multiply it by <i><b>cos(φ)</b></i>, where <i><b>φ(P)</b></i> is an angle between magnetic field intensity vector <i><b><span style="text-decoration: overline;">B(P)</span></b></i> and normal (perpendicular vector) to a surface at point <i><b>P</b></i>.<br /><br />So, the expression for an infinitesimal <i>magnetic field flux</i> around point <i><b>P</b></i> is<br /><br /><i>d<b>Φ = B(P)·cos(φ(P))·</b>d<b>A</b></i><br /><br /><br /><br />The unit of measurement of <i>magnetic field flux</i> is <b>Weber (Wb)</b> that is equal to a flux of a magnetic field of intensity through an area of <br /><br /><nobr><b>1 tesla (1T)</b></nobr><nobr><b>1 square meter (1m²)</b>:</nobr><br /><br /><i><b>1 Wb = 1 T · 1 m²</b></i><br /><br /><br /><br />Once we have determined the <i>magnetic field flux</i> flowing through a<br /> small area around each point of a surface, we just have to summarize <br />all these individual amounts to obtain the total <i>magnetic field flux</i> flowing through an entire surface.<br /><br />Mathematically, it means we have to integrate the above expression for <i>d<b>Φ</b></i> along a surface <i><b>S</b></i>.<br /><br /><br /><br />We will not consider this most general case in this course. Our magnetic fields will be, mostly, uniform and surfaces flat.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-84543243518976171792020-07-16T11:32:00.001-07:002020-07-16T11:32:30.546-07:00Problems on Self-Induction: UNIZOR.COM - Physics4Teens - Electromagnetis...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Problems on Self-Induction</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />Consider a wire of resistance <i><b>R<sub>0</sub></b></i> bent into a circle of a radius <i><b>r</b></i> with its ends connected to a battery producing a voltage <i><b>U<sub>0</sub></b></i>.<br /><br />There is a switch that can turn the flow of electricity in this circuit ON or OFF during a short time period <i><b>T</b></i>.<br /><br />When it turns ON, the resistance of the circuit, as a function of time <i><b>R(t)</b></i>, changes from infinitely large value to the value of <i><b>R<sub>0</sub></b></i> as<br /><br /><i><b>R(t) = R<sub>0</sub>·T/t</b></i><br /><br />When it turns OFF, the resistance changes from the value of <i><b>R<sub>0</sub></b></i> to an infinitely large value as<br /><br /><i><b>R(t) = R<sub>0</sub>·T/(T−t)</b></i><br /><br />What is the EMF of self-induction <i><b>U<sub>i</sub></b></i> during the current switching ON or OFF?<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br /><i><b>Switching ON</b></i><br /><br />According to the Ohm's Law, the electric current in the circuit is<br /><br /><i><b>I(t) = U<sub>0</sub> <span style="font-size: medium;">/</span>R(t) = U<sub>0</sub>·t <span style="font-size: medium;">/</span>(R<sub>0</sub>·T)</b></i><br /><br />The magnetic field intensity inside a loop, as we discussed in the lecture on magnetism of a current in a loop, is<br /><br /><i><b>B(t) = μ<sub>0</sub>·I(t) <span style="font-size: medium;">/</span>(2r) = μ<sub>0</sub>·U<sub>0</sub>·t <span style="font-size: medium;">/</span>(2r·R<sub>0</sub>·T)</b></i><br /><br />The magnetic flux going through the wire loop of the radius <i><b>r</b></i> is a product of the intensity of the magnetic field by the area of a loop<br /><br /><i><b>Φ(t) = B(t)·πr² = μ<sub>0</sub>·U<sub>0</sub>·t·πr² <span style="font-size: medium;">/</span>(2r·R<sub>0</sub>·T) = πμ<sub>0</sub>·U<sub>0</sub>·t·r <span style="font-size: medium;">/</span>(2R<sub>0</sub>·T)</b></i><br /><br />The EMF of self-induction is the first derivative of the magnetic flux <br />by time with a minus sign (minus sign because the induced EMF works to <br />prevent the increase in the current)<br /><br /><i><b>U<sub>i</sub> = −</b>d<b>Φ/</b>d<b>t = −πμ<sub>0</sub>·U<sub>0</sub>·r <span style="font-size: medium;">/</span>(2R<sub>0</sub>·T)</b></i><br /><br />It's negative, so it reduces the overall voltage in the circuit and, <br />therefore, reduces the electric current in it. At the end of switching <br />ON (at <i><b>t=T</b></i>) the current will not reach its maximum value but will continue to rise after the switch is completely ON until it reaches it.<br /><br /><br /><br /><i><b>Switching OFF</b></i><br /><br />According to the Ohm's Law, the electric current in the circuit is<br /><br /><i><b>I(t) = U<sub>0</sub> <span style="font-size: medium;">/</span>R(t) = U<sub>0</sub>·(T−t) <span style="font-size: medium;">/</span>(R<sub>0</sub>·T)</b></i><br /><br />The magnetic field intensity inside a loop, as we discussed in the lecture on magnetism of a current in a loop, is<br /><br /><i><b>B(t) = μ<sub>0</sub>·I(t) <span style="font-size: medium;">/</span>(2r) = μ<sub>0</sub>·U<sub>0</sub>·(T−t) <span style="font-size: medium;">/</span>(2r·R<sub>0</sub>·T)</b></i><br /><br />The magnetic flux going through the wire loop of the radius <i><b>r</b></i> is a product of the intensity of the magnetic field by the area of a loop<br /><br /><i><b>Φ(t) = B(t)·πr² = μ<sub>0</sub>·U<sub>0</sub>·(T−t)·πr² <span style="font-size: medium;">/</span>(2r·R<sub>0</sub>·T) = πμ<sub>0</sub>·U<sub>0</sub>·(T−t)·r <span style="font-size: medium;">/</span>(2R<sub>0</sub>·T)</b></i><br /><br />The EMF of self-induction is the first derivative of the magnetic flux <br />by time with a minus sign (minus sign because the induced EMF works to <br />prevent the increase in the current)<br /><br /><i><b>U<sub>i</sub> = −</b>d<b>Φ/</b>d<b>t = πμ<sub>0</sub>·U<sub>0</sub>·r <span style="font-size: medium;">/</span>(2R<sub>0</sub>·T)</b></i><br /><br />It's positive, so it adds to the overall voltage in the circuit and, <br />therefore, increases the electric current in it, which, potentially, can<br /> be harmful for electronic devices on the circuit.<br /><br /><br /><br /><br /><br /><i>Problem B</i><br /><br /><br /><br />In the same setting as in <i>Problem A</i> calculate minimum time <i><b>T<sub>min</sub></b></i> of switching a circuit OFF in order for the current in a circuit grow no more then 10% from the theoretical value <br /><br /><nobr><i><b>I<sub>0</sub> = U<sub>0</sub> <span style="font-size: medium;">/</span>R<sub>0</sub></b></i>,</nobr><nobr><i><b>I<sub>0</sub> = U<sub>0</sub> <span style="font-size: medium;">/</span>R<sub>0</sub></b></i>.</nobr><br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Induced EMF equals to (from <i>Problem A</i>)<br /><br /><i><b>U<sub>i</sub> = πμ<sub>0</sub>·U<sub>0</sub>·r <span style="font-size: medium;">/</span>(2R<sub>0</sub>·T)</b></i><br /><br />It's a constant and does not depend on time <i><b>t</b></i>.<br /><br />The resistance of a circuit is growing with time from <i><b>R<sub>0</sub></b></i> at <i><b>t=0</b></i> to infinity at <i><b>t=T</b></i>. Therefore, the maximum electric current will be observed at <i><b>t=0</b></i>, when the resistance is minimal, and will equal to<br /><br /><i><b>I<sub>max</sub> = (U<sub>0</sub>+U<sub>i</sub>) <span style="font-size: medium;">/</span>R<sub>0</sub> = I<sub>0</sub> + (U<sub>i</sub> <span style="font-size: medium;">/</span>R<sub>0</sub>)</b></i><br /><br />In order for this current not to exceed <i><b>I<sub>0</sub></b></i> by 10% the following inequality must be satisfied<br /><br /><i><b>U<sub>i</sub> <span style="font-size: medium;">/</span>R<sub>0</sub> ≤ 0.1·U<sub>0</sub> <span style="font-size: medium;">/</span>R<sub>0</sub></b></i><br /><br />The resistance cancels and the remaining inequality is<br /><br /><i><b>U<sub>i</sub> ≤ 0.1·U<sub>0</sub></b></i><br /><br />Using the value of <i><b>U<sub>i</sub></b></i> calculated in <i>Problem A</i>, we obtain the following inequality that should be resolved for time interval <i><b>T</b></i><br /><br /><i><b>πμ<sub>0</sub>·U<sub>0</sub>·r <span style="font-size: medium;">/</span>(2R<sub>0</sub>·T) ≤ 0.1·U<sub>0</sub></b></i><br /><br />Voltage <i><b>U<sub>0</sub></b></i> cancels out and the remaining inequality for time interval <i><b>T</b></i> is<br /><br /><i><b>T ≥ πμ<sub>0</sub>·r <span style="font-size: medium;">/</span>(0.2R<sub>0</sub>)</b></i><br /><br /><br /><br /><b>Practical case</b><br /><br />Assume the following conditions<br /><br /><i><b>μ<sub>0</sub> = 4π·10<sup>−7</sup></b> N/A²</i><br /><br /><i><b>r = 0.1</b> m</i><br /><br /><i><b>R<sub>0</sub> = 200</b> Ω</i><br /><br />Then the time interval of switching OFF must be at least<br /><br /><i><b>T = πμ<sub>0</sub>·r <span style="font-size: medium;">/</span>(0.2R<sub>0</sub>) ≅ 10<sup>−8</sup></b> sec</i><br /><br />That is a very short interval of time. Mechanical switches usually are <br />much slower, which means the electric current under normal usage in <br />homes will not significantly rise when we switch the electricity OFF.<br /><br /><b>IMPORTANT:<br /><br />Check formula units</b><br /><br /><i><b>(N/A²)·m/Ω = N·m/(A²·Ω) =<br /><br /></b></i>[N·m → W → V·A·sec]<i><b><br /><br />= (V·A·sec)/(A²·Ω) =<br /><br /></b></i>[Ω → V/A]<i><b><br /><br />= (V·A·sec·A)/(A²·V) = sec</b></i><br /><br />as it is supposed to be.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-8813063106787559182020-07-13T06:15:00.001-07:002020-07-13T06:15:15.156-07:00Self-Induction: UNIZOR.COM - Physics4Teens - Electromagnetism -<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Self-Induction</u><br /><br /><br /><br />Let's study the process of increasing the direct electric current in a <br />wire loop that is connected to a source of a direct electricity (like <br />battery) that generates electromotive force of voltage <i><b>U<sub>0</sub></b></i>.<br /><br />This increase might be attributed to a decrease in resistance in this <br />loop. For example, if the wire loop was initially disconnected from a <br />source of electricity and we connected it, flipping some switch, the <br />resistance is changing from being practically infinite to some value <i><b>R</b></i> during a fraction of a second and the electric current would grow from zero to some value<br /><br /><i><b>I = U/R</b></i> (Ohm's Law).<br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/MagField_Loop.png" style="height: 200px; width: 200px;" /><br /><br />The first thing that we can observe is that there is a magnetic field <br />generated by an electric current with magnetic field lines going through<br /> a loop, thus effectively making it a magnet.<br /><br /><br /><br />The next thing we note is that the intensity of a magnetic field <br />generated by an increasing direct current is also increasing with the <br />current because of their relationship that we have described in the <br />lecture about magnetism of an electric current loop.<br /><br />In the center of an ideally circular loop this intensity, as we calculated in that lecture, is<br /><br /><i><b>B = μ<sub>0</sub>·I<span style="font-size: medium;">/</span>(2R)</b></i><br /><br />where<br /><br /><i><b>μ<sub>0</sub></b></i> is <i>permeability</i> of space,<br /><br /><i><b>I</b></i> is the current in a wire loop,<br /><br /><i><b>R</b></i> is the radius of a wire loop.<br /><br /><br /><br />At other points around the wire the intensity of a magnetic field also <br />increases proportionally to the electric current in a wire. Even if the <br />shape of our wire loop is not ideally circular, the proportionality of <br />intensity of a magnetic field at any point to an electric current would <br />still be held, because each infinitesimal segment of a wire creates its <br />own magnetic field and the magnetic field intensity is an additive <br />function, that is, produced by two sources, intensities from each are <br />added as vectors, resulting in a combined intensity.<br /><br />As a result, the <i>magnetic flux</i> going through a wire loop increases proportionally to an increase in electric current.<br /><br /><br /><br />Now recall the Faraday's Law of <b>magnetic induction</b>. It states that changing <i>magnetic flux</i> going through an electric circuit generates <i>electromotive force (EMF)</i> proportional to a <b>rate of change</b> of the <i>magnetic flux</i>, where <b>rate of change</b> means a change per unit of time, which, using mathematical language, is the first derivative of a variable <i>magnetic flux</i> by time.<br /><br /><br /><br />Let's ignore for now the signs of the rate of change of the <i>magnetic flux</i> and the <i>EMF</i>, then the Faraday's Law can be represented as<br /><br /><i><b>U = </b>d<b>Φ/</b>d<b>t</b></i><br /><br />where<br /><br /><i><b>U</b></i> is the absolute value of a magnitude of the generated <i>EMF</i>,<br /><br /><i><b>Φ</b></i> is <i>magnetic flux</i>,<br /><br /><i><b>t</b></i> is time,<br /><br /><i>d<b>Φ/</b>d<b>t</b></i> is the rate of change of magnetic flux (first derivative of magnetic flux by time), taken as an absolute positive value.<br /><br /><br /><br />From <br /><br />(1) the proportionality of a magnetic field intensity at any point around a wire to an electric current going through a wire,<br /><br />(2) definition of magnetic flux and<br /><br />(3) the Faraday's Law<br /><br />follows that the increasing electric current, causing an increasing <br />magnetic flux going through a circuit (so, the first derivative of a <br />magnetic flux is <u>positive</u>), generates a <i>secondary EMF</i> in <br />the circuit that is proportional to a rate of change of a magnetic flux,<br /> which, in turn, is proportional to a rate of change of the electric <br />current in a loop.<br /><br />So, <b>generated (induced) secondary <i>EMF</i> is proportional to a rate of change of an electric current in a loop</b>.<br /><br /><br /><br />This induced <i>secondary EMF</i> causes the secondary electric current in a loop that interferes with the original electric current.<br /><br />The process of interference of two electric currents, varying original <br />one and that caused by a change of magnetic field flux, is called <b>self-induction</b>.<br /><br /><br /><br />Now let's talk about the direction of that secondary current from the position of the Law of Energy Conservation.<br /><br />If the secondary current flows in the same direction as the increasing <br />original one, the rate of change of electric current in a loop will <br />increase, causing a greater rate of change of the magnetic flux, causing<br /> even greater rate of change of the electric current flowing through a <br />wire, causing even greater flux etc. This is a perfect source of free <br />energy, so it cannot be true.<br /><br />Indeed, the secondary electric current must be directed opposite to an <br />increasing original one, thus slowing the rate of increase.<br /><br /><b>The <i>EMF</i> generated by <i>self-induction</i> is opposite to the one that originated the increase in the electric current in a loop</b>.<br /><br /><br /><br />As a practical implication of this principle, when we switch a lamp on, <br />and the time a switch actually closes the circuit is, for example 0.01 <br />sec, the electric current in a circuit will not reach its maximum during<br /> this exact time, but later, like during 0.02 sec, because of <br />self-induction that slows the increase of amperage.<br /><br /><br /><br />Let's consider the process of decreasing of the electric current in a <br />loop, like when we disconnect a wire from a primary source of <br />electricity.<br /><br /><br /><br />Decreasing electric current in a wire loop causes proportional decrease <br />of a magnetic field intensity at any point around a wire.<br /><br />This, in turn, causes proportional decrease of <i>magnetic flux</i> flowing through a wire loop with rate of change (first derivative of flux by time) <u>negative</u>.<br /><br />Negative rate of change of magnetic flux causes the generation of <i>secondary (induced) EMF</i> in a wire, that results in a secondary electric current, that interferes with the decreasing primary one.<br /><br /><br /><br />Let's analyze the direction of this secondary electric current.<br /><br />Above, when the primary electric current was increasing and the rate of change of magnetic flux was <u>positive</u>,<br /> we concluded that induced EMF causes the secondary electric current to <br />go in the opposite to primary direction to preserve the Law of Energy <br />Conservation.<br /><br /><br /><br />In a case of decreasing primary electric current we see that the <br />magnetic flux flowing through a wire loop is decreasing, it's rate of <br />change (first derivative) is <u>negative</u>, so the <i>secondary (induced) EMF</i> must be directed opposite to the one generated by an increasing primary electric current.<br /><br /><br /><br />Therefore, if in the case of increasing primary electric current the <i>secondary EMF</i> "worked" <u>against the electric current</u>,<br /> generating a secondary current in a direction opposite to an increasing<br /> primary one, in the case of decreasing primary current the <i>secondary (induced) EMF</i> generates the secondary current <u>in the same direction as a decreasing primary current</u>, "helping" the decreasing primary current, prolonging the decrease.<br /><br /><br /><br />So, in both cases the <i>secondary (induced) EMF</i> works <u>opposite to a rate of change of the magnetic flux</u> going through a wire loop.<br /><br />When the magnetic flux increases, the <i>induced EMF</i> generates an <br />electric current in the opposite direction to a primary one, thus <br />slowing the rate of increase of the electric current and, therefore, <br />against the rate of increase of the magnetic flux.<br /><br />When the magnetic flux decreases, the <i>induced EMF</i> generates an <br />electric current in the same direction as a primary one, thus slowing <br />the decrease of the electric current and, therefore, against the rate of<br /> decrease of the magnetic flux.<br /><br /><br /><br />The above considerations are the primary reason why the Faraday's Law is expressed as<br /><i><b>U = −</b>d<b>Φ/</b>d<b>t</b></i><br /><br />where the minus sign signifies that the <i>induced EMF</i> works against the rate of change of a magnetic flux.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-76034047312410568662020-07-10T17:05:00.001-07:002020-07-10T17:05:02.779-07:00Current in a Loop: UNIZOR.COM - Physics4Teens - Electromagnetism - Magne...<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetism of Electric Current in a Loop</u><br /><br /><br /><br />Let's recall some general considerations of any field produced by a <br />point-source (like gravitational or electrostatic) and the force it <br />exerts at certain distance from its center.<br /><br /><br /><br />In three-dimensional space the field influence is spreading around in <br />all directions from its center, and at any given time it reaches and <br />spreads over a sphere of the area <i><b>4πR²</b></i>, where <i><b>R</b></i> is the radius of this sphere.<br /><br /><br /><br />This important consideration prompted the dependency between the <br />magnitude of the field force at any point in space and inverse of a <br />square of a distance between this point and the source of a field.<br /><br />This dependency exists in all laws related to fields we studied, and it's true for magnetic field as well.<br /><br /><br /><br />There is, however, an important complication with the sources of <br />magnetic field. Strictly speaking, they are not and theoretically cannot<br /> be just points, because they always have a characteristic of the <br />direction. If it's a permanent magnet, there is a North pole to South <br />pole direction. If the source is a direct current, there is a direction <br />of electrons that causes the existence of a magnetic field.<br /><br />So, besides the distance from the source of a magnetic field, another <br />important aspect that influences the magnetic field is how the point of <br />measuring the magnetic field is positioned relatively to a direction of <br />the electric current at the source of a field.<br /><br /><br /><br />Consider measuring a strength of a magnetic field at point <i><b>P</b></i> on the following picture.<br /><br /><img src="http://www.unizor.com/Pictures/Biot_Savart.png" style="height: 160px; width: 200px;" /><br /><br />Here an infinitesimal segment <i><b>AB</b></i> of the length <i>d<b>S</b></i> of a wire with an electric current <i><b>I</b></i> produces a magnetic field that we are measuring at point <i><b>P</b></i>. From the viewpoint of point <i><b>P</b></i> segment <i><b>AB</b></i> is visible at an angle and seems having the size of a segment <i><b>BC</b></i> that is perpendicular to a radius-vector <i><b><span style="text-decoration: overline;">R </span></b></i> from the source of magnetic field segment <i><b>AB</b></i> to the point of measuring <i><b>P</b></i>.<br /><br /><br /><br />Simple geometry gives the visible from point <i><b>P</b></i> length of segment <i><b>AB</b></i> - the length of <i><b>BC</b></i>, which is the length of <i><b>AB</b></i> multiplied by <i><b>sin(α)</b></i>.<br /><br /><br /><br />With this correction to a simple dependency of the strength of magnetic <br />field on the inverse square of a distance from the field source we can <br />represent the magnitude of the vector of magnetic field strength or <br />intensity at point <i><b>P</b></i> as<br /><br /><i>d<b>B = k·I·</b>d<b>S·sin(α) <span style="font-size: medium;">/</span> R²</b></i><br /><br />The constant <i><b>k</b></i> in SI units equals to <i><b>μ<sub>0</sub>/(4π)</b></i>, where <i><b>μ<sub>0</sub></b></i> is the <i>permeability</i> of free space. So the formula for a magnitude of the strength of magnetic field looks like this:<br /><br /><i>d<b>B = μ<sub>0</sub>·I·</b>d<b>S·sin(α) <span style="font-size: medium;">/</span> (4πR²)</b></i><br /><br /><br /><br />The vector of force has magnitude and direction. The magnitude was <br />evaluated above. The direction of this force, based on experimental <br />data, is always perpendicular to both the direction of the current <i><b>I</b></i> and the direction of the radius-vector from the source of magnetic field to a point of measurement <i><b>P</b></i>.<br /><br /><br /><br />This allows us to express the vector of magnetic field intensity <i><b><span style="text-decoration: overline;">B </span></b></i> in a form of a vector product<br /><br /><i><b><span style="text-decoration: overline;">B </span> = μ<sub>0</sub>·I·<span style="text-decoration: overline;">dS </span></b></i>⨯<i><b> <span style="text-decoration: overline;">r </span> <span style="font-size: medium;">/</span> (4πR²)</b></i><br /><br />where <i><b><span style="text-decoration: overline;">r </span></b></i> is a unit vector in the direction from the source of magnetic field segment <i><b>AB</b></i> to the point of measurement <i><b>P</b></i>.<br /><br />Equivalent formula is<br /><br /><i><b><span style="text-decoration: overline;">B </span> = μ<sub>0</sub>·I·<span style="text-decoration: overline;">dS </span></b></i>⨯<i><b> <span style="text-decoration: overline;">R </span> <span style="font-size: medium;">/</span> (4πR³)</b></i><br /><br /><br /><br />Let's apply this formula to calculate the magnetic field intensity and direction in a center of a circular wire loop of radius <i><b>R</b></i> with electric current <i><b>I</b></i> running through it.<br /><br /><br /><br />We divide the circular wire loop into infinitesimal segments of the length <i>d<b>S</b></i>. Fortunately, all radiuses from any such segment to a center of a loop are perpendicular to these segments, so <i><b>sin(α)=1</b></i> in the above formula for a magnitude of the intensity vector.<br /><br /><br /><br />The magnetic field intensity <i><b>B</b></i> from each segment equals to<br /><br /><i>d<b>B = μ<sub>0</sub>·I·</b>d<b>S <span style="font-size: medium;">/</span> (4πR²)</b></i><br /><br />The direction of this force is perpendicular to both radius to a segment <i>d<b>S</b></i><br /> and to a segment itself. That is, it should be perpendicular to a plane<br /> of the wire loop, that is the vector of magnetic field force in a <br />center of a wire loop is always directed perpendicularly to the loop.<br /><br />Since it's true for any segment <i>d<b>S</b></i>, we can just add all the forces from all the segments, that is, we have to integrate the above expression by <i><b>S</b></i> along the whole circle from <i><b>0</b></i> to <i><b>2πR</b></i>.<br /><br />The only variable in the above expression is <i>d<b>S</b></i>, so the result is<br /><br /><i><b>B =</b></i> <span style="font-size: large;">∫</span><sub>[<i>0,2πR</i>]</sub> <i><b>μ<sub>0</sub>·I·</b>d<b>S<span style="font-size: medium;">/</span>(4πR²) =<br /><br />= 2πR·μ<sub>0</sub>·I<span style="font-size: medium;">/</span>(4πR²)= μ<sub>0</sub>·I<span style="font-size: medium;">/</span>(2R)</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-11697056306544150902020-07-01T20:12:00.001-07:002020-07-01T20:12:26.701-07:00UNIZOR.COM - Physics4Teens - Electromagnetism<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Problems on<br />Electromagnetic Induction</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />The following experiment is conducted in the space with Cartesian coordinates.<br /><br />Two infinitely long parallel wires in XY-plane are parallel to X-axis, one at Y-coordinate <i><b>y=a</b></i> and another at Y-coordinate <i><b>y=−a</b></i> (assuming <i><b>a</b></i> is positive).<br /><br />These two wires are connected by a third wire positioned along the Y-coordinate between points <i><b>(0,a)</b></i> and <i><b>(0,−a)</b></i>.<br /><br />The fourth wire, parallel to the third one, also connects the first two,<br /> but slides along the X-axis, always maintaining its parallel position <br />to Y-axis. The X-coordinate of its position is monotonously increasing <br />with time <i><b>t</b></i> according to some rule <i><b>x=x(t)</b></i>.<br /><br />All four wires are made of the same material and the same cross-section with the electrical resistance of a unit length <i><b>r</b></i>.<br /><br /><br /><br />There is a uniform magnetic field of intensity <i><b>B</b></i> with field lines parallel to Z-axis.<br /><br />Initial position of the fourth wire coincides with the third one, that is <i><b>x(0)=0</b></i>.<br /><br /><br /><br />What should the function <i><b>x(t)</b></i> be to assure the generation of the same constant electric current <i><b>I<sub>0</sub></b></i> in the wire loop?<br /><br />What is the speed of the fourth rod at initial time <i><b>t=0</b></i>?<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />The area of a wire frame is changing with time:<br /><br /><i><b>S(t)=2a·x(t)</b></i>.<br /><br /><i>Magnetic flux</i> through this wire frame is<br /><br /><i><b>Φ(t)=B·S(t)</b></i>.<br /><br />Therefore, the magnitude of the generated electromotive force or voltage <i><b>U(t)</b></i>, as a function of time, is<br /><br /><i><b>U(t) = </b>d<b>Φ(t)/</b>d<b>t = 2B·a·x'(t)</b></i><br /><br />The resistance <i><b>R(t)</b></i> of the wire loop, as a function of time, is a product of a resistance of a unit length of a wire <i><b>r</b></i> by the total length of all four sides of a wire rectangle <i><b>L=4a+2x(t)</b></i>.<br /><br /><i><b>R(t) = r·L = 2r·</b></i>[<i><b>2a+x(t)</b></i>]<br /><br />The electric current <i><b>I(t)</b></i> in a wire loop, according to the Ohm's Law, is<br /><br /><i><b>I(t) = U(t)/R(t) =<br /><br />= 2B·a·x'(t) <span style="font-size: medium;">/</span> </b></i>{<i><b>2r·</b></i>[<i><b>2a+x(t)</b></i>]}<i><b> =<br /><br />= B·a·x'(t) <span style="font-size: medium;">/</span> </b></i>{<i><b>r·</b></i>[<i><b>2a+x(t)</b></i>]}<br /><br />This current has to be constant and equal to <i><b>I<sub>0</sub></b></i>. This leads us to a differential equation <i><b>I<sub>0</sub>=I(t)</b></i><br /><br /><i><b>I<sub>0</sub> = B·a·x'(t) <span style="font-size: medium;">/</span> </b></i>{<i><b>r·</b></i>[<i><b>2a+x(t)</b></i>]}<br /><br />Simplifying this equation, obtain<br /><br /><i><b>I<sub>0</sub>·r <span style="font-size: medium;">/</span> (B·a) = x'(t) <span style="font-size: medium;">/</span> </b></i>[<i><b>2a+x(t)</b></i>]<br /><br />[<i><b>I<sub>0</sub>·r <span style="font-size: medium;">/</span> (B·a)</b></i>]<i><b>·</b>d<b>t =<br /><br />= </b>d</i>[<i><b>2a+x(t)</b></i>]<i><b> <span style="font-size: medium;">/</span> </b></i>[<i><b>2a+x(t)</b></i>]<br /><br />Integrating,<br /><br />[<i><b>I<sub>0</sub>·r <span style="font-size: medium;">/</span> (B·a)</b></i>]<i><b>·t + C = ln(2a+x(t))</b></i><br /><br /><i><b>2a + x(t) = C·e<sup>I<sub>0</sub>·r·t <span style="font-size: medium;">/</span> (B·a)</sup></b></i><br /><br />Since <i><b>x(0)=0</b></i>, <i><b>C=2a</b></i><br /><br /><i><b>x(t) = 2a·</b></i>[<i><b>e<sup>I<sub>0</sub>·r·t <span style="font-size: medium;">/</span> (B·a)</sup> − 1</b></i>]<br /><br />Speed of the motion of the fourth wire along the X-axis is<br /><br /><i><b>x'(t) = 2a·e<sup>I<sub>0</sub>·r·t<span style="font-size: medium;">/</span>(B·a)</sup>·</b></i>[<i><b>I<sub>0</sub>·r<span style="font-size: medium;">/</span>(B·a)</b></i>]<br /><br />or<br /><br /><i><b>x'(t) = e<sup>I<sub>0</sub>·r·t<span style="font-size: medium;">/</span>(B·a)</sup>·</b></i>[<i><b>2I<sub>0</sub>·r<span style="font-size: medium;">/</span>B</b></i>]<br /><br />At time <i><b>t=0</b></i> the initial speed is<br /><br /><i><b>x'(0) = 2I<sub>0</sub>·r <span style="font-size: medium;">/</span> B</b></i><br /><br /><br /><br /><br /><br /><i>Problem B</i><br /><br /><br /><br />A rectangular wire frame in a space with Cartesian coordinates rotates with variable angular speed <i><b>ω(t)</b></i> in a uniform magnetic field <i><b>B</b></i>.<br /><br />In the beginning at <i><b>t=0</b></i> the wire frame is at rest, <i><b>ω(0)=0</b></i>.<br /> As the time passes, the angular speed is monotonously increasing from <br />initial value of zero to some maximum. This models turning the rotation <br />on.<br /><br />The axis of rotation is Z-axis.<br /><br />The initial position of a wire frame is that its plane coincides with XZ-plane.<br /><br /><br /><br />The magnetic field lines are parallel to X-axis.<br /><br />So the angle between the magnetic field lines and the wire frame plane <i><b>φ(t)</b></i> at <i><b>t=0</b></i> equals to zero.<br /><br />The sides of a wire frame parallel to Z-axis (those, that cross the magnetic field lines) have length <i><b>a</b></i>, the other two sides have length <i><b>b</b></i>.<br /><br /><br /><br />Determine the generated electromotive force (EMF) in this wire frame as a function of time <i><b>t</b></i>.<br /><br /><br />Solution<br /><br /><br /><br />The angle <i><b>φ(t)</b></i> between the magnetic field lines and the <br />wire frame plane is changing with time. In the initial position, when <br />the wire frame coincides with XZ-plane, this angle is 0°, since magnetic<br /> field lines are stretched along the X-axis.<br /><br />As the wire frame rotates with variable angular speed <i><b>ω(t)</b></i>, the angle between magnetic field lines and the wire frame plane <i><b>φ(t)</b></i> and the angular speed <i><b>ω(t)</b></i> are related as follows<br /><br /><i>d<b>φ/</b>d<b>t = φ'(t) = ω(t)</b></i><br /><br /><br /><br />This is sufficient to determine the value of <i><b>φ(t)</b></i> by integration of the angular speed on a time interval [<i><b>0,t</b></i>]<br /><br /><i><b>φ(t) = <span style="font-size: large;">∫</span><sub>[0,t]</sub> ω(τ)·</b>d<b>τ</b></i><br /><br /><br /><br /><i>Magnetic field flux</i> <i><b>Φ(t)</b></i> flowing through a wire frame depends on the intensity of the magnetic field <i><b>B</b></i>, area of a wire frame <i><b>S=a·b</b></i> and angle <i><b>φ(t)</b></i> the plane of a wire frame makes with magnetic field lines.<br /><br /><i><b>Φ(t) = B·S·sin(φ(t))</b></i><br /><br /><br /><br />The electromotive force (voltage) <i><b>U(t)</b></i> generated by rotating wire frame equals to a rate of change (first derivative) of the magnetic field flux<br /><br /><i><b>U(t) = </b>d<b>Φ(t)/</b>d<b>t =<br /><br />= B·S·cos(φ(t))·φ'(t) =<br /><br />= B·S·cos(φ(t))·ω(t)</b></i><br /><br />where angle <i><b>φ(t)</b></i> can be obtained by an integration of an angular speed presented above.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-79322270840282135122020-06-23T20:01:00.003-07:002020-06-25T06:47:14.708-07:00UNIZOR.COM - Physics4Teens - Electromagnetism - The Laws of Induction - Wire Frame Rotation<br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Wire Frame Rotation </u><br /><br/>Let's start with a summary of our knowledge about electromagnetic induction before embarking on a rotation of a wire frame in a magnetic field.<br/><br/>The foundation of electromagnetic induction is the <i>Lorentz force</i> that is exerted by a magnetic field onto an electrically charged particle moving across the magnetic field lines.<br/><img src='http://www.unizor.com/Pictures/LorentzForce_Problem_1c.png' style='width:200px;height:150px;'><br/>As we demonstrated in previous lectures, this force equals to<br/><i><b>F = q·V·B</b></i><br/>where<br/><i><b>q</b></i> is the particle's electric charge,<br/><i><b>V</b></i> is the speed of this charge in a direction perpendicular to magnetic field lines (if it's not perpendicular, it should be multiplied by <i>sin(φ)</i>, where <i>φ</i> is an angle between a velocity vector of a particle and magnetic field lines),<br/><i><b>B</b></i> is the intensity of a magnetic field.<br/>In vector form, considering general direction of a particle's velocity and magnetic field intensity vector, the formula is<br/><i><b><span style='text-decoration:overline'>F </span>=q·<span style='text-decoration:overline'>V </span></i>⨯<i> <span style='text-decoration:overline'>B </span></b></i> <br/>The important fact about the <i>Lorentz force</i>, acting on an electrically charged moving particle (like an electron) is that it acts perpendicularly to both velocity of a particle and the magnetic field intensity vectors.<br/><br/>When we move a metal rod parallel to itself and perpendicularly to magnetic field lines, "free" electrons are pulled by this motion towards the general direction of a rod (green arrow on a picture below). <img src='http://www.unizor.com/Pictures/LorentzForce_Rod.png' style='width:200px;height:128px;'><br/>This movement of electrons is perpendicular to magnetic field lines and, therefore, is a subject to <i>Lorentz force</i> that pushes electrons perpendicularly to a general direction of a rod's movement and perpendicularly to magnetic field lines, that is electrons are pushed along the rod to its edge (red vertical arrow on a picture above).<br/><br/>This generates the difference of electric potential on the opposite ends of a rod, one will be positive, another - negative. The faster we move the rod - the stronger Lorentz force is - the greater difference of electric potential will be generated.<br/><br/>The Lorentz force that pushes the electrons to the end of a moving rod meets the resistance of electrostatic forces that repel electrons from each other (blue vertical arrow on a picture above). That's why the concentration of "free" electrons at one end of a rod that moves with certain speed reaches its maximum and stays this way. For every speed of movement of a rod and a particular magnetic field intensity there is a specific difference of potentials or <i>electromotive force (EMF)</i> developed at the ends of a rod.<br/><br/>Let's connect the ends of our rod, that accumulated positive and negative charges as a result of the Lorentz force on moving "free" electrons, through any consumer of electricity, like a lamp.<br/><img src='http://www.unizor.com/Pictures/LorentzForce_Circuit.png' style='width:200px;height:128px;'><br/>The accumulated excess of electrons from one end of a rod will go to the connecting wire to a lamp and to the positively charged end of a rod. This will diminish a concentration of electrons on a negative end of a rod. While a rod continues its movement across the magnetic field, the Lorentz force acting on "free" electrons inside a rod will push more electrons to the end to replace those that went through a lamp to a positive end of a wire, thus establishing a constant electric current in an electric circuit, generated by mechanical movement of a rod perpendicularly to magnetic field lines.<br/><br/>From the first glance it looks like movement of a rod with constant speed, which does not require any energy to spend, generates electricity and, therefore, free energy, which contradicts the Law of Conservation of Energy. What do we miss?<br/><br/>Let's make one more step in our analysis of this experiment. The electric current generated by a rod's movement runs around a circuit and, therefore, runs through a rod as well perpendicularly to magnetic field lines. As we know, the wire with electric current perpendicular to magnetic field lines experiences another manifestation of the Lorentz force - the one that pushes the whole rod perpendicularly to its length.<br/><br/>The Lorentz force analyzed first acts on electrons moving with a rod to the right on the picture above and pushes them to one end of a rod (upwards on the picture). This creates an electric current in a rod, and the Lorentz force, acting on electrons moving along the rod pushes them and the whole rod with them to the left against the initial movement of a rod, thus forcing the need for some outside force to move the rod with constant speed.<br/><br/>As we see, a magnetic field acts <b>against</b> the movement of a rod.<br/>It means, to move a rod, we have to overcome the resistance and perform some work. This work, according to the Law of Energy Conservation, is converted in some other type of energy, like into heat in the lamp on a circuit.<br/><br/>If we follow the trajectory of "free" electrons inside a rod, we would see that they move simultaneously in two directions - to the right with rod's movement and upwards because of the Lorentz force acting on moving particles in a magnetic field perpendicularly to the field's intensity vector.<br/><br/>In vector form the velocity vector of an electron can be represented as<br/><i><b><span style='text-decoration:overline'>V<sub>e</sub></span> = <span style='text-decoration:overline'>V<sub>r</sub></span> + <span style='text-decoration:overline'>V<sub>u</sub></span></b></i><br/>where<br/><i><b><span style='text-decoration:overline'>V<sub>e</sub></span></b></i> is velocity of an electron,<br/><i><b><span style='text-decoration:overline'>V<sub>r</sub></span></b></i> is its velocity to the right,<br/><i><b><span style='text-decoration:overline'>V<sub>u</sub></span></b></i> is its velocity upwards.<br/><br/>Therefore, the total Lorentz force is<br/><i><b><span style='text-decoration:overline'>F </span>=q·(<span style='text-decoration:overline'>V<sub>r </sub></span>+<span style='text-decoration:overline'>V<sub>u </sub></span>)</i>⨯<i> <span style='text-decoration:overline'>B </span> =<br/>= q·<span style='text-decoration:overline'>V<sub>r </sub></span></i>⨯<i> <span style='text-decoration:overline'>B </span> + q·<span style='text-decoration:overline'>V<sub>u </sub></span></i>⨯<i> <span style='text-decoration:overline'>B </span> =<br/>= <span style='text-decoration:overline'>F<sub>u </sub></span> + <span style='text-decoration:overline'>F<sub>l </sub></span></b></i> <br/>where<br/><i><b><span style='text-decoration:overline'>F<sub>u</sub></span></b></i> is upward component of the Lorentz force that pushes electrons towards the end of a rod,<br/><i><b><span style='text-decoration:overline'>F<sub>l </sub></span></b></i> is its component to the left against the rod's motion.<br/><br/>The <i>electromotive force (EMF)</i> generated by the movement of a rod was quantitatively evaluated in the previous lecture. As the rod moves to the right, thus increasing the length of two sides of a circuit's rectangular frame, the magnitude of a generated EMF was equal to the rate of a change of a <i>magnetic field flux</i> through the frame:<br/><i><b>|U| = |</b>d<b>Φ/</b>d<b>t|</b></i><br/>where the <i>magnetic field flux</i> was defined as a product of <i>magnetic field intensity <b>B</b></i> and the <i>area</i> of a wire frame evaluated in a direction perpendicular to the magnetic field lines.<br/>This is the <b>Faraday's Law</b></i> of <i>electromagnetic induction</i>.<br/><br/>To better understand a concept of <i>magnetic field flux</i> we can use a concept of the "number" of <i>magnetic field lines</i>. These lines point toward the direction of magnetic field forces and their density represents the strength of these forces. Using this concept, <i>magnetic field flux</i> is analogous to the "number" of magnetic field lines going through a wire frame. Wider wire frame, more intense magnetic field - greater <i>magnetic field flux</i> goes through the wire frame.<br/><br/>That concludes the summary of our knowledge about electromagnetic induction, as presented so far, and we are ready to switch to a more practical way of generating the electricity from mechanical movement - from a rotation of a wire frame.<br/><br/>Considering it's impractical to change the magnetic flux by changing the geometry of a wire frame, a different approach to generate electricity was suggested - to rotate a wire frame. Rotating it, we change the angle between the wire frame plane and the magnetic field lines, thus changing the <i>magnetic flux</i> going through a frame.<br/>Let's examine closer this process.<br/><br/>Consider a rotation of a wire frame with dimensions <i><b>a</b></i> by <i><b>b</b></i>, rotating with a constant angular speed <i><b>ω</b></i> in the uniform magnetic field <i><b>B</b></i>.<br/><br/><img src='http://www.unizor.com/Pictures/InductionFrame.png' style='width:200px;height:200px;'><br/><br/>Consider the initial position of a wire frame to be, as shown on a picture above, with an axis of rotation perpendicular to magnetic field lines and side <i><b>b</b></i> of a frame parallel to magnetic field lines. Then side <i><b>a</b></i> would be perpendicular to magnetic field lines.<br/>As the frame rotates with an angular speed <i><b>ω</b></i>, the angle of a side <i><b>b</b></i> with the lines of a magnetic field, as a function of time, is<br/><i><b>φ(t) = ω·t</b></i><br/><br/>Side <i><b>b</b></i> of a wire frame and its opposite do not sweep across magnetic field lines, only side <i><b>a</b></i> and its opposite do.<br/>Side <i><b>a</b></i> and its opposite are crossing magnetic field lines in opposite directions. Viewing in the direction of the magnetic field lines, if side <i><b>a</b></i> crosses the magnetic field lines from right to left, the opposite side at the same time crosses these magnetic lines left to right.<br/><br/>As the wire frame rotates, the "free" electrons inside side <i><b>a</b></i> and its opposite are pushed by the rotation of a wire across the magnetic field lines in opposite directions and, therefore, are pushed by the Lorentz force, exerted by a magnetic field on these electrons, towards opposite ends of corresponding sides of wire frame. If electrons in side <i><b>a</b></i> are pushed upwards, the electrons in the opposite side are pushed downwards and vice versa.<br/>As a result, there is an electric current generated in a rotating wire frame.<br/><br/>When a wire crosses the magnetic field lines with "free" electrons inside, Lorentz force on electrons inside the wire is proportional to a speed of a wire across the magnetic field lines in a direction perpendicular to these lines. Consequently, the faster a wire is crossing the magnetic field lines in a perpendicular to them direction - the more electric charge inside a wire is separated between positive and negative ends. Therefore, the electric current generated by the movement of a wire across the magnetic field lines is proportional to a speed of perpendicularly crossing these lines.<br/><br/>At its original position at time <i><b>t=0</b></i> side <i><b>a</b></i> crosses more magnetic field lines per unit of time than at position of a wire frame turned by 90°. The "number" of magnetic field lines crossed per unit of time is maximum at angle <i><b>φ(t)=0</b></i> and minimum (actually, zero) at <i><b>φ(t)=π/2</b></i>.<br/>As the wire frame continues its rotation, the "number" of magnetic field lines crossed per unit of time increases and reaches another maximum at <i><b>φ(t)=π</b></i>, then diminishes to zero at <i><b>φ(t)=3π/2</b></i>.<br/><br/>To calculate the speed of crossing the magnetic field lines in a perpendicular to them direction by sides <i><b>a</b></i> and its opposite, consider a top view onto our wire frame.<br/><img src='http://www.unizor.com/Pictures/InductionFrameTop.png' style='width:200px;height:200px;'><br/>Side <i><b>a</b></i> and its opposite are circulating around the axis of rotation on a radius <i><b>b/2</b></i>. During time <i><b>t</b></i> the angle of rotation will be <i><b>φ(t)=ω·t</b></i>.<br/><br/>The speed of crossing the magnetic field lines by side <i><b>a</b></i> or its opposite is a Y-component of a velocity vector of the points <i><b>P</b></i> or its opposite.<br/><br/>The Y-coordinate of point <i><b>P</b></i> is<br/><i><b>PQ = b·sin(φ)/2 = b·sin(ω·t)/2</b></i>.<br/>Therefore, the Y-component of a vector of velocity of point <i><b>P</b></i> is a derivative of this function<br/><i><b>V<sub>Y</sub> = b·ω·cos(ω·t)/2</b></i>.<br/><br/>Analyzing this expression, we quantitatively confirm our considerations about speed of crossing the magnetic field lines presented above.<br/>At <i><b>t=0</b></i> this function is at maximum and equals to <i><b>b·ω/2</b></i> - linear speed of a point-object rotating with angular speed <i><b>ω</b></i> on a radius <i><b>b/2</b></i>.<br/>At <i><b>t=90°</b></i> the value of <i><b>V<sub>Y</sub></b></i> is zero because the point <i><b>P</b></i> moves parallel to X-axis.<br/>Than the Y-component of velocity goes to negative part, crossing the magnetic field lines in an opposite direction etc.<br/>During one full rotation the direction of an electric current in a wire frame will change because each side half of a circle crosses the magnetic field lines in one direction and on another half of a circle crosses them in an opposite direction. This is how <b>alternate current</b> is produced by electric power plants.<br/><br/>As Lorentz force on "free" electrons within side <i><b>a</b></i> separates them from their atoms and pushes to one end of this side, Lorentz force on the opposite side of a frame performs similar action, separating electrons towards the opposite end of a wire, thus assuring the electrical current inside a wire frame <i><b>I=I(t)</b></i>.<br/><br/>Rotation of side <i><b>a</b></i> and its opposite can be represented as simultaneous movement along two coordinates. Movement along X-axis, as presented on a picture above, does not cross magnetic field lines, there is no Lorentz force related to this movement, no separation of electrons from their nuclei, no EMF.<br/>But movement along Y-axis does cross magnetic field lines and, therefore, the Lorentz force pushes electrons to one end of a wire and resists the rotation by acting against the movement along the Y-axis.<br/>As we have an electric current <i><b>I(t)</b></i> inside a wire frame, there is a Lorentz force acting upon side <i><b>a</b></i> and its opposite that needs to be overcome to assure the rotation with constant angular speed <i><b>ω</b></i>.<br/><br/>The force on side <i><b>a</b></i> directed against its movement along the Y-axis is<br/><i><b>F<sub>a</sub>(t) = I(t)·a·B</b></i><br/>and the same by magnitude force, but acting upon a side opposite to <i><b>a</b></i>, also against its movement along Y-axis.<br/>Both these forces are acting <b>against</b> the rotation. When side <i><b>a</b></i> moves in the positive direction of the Y-axis, the force <i><b>F<sub>a</sub>(t)</b></i> is directed towards the negative one and vice versa. Same with the side opposite to side <i><b>a</b></i>.<br/><br/>The key consideration now is the following.<br/><b>The force <i><b>F<sub>a</sub>(t)</b></i> and its opposite must be overcome by a rotation, that is some work must be exerted by a rotating mechanism, which, according to the Law of Energy Conservation, must be equal to amount of work the generated electric current exerts by circulating in the wire frame</b>.<br/>That is, mechanical work of rotation is converted into electrical work - electric power is generated by a rotation.<br/><br/>The Y-coordinate of point <i><b>P</b></i>, rotating around the origin of coordinates with an angular speed <i><b>ω</b></i> on a radius <i><b>b/2</b></i>, as a function of time <i><b>t</b></i>, is<br/><i><b>P<sub>Y</sub>(t) = b·sin(ω·t)/2</b></i><br/>The Y-component of a velocity vector of point <i><b>P</b></i> is a derivative of this function<br/><i><b>V<sub>Y</sub>(t) = b·ω·cos(ω·t)/2</b></i><br/><br/>During the time from <i><b>t</b></i> to <i><b>t+</b>d<b>t</b></i> point <i><b>P</b></i> and, therefore, side <i><b>a</b></i> will be moved by a rotation in the Y-direction by a distance<br/><i>d<b>S<sub>Y</sub> = V<sub>Y</sub>(t)·</b>d<b>t =<br/>= b·ω·cos(ω·t)·</b>d<b>t/2</b></i><br/>Therefore, the work needed to overcome the force <i><b>F<sub>a</sub>(t)</b></i>, acting in the direction of Y-axis against the rotation of side <i><b>a</b></i>, during the time from <i><b>t</b></i> to <i><b>t+</b>d<b>t</b></i> is<br/><i>d<b>W<sub>a</sub> = F<sub>a</sub>(t)·V<sub>Y</sub>(t)·</b>d<b>t =<br/>= </b></i>[<i><b>I(t)·a·B</b></i>]<i><b>·b·ω·cos(ω·t)·</b>d<b>t/2</b></i><br/><br/>The same by magnitude and opposite in direction force acts on a side opposite to <i><b>a</b></i>, also acting against its movement in the direction of the Y-axis. Therefore, the increment of work needed to be exerted by a rotation mechanism to assure the rotation at angular speed <i><b>ω</b></i>, as a function of time <i><b>t</b></i> during the time from <i><b>t</b></i> to <i><b>t+</b>d<b>t</b></i> is<br/><i>d<b>W = </b></i>[<i><b>I(t)·a·B</b></i>]<i><b>·b·ω·cos(ω·t)·</b>d<b>t</b></i><br/><br/>The power exerted by a rotating mechanism at time <i><b>t</b></i> is<br/><i><b>P(t) = </b>d<b>W/</b>d<b>t =<br/>= I(t)·a·B·b·ω·cos(ω·t)</b></i><br/><br/>During this period from <i><b>t</b></i> to <i><b>t+</b>d<b>t</b></i> the electricity generated by the rotation of the wire has the value of current equal to <i><b>I(t)</b></i>. The cause of this current is the <i>electromotive force (EMF)</i> or, simply, <i>voltage</i> <i><b>U(t)</b></i> generated by Lorentz force on "free" electrons that separates them from their atoms.<br/><br/>The power exerted by an electric current <i><b>I</b></i> with voltage <i><b>U</b></i> is <i><b>P=I·U</b></i>.<br/>Equating this power of electric current to a power exerted by a rotating mechanism, as the Law of Energy Conservation dictates, we obtain<br/><i><b>I(t)·a·B·b·ω·cos(ω·t) = I(t)·U(t)</b></i><br/>From this we can find EMF generated by a rotation of a wire frame:<br/><i><b>U(t) = B·a·b·ω·cos(ω·t)</b></i><br/>Applying simple geometry, we can see that the area of a wire frame perpendicular to magnetic lines at time <i><b>t</b></i> equals to<br/><i><b>S(t) = a·b·sin(ω·t)</b></i><br/>Therefore, <i>magnetic field flux</i> through the wire frame at time <i><b>t</b></i> is<br/><i><b>Φ(t) = B·a·b·sin(ω·t)</b></i><br/>The rate of change of this <i>flux</i> is, therefore, a derivative<br/><i><b></b>d<b>Φ(t)/</b>d<b>t = B·a·b·ω·cos(ω·t)</b></i><br/><br/>Comparing this with the above expression for <i>EMF</i>, we see that<br/><i><b>U(t) = </b>d<b>Φ(t)/</b>d<b>t</b></i><br/>which is the same as in a case of a rod moving parallel to itself on rails perpendicularly to the magnetic field lines, discussed in the prior lecture.<br/>This brings us to an important generalization of the <b>Faraday's Law</b>.<br/><b>If the magnetic flux going through a wire loop is changing, the rate of this change equals to an electromotive force generated inside a wire</b>. Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-49882413716692473062020-06-17T09:18:00.001-07:002020-06-17T09:18:57.932-07:00UNIZOR.COM - Physics4Teens - Electromagnetism - The Laws of Induction - ...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/7BCnjx84m-4" width="480"></iframe> <i> </i><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Faraday's Law of<br /><br />Electromagnetic Induction</u><br /><br /><br /><br />Let's consider an experiment we described in a lecture about Lorentz <br />force exerted by a uniform magnetic field onto a wire with an electric <br />current running through it.<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce.jpg" style="height: 145px; width: 200px;" /><br /><br />On this picture the density of magnetic field lines (red arrows) <br />represents the intensity of this field. If, for example, one line per <br />square centimeter of an area perpendicular to the direction of magnetic <br />field lines means the field of intensity 0.001T (i.e. 1/1000 of <i>tesla</i>), the density of 10 such lines per square centimeter signifies the intensity 0.01T (i.e. 1/100 of <i>tesla</i>).<br /><br />In this lecture we will use an expression "the number of magnetic field <br />lines" in a context of a wire crossing them or sweeping across them. It <br />only means to demonstrate that the intensity of a magnetic field should <br />participate in our logic, formulas and calculations as a factor <br />proportional to a perpendicular to magnetic lines area, crossed or swept<br /> across.<br /><br /><br /><br />Recall that the Lorentz force exerted by a uniform magnetic field onto a<br /> wire with an electric current running through it for any angle <i><b>φ</b></i> between the electric current in a wire and magnetic field lines equals to<br /><br /><i><b>F = I·L·B·sin(φ)</b></i><br /><br />where<br /><br /><i><b>I</b></i> is the <i>amperage</i> of an electric current running through a wire,<br /><br /><i><b>L</b></i> is the length of a wire,<br /><br /><i><b>B</b></i> is the strength of an external magnetic field<br /><br /><i><b>φ</b></i> is the angle between the direction of the electric current and lines of an external magnetic field<br /><br /><br /><br />Taking into consideration the direction of the Lorentz force <br />perpendicular to both vectors - electric current and lines of an <br />external uniform magnetic field, the above formula can be represented <br />using a <i>vector product</i><br /><br /><i><b><span style="text-decoration: overline;">F </span> = <span style="text-decoration: overline;">I </span>·L⨯ <span style="text-decoration: overline;">B </span> </b></i><br /><br /><br /><br />In a simple configuration of the magnetic field and wire on a picture above <i><b>φ=90°</b></i> and the formula for a magnitude of a Lorentz force looks like<br /><br /><i><b>F = I·L·B</b></i><br /><br /><br /><br />From the microscopic viewpoint the magnetic field exerts a Lorentz force<br /> onto each electron moving inside the wire as a part of the electric <br />current. This force is perpendicular to both magnetic field lines and <br />the electric current in a wire and, consequently, causes the motion of <br />an entire wire.<br /><br />If an electric current in a wire is produced by a charge <i><b>q</b></i> passing through a wire of length <i><b>L</b></i> during time <i><b>t</b></i>, thus making the electric current equal to <i><b>I=q/t</b></i> with the speed of moving <i><b>V=L/t</b></i>, the formula for Lorentz force acting on this charge becomes<br /><br /><i><b>F = I·L·B = (q/t)·(V·t)·B =<br /><br />= q·V·B</b></i><br /><br /><br /><br />The cause of the electric current in a wire and, therefore, motion of <br />electrons in one direction in the experiment above is some <i>voltage</i> or <i>electromotive force (EMF)</i> applied to wire's ends. So, the voltage or EMF at wire's ends causes the current in it and, consequently, its movement.<br /><br />But what happens if we attempt to reverse the cause and effect and move the wire ourselves without any voltage applied to it?<br /><br />Will the voltage be generated?<br /><br /><br /><br />Consider a similar experiment, but no initial current running through a <br />wire. Instead, the wire is connected to an ammeter to measure the <br />intensity of the electric current running through it, thus making a <br />closed electric circuit.<br /><br /><img src="http://www.unizor.com/Pictures/MagInduction.jpg" style="height: 145px; width: 200px;" /><br /><br />Let's move the wire in the upward direction as indicated on a picture. <br />The experiment shows that, while the wire is moving across the magnetic <br />field lines, there is an electric current in it.<br /><br />Let's examine the reason why it happens.<br /><br /><br /><br />There are positively charged nuclei of atoms relatively fixed in their <br />position by inter-atomic forces and there are electrons rotating around <br />these nuclei in each atom on different orbits. Usually, one or two <br />electrons from the outer orbits are relatively free to drift or exchange<br /> their atom hosts. These "free" electrons are moving together with the <br />wire.<br /><br /><br /><br />As we investigated the motion of charged particles in a magnetic field, <br />we came to a conclusion that there is a force acting on these charged <br />particles (electrons in our case) directed perpendicularly to their <br />trajectory - the Lorentz force.<br /><br /><br /><br />Moving the wire across the magnetic field lines (upwards on a picture <br />above) results in the Lorentz force to act on "free" electrons, which <br />causes electrons to move in the direction of the Lorentz force <br />perpendicularly to their trajectory, while still being inside a wire. <br />So, they will move to one side of a wire, generating the difference in <br />charges on the wire's ends, that is generating the difference in <br />electric potentials or <i>voltage</i> or <i>electromotive force (EMF)</i>.<br />If we have a loop with this wire being a part of it (like on a picture <br />with an ammeter in a loop), there will be an electric current in a wire.<br /><br /><br /><br />As we see, the movement of a wire across the magnetic field lines generates a difference in electric potentials or an <i>electromotive force (EMF)</i> and, if there is a closed loop, an <i>electric current</i> in a wire. This is called <b>electromagnetic induction</b>.<br /><br /><br /><br />Let's quantify the <i>voltage</i> (or <i>electromotive power - EMF</i>) <br />generated by a straight line wire moving in a uniform magnetic field <br />parallel to itself in a plane perpendicular to magnetic field lines.<br /><br /><img src="http://www.unizor.com/Pictures/Faraday.png" style="height: 111px; width: 200px;" /><br /><br />In a setup presented on the above picture the straight line wire <i><b>PQ</b></i> is moving parallel to itself and parallel to Y-axis along the rails <i><b>AP</b></i> and <i><b>OQ</b></i> stretched parallel to X-axis and connected by a resistor <i><b>R</b></i>, making a closed circuit with a wire <i><b>PQ</b></i> being a part of it.<br /><br /><br /><br />The uniform magnetic field of intensity <i><b>B</b></i> is directed along the Z-axis perpendicularly to XY-plane in a direction of viewing the picture.<br /><br /><br /><br />"Free" electrons, carried by a wire's motion along X-axis, experience <br />the Lorentz force directed perpendicularly to their movement inside a <br />wire, that is they will move to one end of a wire. The Lorentz force <br />acting on each such electron equals to (see <i>Problem 1c</i> in "<b>Magnetic Field</b>" topic of this course)<br /><br /><i><b>F<sub>L</sub> = e·V·B</b></i><br /><br />where<br /><br /><i><b>e</b></i> is an electric charge of an electron that is able to "freely" move,<br /><br /><i><b>V</b></i> is the speed a wire moving along the X-axis,<br /><br /><i><b>B</b></i> is the intensity of the magnetic field.<br /><br /><br /><br />As a result, the difference in electric potential (or <i>EMF</i>) is generated and an electric current <i><b>I</b></i> will start running in the closed circuit <i><b>OAPQ</b></i>.<br /><br />So, Lorentz force on electrons moving with a wire along the X-axis <br />causes the electrons to move to one end of a wire, that causes the <br />electric current in the wire. The latter causes the force exerted by the<br /> magnetic field onto a wire in the direction opposite to its movement.<br /><br /><br /><br />Since we are moving the wire ourselves with constant speed <i><b>V</b></i><br /> towards the positive direction of X-axis, we have to exert work against<br /> the force of magnetic field onto a wire directed in the negative <br />direction.<br /><br /><br /><br />This force of magnetic field onto a wire directed in the negative direction equals, as we mentioned above, to<br /><br /><i><b>F = I·L·B</b></i><br /><br />where <i><b>L</b></i> is the length of a wire <i><b>PQ</b></i>.<br /><br />If during time <i><b>t</b></i> the wire moves by a distance <i><b>S=V·t</b></i>, the work we have to do against the force <i><b>F</b></i> equals to<br /><br /><i><b>W = F·S = I·L·B·V·t</b></i><br /><br />To do this work during a time interval <i><b>t</b></i>, we have to exert a power<br /><br /><i><b>P = W/t = F·S = I·L·B·V</b></i><br /><br /><br /><br />Energy does not disappear without a trace. In this case the energy we spend generated voltage (EMF) <i><b>U</b></i> at wire's ends and an electric current <i><b>I</b></i> in a closed circuit. It will be transformed into heat in the resistor <i><b>R</b></i> with a rate <i><b>I·U</b></i>.<br /><br />Therefore, we can equate two powers - the one we exerted to move a wire and the one consumed by a resistor:<br /><br /><i><b>I·L·B·V = I·U</b></i><br /><br />from which we obtain the amount of EMF generated by moving the wire<br /><br /><i><b>U = L·V·B</b></i><br /><br /><br /><br />Let's examine the expression on the right.<br /><br />The value <i><b>L·V</b></i> represents the rate of increase of the area of a circuit <i><b>OAPQ</b></i> per unit of time.<br /><br />The product of the area and the intensity of the magnetic field with <br />magnetic field lines going through this area perpendicularly to it is <br />called <i>magnetic flux</i> <i><b>Φ</b></i>.<br /><br />Since our magnetic field is uniform, an expression <i><b>L·V·B</b></i> represents the rate of increase of <i>magnetic flux</i>, which we can safely denote using the language of calculus as <i>d<b>Φ/</b>d<b>t</b></i>.<br /><br /><br /><br />In this language we can state that the <b><i>electromotive force (EMF)</i> generated by a moving wire - a part of a closed circuit - equals to a rate of increase of <i>magnetic flux</i> going through this closed circuit</b>.<br /><br /><i><b>U = </b>d<b>Φ/</b>d<b>t</b></i>.<br /><br />This is the <b>Faraday's Law</b> of <i>electromagnetic induction</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-1691905051918393832020-06-05T07:05:00.001-07:002020-06-05T07:05:20.579-07:00UNIZOR.COM - Physics4Teens - Electromagnetism - Magnetism and Electric C...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Parallel Currents - Problems 2</u><br /><br /><br /><br />IMPORTANT NOTES<br /><br /><br /><br />Recall the Lorentz force exerted by a magnetic field <i><b><span style="text-decoration: overline;">B </span></b></i> (a vector) onto straight line electric current <i><b><span style="text-decoration: overline;">I </span></b></i> (also a vector) of length <i><b>L</b></i><br /><br /><i><b><span style="text-decoration: overline;">F </span> = <span style="text-decoration: overline;">I </span>·L⨯ <span style="text-decoration: overline;">B </span></b></i><br /><br />The magnitude of this force can be expressed in terms of magnitudes of <br />the vectors of electric current and magnetic field, taking into <br />consideration an angle <i><b>θ</b></i> between them:<br /><br /><i><b>F = I·L·B·sin(θ)</b></i><br /><br />In case electric current and magnetic field vectors are perpendicular to each other it looks as<br /><br /><i><b>F = I·L·B</b></i><br /><br />The last formula will be used in this levture.<br /><br /><br /><br />Also, as we know from previous lectures, the magnetic field force lines <br />around a long thin straight line wire with electric current <i><b>I</b></i><br /> running through it are circular in planes perpendicular to a wire and <br />centered at the points of the wire. Their direction is determined by a <br />corkscrew rule or the rule of the right hand.<br /><br />On a distance <i><b>R</b></i> from the wire the vector of intensity of a magnetic field produced by this wire has a magnitude<br /><br /><i><b>B = μ<sub>0</sub>·I/(2π·R)</b></i><br /><br />where <i><b>μ<sub>0</sub></b></i> is a constant <i>permeability</i> of vacuum.<br /><br /><br /><br />The direction of this force vector at any point of space around a wire <br />corresponds to the direction of the circular magnetic field line going <br />through this point and is tangential to this magnetic line.<br /><br /><br /><br /><i>Problem 2A</i><br /><br />Two ideally long and thin straight wires are parallel to each other and positioned at distance <i><b>d</b></i> from each other.<br /><br />One wire carries an electric current of amperage <i><b>I<sub>1</sub></b></i>.<br /><br />Another wire carries an electric current of amperage <i><b>I<sub>2</sub></b></i> in the opposite direction.<br /><br />Determine the magnitude of the repelling force between these two wires per unit of length of each.<br /><br /><br /><br /><i>Solution</i><br /><br /><img src="http://www.unizor.com/Pictures/TwoCurrents_Problem_2A.png" style="height: 180px; width: 200px;" /><br /><br />Consider the first wire as having an infinite length, while the second one to have a finite length <i><b>L</b></i>.<br /><br />The magnetic field of the first wire at each point of the second wire can be considered as uniform, having a magnitude<br /><br /><i><b>B<sub>1</sub> = μ<sub>0</sub>·I<sub>1</sub>/(2π·d)</b></i><br /><br />According to the formula for a force exerted by a uniform magnetic field<br /> onto a wire with electric current that is perpendicular to the force of<br /> magnetic field, the repelling Lorentz force exerted by the first wire <br />onto the second one is<br /><br /><i><b>F<sub>12</sub> = I<sub>2</sub>·L·B<sub>1</sub> =<br /><br />= μ<sub>0</sub>·I<sub>1</sub>·I<sub>2</sub>·L <span style="font-size: medium;">/</span>(2π·d)</b></i><br /><br />If we calculate this force per unit of length of the second wire, the result is<br /><br /><i><b><span style="text-decoration: overline;">F<sub>12</sub></span> = μ<sub>0</sub>·I<sub>1</sub>·I<sub>2</sub> <span style="font-size: medium;">/</span>(2π·d)</b></i><br /><br /><br /><br />Now consider a second wire as having an infinite length, while the first one to have a length <i><b>L</b></i>.<br /><br />All previous results are applicable to this reverse logic, and we can <br />determine the magnitude of the repelling Lorentz force exerted by the <br />second wire per unit of length of the first<br /><br /><i><b><span style="text-decoration: overline;">F<sub>21</sub></span> = μ<sub>0</sub>·I<sub>1</sub>·I<sub>2</sub> <span style="font-size: medium;">/</span>(2π·d)</b></i><br /><br />It's also a repelling force, so it's direction is opposite to <i><b><span style="text-decoration: overline;">F<sub>12</sub></span></b></i>.<br /><br /><br /><br />It's not coincidental that the forces <i><b><span style="text-decoration: overline;">F<sub>12</sub></span></b></i> and <i><b><span style="text-decoration: overline;">F<sub>21</sub></span></b></i> are equal in magnitude, since it corresponds to the Third Newton's Law.<br /><br /><br /><br /><i>Problem 2B</i><br /><br />Two identical sufficiently long and thin straight wires of mass <i><b>m</b></i> per unit of length are horizontally hanging on threads of the same length <i><b>L</b></i> parallel to each other at the same height. The initial distance between them is <i><b>d</b></i> and they can swing on their corresponding threads parallel to themselves.<br /><br />Initially there is no current in these wires and the threads holding these wires are in vertical position.<br /><br />When we run electric current <i><b>I</b></i> through the first wire and the same in amperage electric current <i><b>I</b></i> through the second one in the opposite direction, the wires swing away from each other on their threads.<br /><br />At equilibrium both threads made an angle <i><b>φ</b></i> with a vertical.<br /><br />Determine the amperage <i><b>I</b></i>.<br /><br /><br /><br /><i>Solution</i><br /><br />The following picture represents the view along the direction of wires <br />before and after the electric currents in both wires are turned on.<br /><img src="http://www.unizor.com/Pictures/TwoCurrents_Problem_2B.png" style="height: 170px; width: 200px;" /><br /><br />The wires are represented as circles. Before the electric current is <br />turned on the circles are white. When the electric current is on, the <br />circles are blue and red to represent the opposite directions of the <br />electric current in these wires.<br /><br /><br /><br />Let <i><b>T</b></i> be the tension of a thread per unit length of a wire.<br /><br />A vertical component of the thread tension per unit length of a wire <i><b>T·cos(φ)</b></i> balances the weight of the unit length of a wire <i><b>m·g</b></i>, which can be expressed as an equation<br /><br /><i><b>T·cos(φ) = m·g</b></i><br /><br />from which we derive<br /><br /><i><b>T = m·g/cos(φ)</b></i><br /><br /><br /><br />A horizontal component of the thread tension per unit length of a wire <i><b>T·sin(φ)</b></i> is supposed to balance the magnetic field repelling force, which has been calculated in the above Problem 2A as<br /><br /><i><b><span style="text-decoration: overline;">F<sub>12</sub></span> = μ<sub>0</sub>·I<sub>1</sub>·I<sub>2</sub> <span style="font-size: medium;">/</span>(2π·d)</b></i><br /><br />except in our case<br /><br /><i><b>I<sub>1</sub> = I<sub>2</sub> = I</b></i><br /><br />and, instead of the distance <i><b>d</b></i> we have to use <i><b>D=d+2L·sin(φ)</b></i> - the final distance between the wires at the point of equilibrium.<br /><br />The result is<br /><br /><i><b>T·sin(φ) = μ<sub>0</sub>·I² <span style="font-size: medium;">/</span>(2π·D)</b></i><br /><br />from which follows<br /><br /><i><b>I² = 2π·D·T·sin(φ)/μ<sub>0</sub> =<br /><br />= 2π·D·m·g·sin(φ)/(μ<sub>0</sub>·cos(φ)) =<br /><br />= 2π·D·m·g·tan(φ)/μ<sub>0</sub></b></i><br /><br />This gives the amperage of the current equal to<br /><br /><i><b>I = √<span style="text-decoration: overline;">2π·D·m·g·tan(φ)/μ<sub>0</sub></span></b></i><br /><br />where <i><b>D = d+2L·sin(φ)</b></i><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-45193156265255669902020-06-03T09:07:00.001-07:002020-06-03T09:07:36.467-07:00Unizor - Physics4Teens - Electromagnetism - Magnetism and Electric Curre...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Two Parallel Straight Line Currents</u><br /><br /><br /><br />Let's examine a behavior of two ideally long and thin parallel wires, each carrying an electric current.<br /><br /><img src="http://www.unizor.com/Pictures/TwoCurrents.png" style="height: 200px; width: 200px;" /><br /><br />Two wires carrying an electric current are blue and brown on this picture.<br /><br />As we know, since the wires carry electric current, there is a magnetic <br />field around each of these wires. It has circular form, as presented on a<br /> picture above in corresponding colors. The direction of the magnetic <br />field lines is determined by the <i>corkscrew rule</i> or the <i>right hand rule</i> and is shown on the picture.<br /><br /><br /><br />Each circular magnetic field line around the blue wire and any <br />tangential to it, that is the vector of magnetic force, belong to a <br />plane perpendicular to this wire. Since the brown wire is parallel to <br />the blue one, any tangential to any magnetic field line of the blue wire<br /> is perpendicular to the brown wire.<br /><br /><br /><br />Since the brown wire's electric current is perpendicular to the <br />direction of the magnetic field lines of the field produced by the blue <br />wire, the brown wire is under the Lorentz force produced by the magnetic<br /> field of the blue wire. Using the <i>right hand rule</i> for the brown <br />wire and blue magnetic field lines, we determine the direction of this <br />Lorentz force. It appears to be directed from the brown wire towards the<br /> blue one.<br /><br /><br /><br />The wires are in symmetrical positions, so exactly the same logic can be<br /> applied to the blue wire in the magnetic field of the brown wire. As a <br />result, the Lorentz force produced by the magnetic field of the brown <br />wire attracts the blue wire towards the brown one.<br /><br /><br /><br />Now we see that both wires attract each other, which is caused by the <br />Lorentz force exerted on each of them from the magnetic field of the <br />other one.<br /><br /><br /><br />Let's arrange a similar experiment having electric currents in the wires<br /> to go in opposite directions. Let the blue wire's current to go the <br />same way as on the picture above, that is upward, but the direction of <br />the electric current in the brown wire to be downward.<br /><br />The logic about circular magnetic field lines around any wire and <br />perpendicularity of the electric current of one wire to the magnetic <br />field lines of the opposite one remains the same. The only difference is<br /> that, when we apply the <i>right hand rule</i> to determine the <br />direction of the Lorentz force, the result will be opposite than before,<br /> the Lorentz force will push the brown wire away from the blue one.<br /><br /><br /><br />Analogous result will be when we consider the electric current in the <br />blue wire in the magnetic field of the brown wire. The Lorentz force in <br />this case will also be repelling.<br /><br /><br /><br />The conclusion of these experiments is that<br /><br />(a) parallel wires carrying electric charge will attract each other if the currents are in the same direction;<br /><br />(b) parallel wires carrying electric charge will repel each other if the currents are in the opposite directions.<br /><br /><br /><br />This effect was studied by Ampere and the mutual attraction or repulsion<br /> of two parallel wires carrying an electric current is often referred to<br /> as the <i>Ampere force</i>.<br /><br /><br /><br />Contemporary definition of an <i>1 ampere (1A)</i> as a unit of electric current is <i>1 coulomb per 1 second (1C/1sec)</i> with a unit of charge <i>1 coulomb</i> defined based on the charge of certain number of electrons.<br /><br /><br /><br />At the time of Ampere's experiments with parallel wires scientists did <br />not know about electrons and could not measure the electric current in <br />the same terms as we do now.<br /><br />But Ampere's experiments have opened the way to measure the electric <br />current by measuring the purely mechanical characteristic of the force <br />of attracting or repelling between the wires carrying the electric <br />current. So, the first definition of a unit of current was based on <br />measuring the <i>Ampere force</i> between the two wires.<br /><br /><br /><br />Here is a simple description of how the electric current can be measured (not necessarily the method used by Ampere).<br /><br /><br /><br />Let two identical wires hang horizontally parallel to each other on threads of the same length from the same height.<br /><br />Without an electric current running through them these wires hang on vertical threads.<br /><br /><br /><br />When we apply voltage to both of them from the same battery, the <br />electric current in each of them produces the magnetic field that will <br />cause their mutual attraction or repelling, and the threads holding the <br />wires would deviate from their vertical position. The angle of deviation<br /> can be measured and it can be used to determine the force of attraction<br /> or repelling between the wires.<br /><br /><br /><br />So, knowing the unit of length and the unit of force, <i>1 ampere (1A)</i>,<br /> as a unit of electric current, can be defined as the electric current <br />that produces a unit of force per unit of length of a wire, when this <br />wire is positioned at a unit of length distance from another wire with <br />the same electric current.<br /><br /><br /><br />That was the original method of definition of the unit of electrical <br />current. At the time it was using old units of length and force.<br /><br />Later on the definition was slightly changed using concepts of <br />infinitely long and infinitesimally thin wires (to avoid irregularities <br />of the magnetic field) and contemporary measures of length (<i>meter</i>) and force (<i>newton</i>).Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-30443248570261692232020-05-31T19:52:00.001-07:002020-05-31T19:52:07.086-07:00Unizor - Physics4Teens - Electromagnetism - Magnetism and Electric Curre...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Two Currents - Problems 1</u><br /><br /><br /><br />IMPORTANT NOTES<br /><br /><br /><br />As we know from a previous lecture, the magnetic field force lines <br />around a long thin straight line wire with electric current running <br />through it are circular in planes perpendicular to a wire and centered <br />at the points of the wire.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticFieldOfCurrent.png" style="height: 200px; width: 200px;" /><br /><br /><br /><br />These lines are attributed a direction that can be determined using the <i>rule of the right hand</i> or Maxwell's <i>corkscrew rule</i>.<br /><br /><br /><br />The <i>rule of the right hand</i> states that, if you wrap your right <br />hand around a wire such that your thumb points to a direction of an <br />electric current in the wire, which is, by definition, from positive to <br />negative, then your fingers will point to a direction of the magnetic <br />field lines. This direction is the direction of any tangential to a line<br /> vector of force.<br /><br /><br /><br />The <i>corkscrew rule</i> states that, if we imagine a corked bottle <br />along the wire such that the direction of an electric current enters the<br /> bottle through a cork, to open the bottle we need to rotate the regular<br /> cork opener in the direction of the magnetic lines around the wire.<br /><br /><br /><br />The magnitude of the intensity of a magnetic field (a vector of force) depends on the electric current's amperage <i><b>I</b></i> and the distance <i><b>R</b></i> to the wire as follows<br /><br /><i><b>B = μ<sub>0</sub>I/(2π·R)</b></i><br /><br />where <i><b>μ<sub>0</sub></b></i> is a constant called <i>permeability</i> of vacuum.<br /><br /><br /><br />The direction of this force vector at any point of space around a wire <br />is always in the same plane as a circular magnetic line going through <br />the same point and is tangential to this magnetic line.<br /><br />The direction of the vector corresponds to the direction of the magnetic line.<br /><br /><br /><br /><i>Problem 1A</i><br /><br />Two ideally long and thin wires are positioned in space with Cartesian coordinates.<br /><br />One goes through point <i><b>A(0,0,a)</b></i> on Z-axis (<i><b>a > 0</b></i>) parallel to X-axis and carries an electric current of amperage <i><b>I</b></i> in the positive direction of X-axis.<br /><br />Another wire goes through point <i><b>B(0,0,−a)</b></i> on Z-axis parallel to Y-axis and carries an electric current of the same amperage <i><b>I</b></i> in the positive direction of Y-axis.<br /><br />Determine the X-, Y- and Z-components and magnitude of the vector of magnetic field intensity at the origin of coordinates <i><b>(0,0,0)</b></i>.<br /><br /><br /><br /><i>Solution</i><br /><br /><img src="http://www.unizor.com/Pictures/TwoCurrents_Problem_1A.png" style="height: 200px; width: 200px;" /><br /><br />The magnetic field at point <i><b>(0,0,0)</b></i> is a combination of two fields - one with intensity vector <i><b>B<sub>1</sub></b></i>, the source in the first wire that carries electric charge <i><b>I</b></i> parallel to X-axis, from which the origin of coordinates is at distance <i><b>a</b></i>, and another with intensity vector <i><b>B<sub>2</sub></b></i>, the source in the second wire that carries electric current <i><b>I</b></i> parallel to Y-axis, from which the origin of coordinates is at distance <i><b>a</b></i>.<br /><br />The resulting field intensity vector is a vector sum of vectors <i><b>B<sub>1</sub></b></i> and <i><b>B<sub>2</sub></b></i>.<br /><br /><br /><br />The first magnetic field has its force lines forming a cylindrical <br />surface with an axis being the first wire parallel to X-axis. At point <i><b>(0,0,0)</b></i> the direction of this magnetic field intensity vector is perpendicular to a radius from the first wire towards point <i><b>(0,0,0)</b></i>, that is along the Y-axis towards its positive direction.<br /><br />Considering the values of electric current <i><b>I</b></i> and the distance to the source (the first wire) equaled to <i><b>a</b></i>, the magnitude of this vector is<br /><br /><i><b>|B<sub>1</sub>| = μ<sub>0</sub>·I/(2π·a)</b></i><br /><br />So, this vector in Cartesian coordinates is<br /><br />{<i><b>0; μ<sub>0</sub>·I/(2π·a); 0</b></i>}<br /><br /><br /><br />The second magnetic field has its force lines forming a cylindrical <br />surface with an axis being the second wire parallel to Y-axis. At point <i><b>(0,0,0)</b></i> the direction of this magnetic field intensity vector is perpendicular to a radius from the second wire towards point <i><b>(0,0,0)</b></i>, that is along the X-axis towards its positive direction.<br /><br />Considering the values of electric current <i><b>I</b></i> and the distance to the source (the second wire) equaled to <i><b>a</b></i>, the magnitude of this vector is<br /><br /><i><b>|B<sub>2</sub>| = μ<sub>0</sub>·I/(2π·a)</b></i><br /><br />So, this vector in Cartesian coordinates is<br /><br />{<i><b>μ<sub>0</sub>·I/(2π·a); 0; 0</b></i>}<br /><br /><br /><br />Therefore, the combined vector of magnetic field intensity has its three coordinates<br /><br />{<i><b>μ<sub>0</sub>·I/(2π·a); μ<sub>0</sub>·I/(2π·a); 0</b></i>}<br /><br /><br /><br />The magnitude of this vector is<br /><br /><i><b>μ<sub>0</sub>·I·√<span style="text-decoration: overline;">2</span>/(2π·a)</b></i><br /><br /><br /><br /><i>Problem 1B</i><br /><br />Two ideally long and thin wires are positioned in space parallel to Z-axis.<br /><br />One goes through point <i><b>A(a,0,0)</b></i> (<i><b>a > 0</b></i>) and carries an electric current of amperage <i><b>I<sub>1</sub></b></i> in the negative direction of the Z-axis.<br /><br />Another wire goes through point <i><b>B(0,b,0)</b></i> (<i><b>b > 0</b></i>) and carries an electric current of amperage <i><b>I<sub>2</sub></b></i> in the positive direction of the Z-axis.<br /><br />Determine the X-, Y- and Z-components and magnitude of the vector of magnetic field intensity at the origin of coordinates <i><b>(0,0,0)</b></i>.<br /><br /><br /><br /><i>Solution</i><br /><br /><img src="http://www.unizor.com/Pictures/TwoCurrents_Problem_1B.png" style="height: 200px; width: 200px;" /><br /><br />The magnetic field at point <i><b>(0,0,0)</b></i> is a combination of two fields - one with intensity vector <i><b>B<sub>1</sub></b></i>, the source in the first wire that carries electric charge <i><b>I<sub>1</sub></b></i>, from which the origin of coordinates is at distance <i><b>a</b></i>, and another with intensity vector <i><b>B<sub>2</sub></b></i>, the source in the second wire that carries electric current <i><b>I<sub>2</sub></b></i>, from which the origin of coordinates is at distance <i><b>b</b></i>.<br /><br />The resulting field intensity vector is a vector sum of vectors <i><b>B<sub>1</sub></b></i> and <i><b>B<sub>2</sub></b></i>.<br /><br /><br /><br />The Z-coordinate of both vectors is zero.<br /><br /><br /><br />The first magnetic field has its force lines forming a cylindrical surface with an axis being the first wire. At point <i><b>(0,0,0)</b></i> the direction of this magnetic field intensity vector is perpendicular to a radius from the first wire towards point <i><b>(0,0,0)</b></i>, that is along the Y-axis towards its positive direction.<br /><br />Considering the values of electric current <i><b>I<sub>1</sub></b></i> and the distance to the source (the first wire) equaled to <i><b>a</b></i>, the magnitude of this vector is<br /><br /><i><b>|B<sub>1</sub>| = μ<sub>0</sub>·I<sub>1</sub>/(2π·a)</b></i><br /><br />So, this vector in Cartesian coordinates is<br /><br />{<i><b>0; μ<sub>0</sub>·I<sub>1</sub>/(2π·a); 0</b></i>}<br /><br /><br /><br />The second magnetic field has its force lines forming a cylindrical surface with an axis being the second wire. At point <i><b>(0,0,0)</b></i> the direction of this magnetic field intensity vector is perpendicular to a radius from the second wire towards point <i><b>(0,0,0)</b></i>, that is along the X-axis towards its positive direction.<br /><br />Considering the values of electric current <i><b>I<sub>2</sub></b></i> and the distance to the source (the second wire) equaled to <i><b>b</b></i>, the magnitude of this vector is<br /><br /><i><b>|B<sub>2</sub>| = μ<sub>0</sub>·I<sub>2</sub>/(2π·b)</b></i><br /><br />So, this vector in Cartesian coordinates is<br /><br />{<i><b>μ<sub>0</sub>·I<sub>2</sub>/(2π·b); 0; 0</b></i>}<br /><br /><br /><br />Therefore, the combined vector of magnetic field intensity has its three coordinates<br /><br />{<i><b>μ<sub>0</sub>·I<sub>2</sub>/(2π·b); μ<sub>0</sub>·I<sub>1</sub>/(2π·a); 0</b></i>}<br /><br /><br /><br />The magnitude of this vector is<br /><br />[<i><b>μ<sub>0</sub>/(2π)</b></i>]<i><b>√<span style="text-decoration: overline;">(I<sub>1</sub>/a)²+(I<sub>2</sub>/b)²</span></b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-63018567967996581292020-05-30T12:08:00.001-07:002020-05-30T12:08:32.504-07:00Unizor - Physics4Teens - Electromagnetism - Magnetism and Electric Current<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetism of Straight Line Current</u><br /><br /><br /><br />Magnetic properties of permanent magnets are attributed to parallel <br />orientation of all axes of rotation of electrons around corresponding <br />nuclei and the same direction of this rotation.<br /><br /><br /><br />Consider an Ampere model of magnetism that we have addressed in one of <br />the previous lectures and, in particular, all electrons rotating in the <br />same plane.<br /><br />If all electrons rotate in the same direction within the same plane <br />around parallel axes, electrons moving near each other are moving in <br />opposite directions and neutralize each other, as if there is no current<br /> there at all.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticDipole_AmpereModel.png" style="height: 100px; width: 200px;" /><br /><br />So, within every plane perpendicular to the North-South axis of a magnet<br /> all inner currents are neutralized, and the only really present current<br /> is around the outer boundary of a magnet.<br /><br /><br /><br />This makes the magnetic properties of permanent magnet equivalent to <br />properties of an electric current in a loop. The flow of electrons, <br />constituting this electric current, occurs within one plane with a <br />perpendicular to this plane making the North-South line of this <br />artificially made magnet.<br /><br /><img src="http://www.unizor.com/Pictures/LoopMagnet.jpg" style="height: 200px; width: 200px;" /><br /><br /><br /><br />Simple experiment with iron filings confirms the similarities of <br />magnetic properties of permanent magnet and a loop of wire with electric<br /> current running through it.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticLinesCurrentLoop.jpg" style="height: 150px; width: 200px;" /><br /><br /><br /><br />In the previous lectures we described the Lorentz force that acts on the<br /> electric current in the magnetic field of a permanent magnet. Now we <br />will use the electric current as the source of the magnetic field and <br />will talk about the properties of this magnetic field in relation to <br />electric characteristics of the current.<br /><br /><br /><br />First of all, we will switch from an electric current in a loop to a straight line current.<br /><br />Consider the magnetic field lines around the wire carrying the electric <br />current in a loop. Inside a loop they go in the direction from the South<br /> pole towards the North along the axis, then circle around the wire from<br /> the North pole back to the South.<br /><br />The round shape of a wire causes the polarization of the magnetic field.<br /> Polarity is determined by the high density of the magnetic field lines <br />inside the loop, all pointing to the North, while the opposite direction<br /> of the lines outside the wire loop is less dense, representing a weaker<br /> magnetic field.<br /><br /><br /><br />Now let's open up a loop into a straight line electric current.<br /><br />Magnetic field will not disappear and magnetic field lines will still go<br /> around the wire that carries an electric current, just more <br />symmetrically than in case of a wire in a loop.<br /><br />Simple experiment with iron filings confirms the circular shape of <br />magnetic field lines around a straight line wire carrying an electric <br />current.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticLinesCurrentStraight.jpg" style="height: 200px; width: 200px;" /><br /><br /><br /><br />The following picture represents a straight line wire carrying the electric current and magnetic field lines around it.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticFieldOfCurrent.png" style="height: 200px; width: 200px;" /><br /><br />The current in the wire causes the magnetic field to be formed around it.<br /><br />The magnetic field lines are now completely symmetrical relative to the <br />wire, thus the magnetic field has no polarity. All lines are perfectly <br />circular, each forms a circle of certain radius around a wire, lying in <br />the plane perpendicular to the wire and representing the points of the <br />same strength of the intensity vector of the magnetic field.<br /><br /><br /><br />Let's assume that our wire is ideally straight, infinitely long, and infinitesimally thin.<br /><br />Our task is to relate the electric current running through it with the intensity of the magnetic field <i><b>B</b></i> around it at distance <i><b>R</b></i> from the wire.<br /><br /><br /><br />The intensity of the magnetic field <i><b>B</b></i> is a vector, whose <br />magnitude we want to determine. The direction of this vector is always <br />tangential to the circular magnetic line lying in the plane <br />perpendicular to the wire and going through a point where we want to <br />measure this magnetic field intensity and, therefore, always <br />perpendicular to the wire.<br /><br /><br /><br />From considerations of symmetry, the distance <i><b>R</b></i> should be <br />the only variable needed to characterize a point in space around the <br />wire, where we want to determine the intensity of the magnetic field.<br /><br /><br /><br />Since the magnetism of an electric current running through a straight <br />wire depends on existence of the current in a wire, it's reasonable to <br />assume that the more electrons participate in the current (that is, the <br />greater amount of electricity goes through a wire per unit of time, that<br /> is, the greater <i>amperage</i> of an electric current <i><b>I</b></i>) - the stronger magnetic effect it causes. So, the intensity of a magnetic field <i><b>B</b></i> around a wire with electric current <i><b>I</b></i> running through it should be proportional to the amperage of the electric current in a wire:<br /><br /><i><b>B ∝ I</b></i><br /><br /><br /><br />If we consider a field, including a magnetic field, as some form of <br />energy, emitting by a source of this field and spreading into space all <br />around this source with certain speed, at any given moment of time it <br />reaches new "frontier" and spreads over this <i>surface of equal timing</i><br /> (this is not a generally used terminology, but is appropriate to better<br /> understand the concept of a field). Obviously, the farther we are from a<br /> source - the greater "frontier" area is covered by a field and less of a<br /> field energy falls on a unit of area of this surface.<br /><br />Hence, the field intensity, which can be viewed as amount of energy <br />falling on a unit of area per unit of time should diminish as the <br />distance from the source of a field is increasing because the area of a <br />surface of equal timing increases with time.<br /><br /><br /><br />These considerations were a basis for deriving the intensity of an <br />electrostatic field of a point charge as being inversely proportional to<br /> a square of a distance from this point charge and related to the fact <br />that all points at a distance <i><b>R</b></i> from a source of a filed form a sphere and the area of a sphere of radius <i><b>R</b></i> around a source of a field is <i><b>4πR²</b></i>.<br /> The same amount of energy going through a sphere of one radius goes <br />through a sphere of a radius twice as big and, therefore, "covers" the <br />area four times bigger.<br /><br /><br /><br />Let's examine the magnetic field of a straight line current using the same logic.<br /><br />In this case the field source is a straight line. All points on the same<br /> distance from it form a cylinder. The side area of a cylinder of a <br />radius <i><b>R</b></i> and height <i><b>H</b></i> is <i><b>2πR·H</b></i>, that is proportional to a radius <i><b>R</b></i>.<br /> The height is not important in our case since we assumed that the wire <br />carrying the electric current is infinite, but the factor <i><b>2πR</b></i> must be in the denominator of a the formula for intensity of a magnetic field of a straight line electric current.<br /><br /><br /><br />So, we have logically came to a conclusion of proportionality of the <br />intensity of a magnetic field to the amperage of the current and inverse<br /> proportionality to the distance from the wire:<br /><br /><i><b>B ∝ I/(2πR)</b></i><br /><br /><br /><br />Coefficient of proportionality in this formula is called <i>the permeability of free space</i> and is denoted <i><b>μ<sub>0</sub></b></i>. So, the final formula for intensity of a magnetic field of a straight wire carrying electric current <i><b>I</b></i> at a distance <i><b>R</b></i> from a wire is:<br /><br /><i><b>B = μ<sub>0</sub>I/(2πR)</b></i><br /><br />The value of the permeability of free space constant depends on the <br />units of measurement and in SI units, according to the above formula, <br />it's supposed to be measured in<br /><br /><i>T·m/A = N·m/(A·m·A) = N/A²</i><br /><br />Its value is, approximately,<br /><br /><i><b>μ<sub>0</sub> ≅ 4π·10<sup>−7</sup> N/A²</b></i><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-34333115125631702822020-05-26T06:59:00.001-07:002020-05-26T06:59:33.225-07:00Unizor - Physics4Teens - Electromagnetism - Magnetic Field - Magnetic Fi...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetic Field Lines</u><br /><br /><br /><br /><i>Magnetic field</i> is a <b>force field</b>, which means that there is<br /> a force, acting on a probe object positioned at a distance from the <br />source of this field, and a force is a <b>vector</b> that has a <b>direction</b> and a <b>magnitude</b>.<br /><br />Let's examine this force and attempt to determine a direction and a <br />magnitude of vectors of force at different locations around a magnet, <br />acting as a source of a field.<br /><br /><br /><br />Our first complication is a kind of a magnet at the source of a magnetic<br /> field. Different shapes (bar, ring, horseshoe etc.) result in <br />differently arranged fields.<br /><br />Recall that we model the permanent magnet as a set of electrons <br />circulating around parallel axes in the same direction on parallel <br />planes. Each of these rotating electrons we considered as a tiny bar <br />magnet with two poles located on an axis on two opposite sides of a <br />plane of rotation. This construction is called a <b>magnetic dipole</b>.<br /><br />This <b>magnetic dipole</b> is the most elementary magnet possible, and we can use it in our study of the properties of a magnetic field.<br /><br /><br /><br />Bar magnet would be the best choice for a source of a magnetic field for<br /> our study, since it closely resembles each elementary magnetic dipole <br />created by one rotating electron.<br /><br /><br /><br />The next decision we have to make is the shape of a probe object. It is <br />important since different shapes would behave differently in the same <br />magnetic field.<br /><br />Here, again, we choose the bar magnet, as the simplest. Note that a <br />magnet is not a point-object because it has two poles. Therefore, we <br />have to consider two types of its motion - translational motion of its <br />center and rotation around its midpoint.<br /><br /><br /><br />In the previous lecture we presented a two-dimensional picture of iron <br />filings dropped around a bar magnet. Schematically, it's represented as<br /><br /><img src="http://www.unizor.com/Pictures/MagneticField.jpg" style="height: 120px; width: 200px;" /><br /><br />Lines around this magnet represent the <i>magnetic field lines</i>, <br />along which the filings link to each other, and the direction of the <br />compass needle, if positioned at any point in this magnetic field.<br /><br /><br /><br />The designation of which pole of a magnet is North and which is South is<br /> traditional - if hanging freely, North pole of a permanent magnet is <br />the one pointing to geographical North of the Earth. After one magnet's <br />poles are defined, all other magnets' poles can be determined using <br />their interaction with previously defined, according to the rule <br />"similar repel, different attract".<br /><br />By the way, it means that the magnetic pole of the Earth that is close <br />to its geographical North pole is, technically speaking, the South <br />magnetic pole of the Earth. So, when someone says "North magnetic pole <br />of the Earth", it, most likely, means "Magnetic pole of the Earth that <br />is close to its geographical North pole". Not always, though, so it <br />might lead to confusion.<br /><br /><br /><br />Arrows on each line from North pole of a magnet towards its South pole <br />are the traditional definition of the magnetic force direction. It's <br />just the agreement among people similar to an agreement about the <br />definition of the flow of electricity from positive terminal of the <br />source of electricity (where, in reality, there is a deficiency of <br />electrons) to its negative terminal (with excess of electrons), in spite<br /> of the real moving of electrons in the opposite direction.<br /><br /><br /><br />Another important quality of these <i>magnetic field lines</i> is that, <br />if the source of a magnetic field (a bar magnet on the picture above) is<br /> fixed on a flat surface and another very small and light probe magnet <br />could freely move without friction in the magnetic field on that <br />surface, its center would move along the <i>magnetic field lines</i> in <br />the direction of the arrows on the picture above, always oriented <br />tangential to a magnetic line it's moving along, pointing its North pole<br /> towards the South pole of a magnet that is the source of the magnetic <br />field.<br /><br /><br /><br /><i>Magnetic field lines</i> never cross, as they represent the <br />trajectories. If they cross at any point, the probe magnet would have <br />ambiguous dynamics at this point.<br /><br /><br /><br />The density of the <i>magnetic field lines</i> visually represents the <br />strength of the magnetic forces at each point. The lines close to the <br />poles of a magnet are the most dense, as the field is stronger there.<br /><br /><br /><br />Let's position our probe bar magnet on any line around a source of this <br />magnetic field. If we let it turn freely, as if this probe magnet is an <br />arrow of a compass, it will align along the tangential to a <i>magnetic field line</i> it is on.<br /><br /><br /><br />The North pole of a probe object in this position will point towards the<br /> South pole of a bar magnet in the center of the field and the South <br />pole of a probe magnet will point towards the North pole of a center <br />magnet.<br /><br /><br /><br />Two attracting and two repelling forces from two poles of a center magnet, acting on a probe object, represent the <b>torque</b> that turns the probe magnet and holds it in a position along the magnetic field line.<br /><br />The pole of a center magnet that is closer to a probe object forces the <br />probe object to turn the opposite pole towards it to a greater degree.<br /><br /><br /><br />Obviously, there is a resultant force that ultimately moves the center of a probe object closer to a center magnet.<br /><br /><br /><br />Also, in a special case of a probe object positioned exactly on the <br />continuation of the North-South line between the poles of the center <br />magnet both forces from two poles of a center magnet act along this same<br /> line and can be added easily.<br /><br /><br /><br />The above considerations are related to a <b>direction</b> of the forces acting on a probe magnet in the magnetic field of a bar magnet.<br /><br />The <b>magnitude</b> of these forces is a more involved subject and is <br />related to techniques of measurement of the strength of a magnetic field<br /> at different points. This will be a subject of the next topic.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-54036484901063364732020-05-25T12:04:00.001-07:002020-05-25T12:04:25.709-07:00Unizor - Physics4Teens - Electromagnetism - Magnetic Field - Problems 1<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetism - Lorentz Force - Problems 1</u><br /><br /><br /><br /><i>Problem 1a</i><br /><br />Consider the experiment pictured below.<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce_Problem_1.jpg" style="height: 100px; width: 200px;" /><br /><br />A copper wire (yellow) of resistance <i><b>R</b></i> is connected to a battery with voltage <i><b>U</b></i> and is swinging on two connecting wires (green) in a magnetic field of a permanent magnet.<br /><br />All green connections are assumed light and their weight can be ignored.<br /> Also ignored should be their electric resistance. Assume the uniformity<br /> of the magnetic field of a magnet with magnetic field lines directed <br />vertically and perpendicularly to a copper wire.<br /><br />The mass of a copper wire is <i><b>M</b></i> and its length is <i><b>L</b></i>.<br /><br />The experiment is conducted in the gravitational field with a free fall acceleration <i><b>g</b></i>.<br /><br />The magnetic field exerts the Lorentz force onto a wire pushing it <br />horizontally out from the field space, so green vertical connectors to a<br /> copper wire make angle <i><b>φ</b></i> with vertical.<br /><br />What is the intensity of a magnetic field <i><b>B</b></i>?<br /><br /><br /><br /><i>Solution</i><br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce_Problem_1a.png" style="height: 200px; width: 200px;" /><br /><br /><i><b>T·cos(φ) = M·g</b></i><br /><br /><i><b>T = M·g/cos(φ)</b></i><br /><br /><i><b>F = T·sin(φ) = M·g·tan(φ)</b></i><br /><br /><i><b>I = U/R</b></i><br /><br /><i><b>F = I·L·B = U·L·B/R</b></i><br /><br /><i><b>M·g·tan(φ) = U·L·B/R</b></i><br /><br /><i><b>B = M·R·g·tan(φ)/(U·L)</b></i><br /><br /><br /><br /><i>Problem 1b</i><br /><br />An electric point-charge <i><b>q</b></i> travels with a speed <i><b>v</b></i> along a wire of length <i><b>L</b></i>.<br /><br />What is the value of the equivalent direct electric current <i><b>I</b></i> in the wire that moves the same amount of electricity per unit of time?<br /><br />What is the Lorentz force exerted onto a charge <i><b>q</b></i>, if it moves in a uniform magnetic field of intensity <i><b>B</b></i> perpendicularly to the field lines with a speed <i><b>v</b></i>.<br /><br /><br /><br /><i>Solution</i><br /><br />Let <i><b>T</b></i> be the time of traveling from the beginning to the end of a wire.<br /><br /><i><b>T = L/v</b></i><br /><br /><i><b>I = q/T = q·v/L</b></i><br /><br /><i><b>F = I·L·B = q·v·B</b></i><br /><br />Notice, the Lorentz force onto a wire in case of only a point-charge <br />running through it does not depend on the length of a wire, as it is <br />applied only locally to a point-charge, not an entire wire. Would be the<br /> same if a particle travels in vacuum with a magnetic field present.<br /><br /><br /><br /><i>Problem 1c</i><br /><br />An electric point-charge <i><b>q</b></i> of mass <i><b>m</b></i> enters a uniform magnetic field of intensity <i><b>B</b></i> perpendicularly to the field lines with a speed <i><b>v</b></i>.<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce_Problem_1c.png" style="height: 150px; width: 200px;" /><br /><br />Suggest some reasoning (rigorous proof is difficult) that the trajectory<br /> of this charge should be a circle and determine the radius of this <br />circle.<br /><br /><br /><br /><i>Solution</i><br /><br />The Lorentz force exerted on a point-charge <i><b>q</b></i>, moving with speed <i><b>v</b></i> perpendicularly to force lines of a permanent magnetic field of intensity <i><b>B</b></i>, is directed always perpendicularly to a trajectory of a charge and equals to <i><b>F=q·v·B</b></i> (see previous problem).<br /><br /><br /><br />Since the Lorentz force is always perpendicular to trajectory, the linear speed <i><b>v</b></i> of a point-charge remains constant, while its direction always curves toward the direction of the force. Constant linear speed <i><b>v</b></i><br /> means that the magnitude of the Lorentz force is also constant and only<br /> direction changes to be perpendicular to a trajectory of a charge.<br /><br /><br /><br />According to the Newton's Second Law, this force causes acceleration <i><b>a=F/m</b></i>,<br /> which is a vector of constant magnitude, since the Lorentz force has <br />constant magnitude and always perpendicular to a trajectory, since the <br />force causing this acceleration is always perpendicular to a trajectory.<br /><br />So, the charge moves along a trajectory with constant linear speed and <br />constant acceleration always directed perpendicularly to a trajectory.<br /><br /><br /><br />Every smooth curve at any point on an infinitesimal segment around this <br />point can be approximated by a small circular arc of some radius (<i>radius of curvature</i>) with a center at some point (<i>center of curvature</i>). If a curve of a trajectory on an infinitesimal segment is approximated by a circle of some radius <i><b>R</b></i>, the relationship between a radius, linear speed and acceleration towards a center of this circle (<i>centripetal acceleration</i>), according to kinematics of rotational motion, is<br /><br /><i><b>a = v²/R</b></i><br /><br />Therefore, <i><b>R = v²/a</b></i><br /><br /><br /><br />Since <i><b>v</b></i> and <i><b>a</b></i> are constant, the radius of a curvature <i><b>R</b></i><br /> is constant, which is a good reason towards locally circular character <br />of the motion of a charge. It remains to be proven that the center of <br />the locally circular motion does not change its location, but this is a <br />more difficult task, which we will omit.<br /><br />Hence,<br /><br /><i><b>R = v²/a = m·v²/F =<br /><br />= m·v²/q·v·B = m·v/q·B</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-46269171326549991662020-05-19T09:38:00.001-07:002020-05-19T09:38:24.783-07:00Unizor - Physics4Teens - Electromagnetism - Magnetic Field - Lorentz Force<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetism - Lorentz Force</u><br /><br /><br /><br />In this lecture we will look at the interaction between an electric current and a magnetic field.<br /><br /><br /><br />We start with an analogy between magnetic properties of a wire loop with<br /> electric current running through it and those of a permanent magnet. <br />This have been explained in the previous session from the position of <b>Ampere model</b> of magnetism.<br /><br />The picture below illustrates this analogy.<br /><br /><img src="http://www.unizor.com/Pictures/LoopMagnet.jpg" style="height: 200px; width: 200px;" /><br /><br />The wire loop with electric current running through it (thin black arrow<br /> from left to right) creates a magnetic field around it. The lines of <br />this magnetic field (thin dark blue arrows from bottom up) go through <br />the wire loop and around it, closing on themselves, forming their own <br />loops. Inside the wire loop the direction of magnetic lines is from <br />South pole to North, while outside the wire loop they go from North pole<br /> to South. Those magnetic filed line loops that are on the same distance<br /> from the wire make up a tubular surface (a torus) around the wire.<br /><br /><br /><br />This wire loop with electric current running through it and a magnetic <br />field around it would behave like a magnet, like a compass arrow, for <br />example.<br /><br /><br /><br />In particular, positioned inside some external magnetic field, like in <br />the magnetic field of the Earth, and allowed to turn free, it will <br />orient itself in such a way that its North pole will point to a South <br />pole of an external magnetic field, which, in case of the magnetic field<br /> of the Earth, is located not far from its geographical North pole.<br /><br /><br /><br />It should be noted that circular form of a wire loop is not essential. <br />If it's rectangular, the magnetic behavior will be the same. It is, <br />actually, more convenient to work with a rectangular frame to illustrate<br /> the interaction of magnetic field and electric current.<br /><br /><br /><br />Let's start the experiment with a rectangular wire loop, that can rotate<br /> around a vertical axis in the external magnetic field. Position it such<br /> that one vertical segment of a wire is close to one pole of an external<br /> magnet, while an opposite side is close to another pole. Let the <br />electric current run through it. <br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce1.png" style="height: 200px; width: 200px;" /><br /><br />If this wire frame with electric current running through it is allowed <br />to rotate around a vertical axis, it will reorient itself with its North<br /> pole directed to the left towards the South pole of an external magnet,<br /> and its South pole directed to the right towards the North pole of an <br />external magnet, as shown on a picture below. The distance from the <br />North pole of an external magnetic field to both vertical sides of a <br />wire will be the same. Same about the South pole of the external <br />magnetic field.<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce2.png" style="height: 200px; width: 200px;" /><br /><br />This turn of a wire is, obviously, the result of forces of interaction <br />between external magnetic field and electric current with its own <br />magnetic field around it.<br /><br /><br /><br />Consider the same two states of a wire (before and after the turn) viewed from above.<br /><br />The initial position of a wire, viewed from above with rotating forces acting on it (blue arrows) is<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce3.png" style="height: 200px; width: 200px;" /><br /><br />Magnetic field lines of a wire with electric current running through it <br />are oriented along vertical direction on this picture, while the <br />magnetic field lines of an external magnetic field are horizontal.<br /><br /><br /><br />As in the case of a compass arrow, aligning itself along the magnetic <br />field lines of the Earth, the external magnetic field forces, acting on <br />the magnetic field of a wire (blue arrows), will turn the wire to orient<br /> its magnetic field lines along the magnetic field lines of an external <br />magnetic field, as shown on the following picture<br /><br /><img src="http://www.unizor.com/Pictures/LorentzForce4.png" style="height: 200px; width: 200px;" /><br /><br /><br /><br />At this final position external magnetic field forces (blue arrows) are <br />balancing each other and the rotation of a wire (after a short wobbling)<br /> will stop.<br /><br /><br /><br />Let's analyze the forces acting on a wire to turn it this way.<br /><br />For this we don't really need a wire loop of any shape, it's sufficient <br />to have a linear wire with electric current running through it <br />positioned in an external magnetic field.<br /><br /><br /><br />If we open up a wire loop into a straight line with electric current <br />running through it, the magnetic field around a wire will still exist, <br />and its lines will be positioned around a wire. Magnetic lines located <br />on the same distance from a wire with electric current will form a <br />cylinder with the line of electric current being its axis.<br /><br /><br /><br />The picture below illustrates the force acting on a straight wire with <br />electric current running through it (straight black line) and its own <br />magnetic field (thin orange ovals around a wire) when it's positioned in<br /> the external magnetic field. In this case the lines of the external <br />magnetic field (light blue arrows going left to right) are perpendicular<br /> to the wire and the direction of the force is perpendicular to both, <br />the direction of the current in the wire and the direction of the <br />magnetic lines of the external magnetic field.<br /><br /><img src="http://www.unizor.com/Pictures/WireInMagField.jpg" style="height: 120px; width: 200px;" /><br /><br />The direction of the force can be determined by the "rule of the right <br />hand", which states that, if the magnetic lines of the external magnetic<br /> field are perpendicularly entering the right hand, while the thumb is <br />directed towards the electric current in the wire, fingers will show the<br /> direction of force.<br /><br />A different formulation of the "right hand rule" that results in the <br />same configuration states that, if magnetic field lines of the external <br />magnetic field are positioned along the fingers in the direction pointed<br /> by them and the electric current in the wire is running in the <br />direction of the thumb, then the force exerted by the magnetic field on <br />the wire is perpendicular to the hand going outside of it.<br /><br /><br /><br />When the wire has a rectangular shape, as on the picture in the <br />beginning of this lecture, two side of a rectangle are perpendicular to <br />the magnetic lines of an external magnetic field. The current in these <br />wire segments is running in opposite directions. As a result, the force <br />of magnetic field pushes these sides of a wire in opposite directions, <br />and the wire will turn until these two opposite forces balance each <br />other.<br /><br /><br /><br />The force of an external magnetic field exerted on the wire with electric current running through it is called <b>Lorentz force</b>.<br /><br /><br /><br />All the above considerations on interaction between an external magnetic field and an electric current are of <i>qualitative</i> character.<br /><br />Let's address <i>quantitative</i> character of this interaction.<br /><br /><br /><br />For starter, we will reduce our interest only to a case of a uniform <br />magnetic field and an infinitesimally thin straight wire running <br />perpendicularly to the magnetic field lines, like on the picture above.<br /><br /><br /><br />It is reasonable to assume that the force exerted by a magnetic field acts on each moving electron within a wire.<br /><br /><br /><br />Considering the force does not exist, if there is no electric current in<br /> a wire (electrons are not moving), but can be observed only when there <br />is an electric current in a wire, another reasonable assumption is that <br />the force depends on the speed of moving electrons, which can be <br />measured as <i>amperage</i> of the electric current.<br /><br /><br /><br />Experiment shows that the force is proportional to an <i>amperage</i>, <br />which can be intuitively explained by the idea that the higher the <br />amperage - the greater "number" of magnetic field lines of an external <br />magnetic field, crossed by electrons per unit of time, and each such <br />crossing results in certain incremental increase in the force exerted by<br /> a field.<br /><br /><br /><br />One more natural assumption is that the longer the wire - the <br />proportionally greater is the force exerted on it by a magnetic field. <br />This also is related to the above mentioned idea of a magnetic field <br />exerting a force on each electron crossing its magnetic field lines.<br /><br /><br /><br />As a result, we come to a conclusion that the force is proportional to a product of electric current and the length of a wire:<br /><br /><i><b>F = b·I·L</b></i><br /><br />where<br /><br /><i><b>b</b></i> is a coefficient of proportionality that characterizes the strength of an external magnetic field,<br /><br /><i><b>I</b></i> is the <i>amperage</i> of an electric current running through a wire,<br /><br /><i><b>L</b></i> is the length of a wire.<br /><br /><br /><br />What's interesting about this formula is that it allows to establish the<br /> units of measurement of the strength of a magnetic field in terms of <br />units of measurement of force (F), electric current (I) and length (L).<br /><br /><br /><br />DEFINITION<br /><br />A uniform magnetic field that exerts a strength of (<i><b>1N</b></i>) on a wire of <br /><br /><nobr><i><b>1 newton</b></i></nobr><nobr><i><b>1 meter</b></i></nobr> (<i><b>1m</b></i>) length with a current running through it perpendicularly to the magnetic lines of a field of <nobr><i><b>1 ampere</b></i></nobr> (<i><b>1A</b></i>) has a strength of <nobr><i><b>1 tesla</b></i></nobr> (<i><b>1T</b></i>).<br /><br /><i><b>Tesla</b></i> is a unit of measurement of the strength of a magnetic field.<br /><br />The strength of a magnetic field is denoted by a symbol <i><b>B</b></i>. The Lorentz force is, therefore, expressed as<br /><br /><i><b>F = I·L·B</b></i><br /><br />where<br /><br /><i><b>I</b></i> is the <i>amperage</i> of an electric current running through a wire,<br /><br /><i><b>L</b></i> is the length of a wire,<br /><br /><i><b>B</b></i> is the strength of an external magnetic field<br /><br /><br /><br />All the above considerations are valid for a case of an electric current<br /> running perpendicularly to lines of a uniform magnetic field.<br /><br /><br /><br />As mentioned above, the <i>Lorentz force</i> exerted on a wire depends on the movement of electrons in the wire crossing the magnetic lines of an external magnetic field. <br /><br />Simple geometry prompts us to conclude that, if the direction of the <br />current is not perpendicular to magnetic lines of an external magnetic <br />field, but at angle <i><b>φ</b></i> with them, the number of magnetic <br />lines crossed by electrons in a unit of time is smaller and, actually, <br />is smaller by a factor <i><b>sin(φ)</b></i>.<br /><br /><br /><br />So, for any angle <i><b>φ</b></i> between the electric current and magnetic field lines of an external magnetic field the formula for Lorentz force would be<br /><br /><i><b>F = I·L·B·sin(φ)</b></i><br /><br />where<br /><br /><i><b>I</b></i> is the <i>amperage</i> of an electric current running through a wire,<br /><br /><i><b>L</b></i> is the length of a wire,<br /><br /><i><b>B</b></i> is the strength of an external magnetic field<br /><br /><i><b>φ</b></i> is the angle between the direction of the electric current and lines of an external magnetic field<br /><br /><br /><br />Taking into consideration the direction of the Lorentz force <br />perpendicular to both vectors - electric current (from plus to minus) <br />and lines of an external uniform magnetic field (from South to North), <br />the above formula can be represented using a <i>vector product</i><br /><br /><i><b><span style="text-decoration: overline;">F </span> = <span style="text-decoration: overline;">I </span>·L⨯ <span style="text-decoration: overline;">B </span> </b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-71074135484090160712020-05-11T09:46:00.001-07:002020-05-11T09:46:44.333-07:00Unizor - Physics4Teens - Electromagnetism - Magnetic Field - Inside Magnet<br /><br /><br /><br /><br /><br /><iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/uDGHU_MZB5A" width="480"></iframe> <i> </i><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetism - Internal Structure<br /><br />of Magnets</u><br /><br /><br /><br />Let's look inside a permanent bar magnet with two poles, North and South.<br /><br />We model its magnetic properties as a result of a cumulative properties <br />of individual electrons rotating along parallel axes within parallel <br />planes in the same direction.<br /><br />Each such rotating electron represent a tiny <i>magnetic dipole</i> with<br /> its own North and South poles with attracting force between opposite <br />poles (North and South) and repelling force between the same poles <br />(North to North or South to South).<br /><br /><img src="http://www.unizor.com/Pictures/MagneticDipoleElectron.png" style="height: 160px; width: 200px;" /><br /><br />The attraction between two rotating electrons that face each other by <br />opposite poles we have explained by the fact that in this case electrons<br /> rotate in the same direction and "help" each other. The repelling of <br />two rotating electrons that face each other by the same poles is <br />explained by the fact that they rotate in opposite directions and <br />"disturb" each other.<br /><br /><br /><br />Since we are talking about permanent magnet, all axes of rotation of <br />electrons are always parallel to each other and planes of rotation are <br />always parallel as well.<br /><br /><br /><br />Consider a situation of two electrons rotating on parallel planes around the same axis on the same radius.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticDipole_GilbertModel.png" style="height: 340px; width: 200px;" /><br /><br />In this case the magnetic properties of the South pole of the upper (on <br />this picture) electron are neutralized by properties of the North pole <br />of an electron under it.<br /><br />So, the magnetic field of a pair of electrons in this position is the <br />same as for one electron with poles located on a greater distance from <br />each other.<br /><br /><br /><br />Now expand this logic to a full size of a bar magnet. The result is that<br /> all internal connections between South and North poles will neutralize <br />each other and the only significant magnetic properties are of those <br />electrons concentrated on two opposite surfaces of a magnet where its <br />North and South poles are located.<br /><br /><br /><br />This looks like some <i>magnetic charges</i> of opposite types, that we called <b>North</b> and <b>South</b>, are concentrated on two opposite ends of a magnet.<br /><br />These <i>magnetic charges</i> behave similarly to electric charges, <br />except magnetic ones always come in pairs. We can even think about <br />magnetic equivalent of the Coulomb Law. The only complication is that we<br /> always have a superposition of two magnetic fields coming from two ends<br /> of magnetic dipole.<br /><br />This is the <b>Gilbert model</b> of magnetic properties, attributed to <br />William Gilbert, an English physician (including a physician for English<br /> royalty), who published in 1600 a six volume treatise that contained <br />all the information about electricity and magnetism known at that time. <br />Gilbert was the one who discovered magnetic properties of Earth and came<br /> up with formulation of properties of magnets and terminology that <br />describes them (like <i>magnetic poles</i>).<br /><br /><br /><br />Consider a different approach - two electrons rotated within the same <br />plane around parallel axes and immediately near each other. The common <br />plane of rotation is, of course, perpendicular to the magnet's <br />North-South axis and axes of rotation of these electrons are parallel to<br /> the magnet's North-South axis.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticDipole_AmpereModel.png" style="height: 100px; width: 200px;" /><br /><br />Electrons moving near each other are moving in opposite directions and <br />neutralize each other, as if there is no current there at all. So, <br />within every plane perpendicular to the North-South axis of a magnet all<br /> inner currents are neutralized, and the only really present current is <br />around the outer boundary of a magnet.<br /><br /><br /><br />This is the <b>Ampere model</b> of magnetism. It makes the magnetic <br />properties of permanent magnet equivalent to properties of an electric <br />current in a loop around the side surface of a magnet with each electron<br /> moving within a plane perpendicular to a magnet's North-South axis.<br /><br /><br />This model of magnetism is extremely important, as it connects the <br />magnetic properties to those of properties of electric current and shows<br /> inherent connection between electricity and magnetism.<br /><br /><br /><br />It also opens the door to <i>electromagnetism</i> - generating magnetic field using electricity.<br /><br />A loop of electric current acts similar to each electron inside a <br />permanent magnet, just on a larger scale. A number of electric current <br />loops of the same radius around the same axis parallel to each other <br />makes the magnetic field even stronger.<br /><br /><img src="http://www.unizor.com/Pictures/Electromagnetism.jpg" style="height: 170px; width: 200px;" /><br /><br /><br /><br />If we make a loop of electric current and put an iron cylinder (which by<br /> itself does not have magnetic properties) inside this loop, the iron <br />cylinder will become magnetic, and the more loops the electric current <br />makes around this cylinder - the stronger the magnetic properties of an <br />iron cylinder will be, and it will act exactly as the permanent magnet, <br />becoming <i>electromagnet</i>.<br /><br />But, as soon as we stop the flow of electric current around this cylinder, it will lose its magnetic properties.<br /><br /><br /><br />Another important feature of the <b>Ampere model</b> is that it allows <br />to measure the strength of the magnetic field produced by an <br />electromagnet by such known physical quantities as <i>amperage</i> of the current circulating in the wire loops, producing the magnetic field, and some geometric properties of the wire loops.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0