tag:blogger.com,1999:blog-37414104180967168272020-07-03T00:52:39.902-07:00Unizor - 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.comBlogger423125tag: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.com0tag:blogger.com,1999:blog-3741410418096716827.post-91338092794221978672020-05-03T07:35:00.001-07:002020-05-03T07:35:58.487-07:00Unizor - Physics4Teens - Electromagnetism - Magnetic Field<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetism - Magnetic Field</u><br /><br /><br /><br />Magnetic forces act on a distance, so there must be a field that exists around each magnet - <i>magnetic field</i>.<br /><br /><br /><br />When studying <i>Electrostatics</i>, we started with the simplest <br />electrically charged object - a point-object with certain excess or <br />deficiency of electrons to make it negatively or positively electrically<br /> charged.<br /><br />The electrostatic field around it was spherical in shape and the only <br />important parameter that determined the relative position of a probe <br />object (also a point-object, positively charged with one coulomb of <br />electricity) was a distance of this probe object from the source of <br />electrostatic field.<br /><br /><br /><br />If we wanted to analyze the electric properties of a more complex source<br /> of electricity, like a rod or a sphere, we could always resort to some <br />relatively simple geometry and calculus to achieve our goal by breaking a<br /> larger object into smaller parts.<br /><br /><br /><br />Looking at the electrostatic field, the forces acting on a probe object <br />are always unidirectional, either attracting or repelling. To calculate <br />the resultant force, we used a vector sum of them, and in simple cases <br />of sources of electrostatic fields (a point-object, a rod, a sphere) it <br />was relatively simple task.<br /><br /><br /><br />With magnets the situation is much more complex. We cannot have a <br />point-object because each magnet has two poles and each of them act in <br />some way. Even more, the magnetic force exhorted by a magnet is changing<br /> as we move from North pole to South, first diminishing to zero in the <br />middle and rising again at the other pole.<br /><br /><br /><br />Probably, the simplest magnet we can deal with, as a source of magnetic <br />field and a probe object, is a thin rod with poles at its ends. But even<br /> in this case we have to take into account all the different forces, <br />attracting and repelling, of different magnitude and directions that act<br /> on probe magnet.<br /><br /><br /><br />Let's experimentally visualize the magnetic field in this simplest case.<br /><br />For this experiment we need a bar magnet and iron shavings. Each shaving<br /> is a little temporary magnet that forms its poles based on the forces <br />of the magnetic field. Then different shavings will attach to each other<br /> by opposite poles and form lines. The picture obtained will represent <br />the vectors of forces in the magnetic field of a bar magnet. Along each <br />line on a picture below lie iron shavings, each with North and South <br />magnetic poles, linked by opposite poles and directed with their South <br />pole closer to North pole of a bar magnet.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticFieldReal.jpg" style="height: 120px; width: 200px;" /><br /><br />The picture below schematically represents these field forces.<br /><br /><img src="http://www.unizor.com/Pictures/MagneticField.jpg" style="height: 120px; width: 200px;" /><br /><br />Direction of forces from North magnetic pole to South was traditionally <br />chosen, similarly to a direction of the electric current was chosen from<br /> a positive terminal to a negative one, regardless of the flow of <br />electrons, unknown at the time of early experiments with electricity.<br /><br /><br /><br />Let's discuss a concept of a magnetic field.<br /><br />As usually, the following explanation is the <u>model</u>, which to some<br /> degree corresponds to experimental and theoretical data, but we do not <br />claim that in reality the things are arranged in exactly this way.<br /><br />However, we offer it as an aid to understanding the concept of a magnetic field.<br /><br /><br /><br />Imagine a small particle rotating within a plane with certain speed on a<br /> certain radius around an axis that is perpendicular to this plane. On <br />the same axis in a plane parallel to the first one another particle is <br />rotating. Generally speaking, it might rotate on a different radius, in <br />the same or opposite direction and with a different speed.<br /><br /><br /><br />Consider a distance between these two parallel planes of rotation of <br />these two particles. When it's large, particles don't really have any <br />interaction. But, when we make this distance small enough, the particles<br /> will "feel" each other.<br /><br /><br /><br />If the particles rotate in the same direction, there will be some <br />attracting force between them and the planes of rotation tend to get <br />closer to each other.<br /><br />If the particles rotate in the opposite directions, there will be some <br />repelling force between them and the planes of rotation tend to increase<br /> the distance between them.<br /><br /><br /><br />The rotating particles in this model behaves like a bar magnets <br />positioned along the axis of rotation with poles determined by a <br />direction of rotation. We can assume that the North pole is defined by a<br /> rule, that, looking from it towards the rotating particle, this <br />particle rotates counterclockwise.<br /><br />If two particles rotate around the same axis in the same direction, each<br /> behaving like a bar magnet, the magnets will attract to each other <br />because they will be facing by opposite poles.<br /><br />If the particles rotate in opposite direction, the corresponding magnets<br /> will be facing each other by the same pole and will repel each other.<br /><br /><br /><br />Now let's assume that these particles are electrons rotating around a <br />nucleus in the atoms inside some object. According to this model, if all<br /> planes of rotation of electrons inside all atoms are parallel and the <br />rotation is such that all North poles of all atoms are directed in the <br />same way, we have a perfect permanent magnet.<br /><br /><br /><br />If planes of electron rotation in some object are randomly directed and <br />are insensitive to outside forces exhorted by magnets, we have a <br />diamagnetic object - the one that cannot be magnetized.<br /><br /><br /><br />If planes of electron rotation in some object are randomly directed, but<br /> outside magnetic forces, interacting with atoms of this object, align <br />all planes of rotation in a parallel fashion with the same direction of <br />the poles, we have an object that can be a temporary magnet.<br /><br /><br /><br />In all those cases our model of rotating electrons, producing a force <br />that in some way acts like a bar magnet positioned along the axis of <br />rotation, seems to explain all the magnetic properties.<br /><br />It also prompts to a connection between the electricity and magnetism <br />because both are caused by position and movement of electrons.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-87538772751768900072020-05-02T08:32:00.001-07:002020-05-02T08:32:42.083-07:00Unizor - Physics4Teens - Magnets - Practical Aspects<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Magnetism - Practical Aspects</u><br /><br /><br /><br />Here is a list of a few cases where permanent magnets of different kind are used.<br /><br /><br /><br />A magnetic screwdriver.<br /><br />Refrigerator door.<br /><br />Magnetic letters.<br /><br />Bracelets with magnetic closure.<br /><br />Compass.<br /><br />Magnetic toys.<br /><br />Different machinery uses magnets (e.g. large disk magnets to pick up metal junk).<br /><br />Medical instruments (e.g. MRI - Magnetic Resonance Imaging).<br /><br />In phones and headphones.<br /><br />In all kinds of motors and generators.<br /><br />In televisions.<br /><br />In computers.<br /><br /><br /><br />Permanent magnets occur in nature. A mineral <i>magnetite</i> is such a permanent magnet. It's black and shiny, when polished. The chemical composition of <i>magnetite</i> is iron oxide, which means that its molecule contains only atoms of iron and oxygen, the formula is Fe<sub>3</sub>O<sub>4</sub>.<br /><br />Two pieces of <i>magnetite</i> will attract each other if faced by <br />opposite poles (North of one to South of another) and repel each other <br />when positioned facing each other by the same pole.<br /><br /><br /><br /><i>Magnetite</i> attracts certain metals, like iron.<br /><br /><br /><br />A piece of <i>magnetite</i>, if hanging on a thread or free floating in <br />water on a wooden plate will always turn along a meridian, thus people <br />used it in navigation before they invented a compass.<br /><br /><br /><br />The naturally occurring <i>magnetite</i> is quite a weak magnet. Much <br />stronger permanent magnets are created artificially by melting together <br />certain metals. One of the strongest artificial magnets are made by <br />combining iron, neodymium and boron into an alloy called <i>neodymium magnets</i>.<br /><br /><br /><br />Artificial magnets can have any shape with location of the poles chosen <br />by a manufacturer. For example, a ring magnet can be created with a <br />North pole on the outside circumference of a ring and South pole on the <br />inside circumference.<br /><br /><br /><br />Consider a bar magnet with poles on its opposite sides.<br /><br />Let's use an iron nail as a probe object to measure the force of attraction at different points on a magnet.<br /><br />The strongest force of attraction will be at the ends of a bar magnet, <br />at its poles. If we move a nail from one end of the magnet to another, <br />the strength of attraction will diminish to zero at the middle, then <br />again will rise to its maximum on an opposite pole.<br /><br />The midpoint of a bar magnet has no attracting strength at all.<br /><br /><br /><br />Interestingly, if we break a bar magnet in the middle, each half will <br />not be a single pole magnet, single pole magnets don't exist. Each half <br />will have two poles, one old and the new one at the point we broke the <br />original magnet. Each new smaller magnet will be weaker than original.<br /><br /><img src="http://www.unizor.com/Pictures/MagnetBar_Split.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br />Artificial magnets can be <i>permanent</i> or <i>temporary</i>. <br />Permanent magnets do not change their magnetic properties. Temporary <br />ones can be magnetized, when are near another permanent magnet, or <br />demagnetized, when there is no other magnets nearby. Plain nails and <br />paperclips have such a property. Their polarity depends on the position <br />of the other permanent magnet. Another example is electromagnets that <br />will be discussed later on in the contents of a connection between <br />electricity and magnetism.<br /><br /><br /><br />There are artificial magnets that, after being magnetized by electric <br />impulse, retain their magnetic properties, including the polarity. At <br />the same time, exposed to another impulse, they can lose their magnetic <br />properties or change the polarity. This type of magnets were used in the<br /> memory of the first computers with each bit of information (1 or 0) <br />stored as a magnetization of one ferrite ring.<br /><br /><img src="http://www.unizor.com/Pictures/FerriteRingsMemory.jpg" style="height: 100px; width: 200px;" /><br /><br /><br /><br />Artificial magnets can be not only in some solid shape, but also form a <br />thin magnetic layer on a surface of some other material. This is how <br />magnetic tape and magnetic disks were created, both used extensively in <br />the computers. The principle of work of these memory devices is based on<br /> the properties of the thin magnetic layer to retain the state of tiny <br />magnets this layer contains. Special devices can "write" information on <br />the magnetic layer by temporary magnetizing the tiny magnets of the <br />layer and "read" this information.<br /><br /><br /><br />Another example is injecting a magnetically-sensitive liquid into a <br />tumor and, using a powerful magnet, heat it up to destroy the tumor.<br /><br /><br /><br />The fact that permanent magnets are permanent and, seemingly, represent <br />an unlimited source of energy, prompted many people to construct a <br />device that would exploit this property to create a perpetual movement <br />and use it to generate unlimited amount of power.<br /><br /><br /><br />Some of these devices did move for a long time. However, they contradict<br /> the principle of energy conservation for a closed system and, sooner or<br /> later, friction and other forces would stop it.<br /><br />Here is a good example of such a device<br /><br /><img src="http://www.unizor.com/Pictures/MagnetMotor_Walden.png" style="height: 120px; width: 200px;" /><br /><br />https://www.youtube.com/watch?v=UqlPePYgdg4 <br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-25324394557748780982020-04-29T09:42:00.001-07:002020-04-29T09:42:56.443-07:00Unizor - Physics4Teens - Electromagnetism - Magnets<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/dFpgiLMn_fE" 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 - Magnets</u><br /><br /><br /><br />We usually don't have a lot of questions about forces that act on a contact. There is an obvious cause and observable effect.<br /><br /><br /><br />Forces that act on a distance cause a logical question "Why?" since <br />there is no obvious cause for observable actions, like attraction or <br />repelling of objects without immediate contact between them.<br /><br /><br /><br />So far, we studied <i>gravitational</i> force, acting on a distance, and <i>electrostatic</i> force, also acting on a distance.<br /><br /><br /><br />We have introduced a concept of a <i>field</i> to explain these actions on a distance and quantitatively study them.<br /><br /><br /><br />In case of <i>gravity</i> we just stated that all objects have certain characteristic, <b>mass</b>, and any object has a <i>gravitational field</i><br /> around it that acts on any other object in this field, according to the<br /> Law of Gravity. The force is always attracting and proportional to a <b>mass</b>. The gravitational force always exists around any object.<br /><br /><br /><br />In case of <i>electrostatics</i> we explained the existence of a force on a distance and <i>electrostatic field</i> by an excess or deficiency of <b>electrons</b>, which we called an <i>electric charge</i> and assigned <b>negative</b> charge (<b>−</b>) to an excess of electrons and <b>positive</b> charge (<b>+</b>)<br /> to their deficiency. In this case the force between two objects <br />positioned at a distance from each other can be either attracting (if <br />one object is positively charged and another - negatively, opposite <br />attracts) or repelling (for similarly charged objects, similar repels) <br />and is proportional to the number of electrons in excess or deficiency.<br /><br /><br /><br />The electrostatic force exists only if there is an excess or deficiency <br />of electrons that usually requires certain effort to create by <br />separating certain electrons from their atoms. Electrostatic forces are <br />usually much stronger than gravitational.<br /><br /><br /><br />There is another observable force on a distance. Certain objects, which are called <i>permanent magnets</i>,<br /> have a force field around them that does not require any efforts to <br />create. This makes them similar to gravitation. But, while the force of <br />gravity is always attracting, magnets attract only some objects - those <br />that have certain <b>magnetic properties</b>.<br /><br /><br /><br />The force field around <i>magnets</i> always exists and has an interesting property - <i>polarity</i>. Each magnet has two <b>poles</b>, which are conditionally called <i>North</i> and <i>South</i>.<br /><br /><br /><br />Two magnets attract each other, if they are facing each other with <br />different poles, that is North pole of one magnet attracts South pole of<br /> another (again, opposite attracts). Positioned to face each other by <br />the same pole (North to North or South to South), they repel each other.<br /><br /><br /><br />Another interesting fact that differentiates magnets from electrically <br />charged objects is that magnets always have two poles - North and South.<br /> There is no magnet with only North or only South pole. In the world of <br />electricity objects are usually either positively charged (deficiency of<br /> electrons) or negatively charged (excess of electrons), not both at the<br /> same time, but in the world of magnetism existence of two poles, North <br />and South, is a necessary property of every magnet.<br /><br /><br /><br />Interestingly, our planet Earth is a giant magnet with North magnetic <br />pole positioned close to the geographical North pole (but not exactly at<br /> it) and South magnetic pole positioned close to South geographical <br />pole. North and South magnetic poles of Earth are, actually, moving <br />because of movements of the planet core.<br /><br /><br /><br />Compass is a device with a small magnet in a form a free rotating arrow,<br /> also having North (usually, blue) and South (usually, red) poles. Since<br /> opposite poles attract, the red part of a compass arrow (South pole of <br />an arrow magnet) points to the North magnetic pole of Earth.<br /><br /><img src="http://www.unizor.com/Pictures/Compass.png" style="height: 200px; width: 200px;" /><br /><br /><br /><br />In places far from Earth poles we can rely on a compass to determine the<br /> general direction to the North. Close to the geographical poles a <br />compass is not a reliable tool, because the North and South magnetic <br />poles of Earth do not coincide with its geographical poles.<br /><br /><br /><br />We do know that electric charge is caused by excess or deficiency of <br />electrons. This is a relatively superficial explanation. We never <br />discussed what caused two electrons to repel each other or proton and <br />electron to attract each other. These are much deeper issues that <br />require significantly more study.<br /><br />With magnetism we also need some, however incomplete, explanation for this phenomenon.<br /><br /><br /><br />Here is what might serve as a superficial explanation.<br /><br /><br /><br />Electrons inside the atoms have their own magnetic properties, derived from their movement and so-called <i>spin</i>,<br /> studied in Quantum Theory. Pair of electrons usually has electrons with<br /> opposite spins and their magnetic poles neutralize each other. But in <br />case of odd number of electrons this balance is distorted. Atom with <br />such a magnetic unbalance becomes a source of magnetic field.<br /><br /><br /><br />Another necessary property of the magnet is the relatively <br />unidirectional position of these atoms (all North poles are pointing to <br />the same direction), thus making the effect of magnetism much stronger.<br /><br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-41157960854891237142020-04-24T09:14:00.001-07:002020-04-24T09:14:28.695-07:00Unizor - Physics4Teens - Electromagnetism - Direct Current - Electric He...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/5HAzkNQ29oU" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Direct Current - Electric Heat - Problems 1</u><br /><br /><br /><br /><i>Problem A</i><br /><br />A lamp with resistance <b>500Ω</b> (<i>ohm</i>) is powered by a source of electricity producing a voltage of <b>100V</b> (<i>volt</i>).<br /><br />What is the power consumption of this lamp?<br /><br /><br /><br /><i>Solution</i><br /><br />Power is work per unit of time. It can be expressed as<br /><br /><i><b>P = U·I</b></i> or<br /><br /><i><b>P = U² <span style="font-size: medium;">/</span> R</b></i> or<br /><br /><i><b>P = I²·R</b></i><br /><br />Substituting the real numbers,<br /><br /><i><b>P = 100² <span style="font-size: medium;">/</span> 500 = 20</b> (watt)</i><br /><br /><br /><br /><i>Problem B</i><br /><br />Express the work performed by a source of electricity in Problem A during 8 hours in <i>joules</i> and in <i>kilowatt-hours</i>.<br /><br /><br /><br /><i>Solution</i><br /><br />By definition,<br /><br /><i>1 watt = 1 joule/sec</i><br /><br />Power consumed by a lamp is <b>20 watt</b> (that is, <b>20 joules per second</b>).<br /><br />During <b>8 hours</b> (that is, <b>8·3600 seconds</b>) the work performed by a source of electricity will be equal to<br /><br /><i><b>W = 20·8·3600 = 576000 J</b> (joules)</i><br /><br />In <i>kilowatt-hours</i> (<i>kWh</i>) the work will be a product of power in <i>kilowatts</i> and time in <i>hours</i>, that is<br /><br /><i><b>W = 0.020·8 = 0.160 kWh</b></i><br /><br /><br /><br /><i>Problem C</i><br /><br />There are usually two characteristics written on an incandescent <br />electric lamp: the difference in electric potential on the terminals of a<br /> source of electricity the lamp is supposed to be connected to in <i>volts</i> (or <i>voltage</i>) and the power it consumes in <i>watts</i> (or <i>wattage</i>).<br /><br />Let the <i>voltage</i> of a lamp be <b>U=220V</b> and the wattage - <b>P=55W</b>.<br /><br />Determine the electric current <b>I</b> going through a lamp and its electrical resistance <b>R</b>.<br /><br /><br /><br /><i>Solution</i><br /><br />Power is work per unit of time. It can be expressed as<br /><br /><i><b>P = U·I</b></i> or<br /><br /><i><b>P = U² <span style="font-size: medium;">/</span> R</b></i> or<br /><br /><i><b>P = I²·R</b></i><br /><br />From this, knowing <i><b>U</b></i> and <i><b>P</b></i>, we derive <i><b>I</b></i> and <i><b>R</b></i>.<br /><br /><i><b>I = P <span style="font-size: medium;">/</span> U</b></i><br /><br /><i><b>R = U² <span style="font-size: medium;">/</span> P</b></i><br /><br />Substituting the real numbers,<br /><br /><i><b>I = 55 <span style="font-size: medium;">/</span> 220 ≅ 0.25</b></i>A (<i>amperes</i>)<br /><br /><i><b>R = 220²<span style="font-size: medium;">/</span>55 = 880</b></i>Ω (<i>ohms</i>)<br /><br /><br /><br /><i>Problem D</i><br /><br />What electric current is supposed to go through an incandescent lamp with characteristics <b>110V</b>, <b>55W</b>, and what electric current will be going through it, if we connect it to a source of electricity with voltage <b>220V</b>?<br /><br />How much heat will be generated by this lamp in this case of abnormally high voltage?<br /><br /><br /><br /><i>Solution</i><br /><br /><i><b>P = U·I = I²·R = U²/R</b></i><br /><br /><i><b>R = U²/P</b></i><br /><br /><i><b>R = (110V)²/(55V) = 220Ω</b></i><br /><br /><i><b>I<sub>norm</sub> = (55W)/(110V) = 0.5A</b></i><br /><br /><i><b>I<sub>high</sub> = (220V)/(220Ω) = 1A</b></i><br /><br /><i><b>P<sub>high</sub> = (220V)²/(220Ω) = 220W</b></i><br /><br />In this case of high voltage the lamp will generate <b>220W</b> of heat, which is 4 times greater amount of heat than normal. It will burn very soon.<br /><br /><br /><br /><i>Problem E</i><br /><br />Two incandescent lamps are connected in <b><u>series</u></b> to a source of electricity with voltage <b>220V</b>.<br /><br />The first lamp has characteristic <b>220V</b> and <b>55W</b>.<br /><br />The second lamp has characteristic <b>110V</b> and <b>55W</b>.<br /><br />What electric current will be going through these lamps along a circuit?<br /><br />Will the amount of heat produced by each lamp be greater or less than the one it's designed to produce?<br /><br /><br /><br /><i>Solution</i><br /><br /><i><b>P = U·I = I²·R = U²/R</b></i><br /><br /><i><b>R = U²/P</b></i><br /><br /><i><b>R<sub>1</sub> = 220²/55 = 880Ω</b></i><br /><br /><i><b>R<sub>2</sub> = 110²/55 = 220Ω</b></i><br /><br /><i><b>R<sub>total</sub> = R<sub>1</sub> + R<sub>2</sub> = 1100Ω</b></i><br /><br /><i><b>I = U/R<sub>total</sub> = 220/1100 = 0.2A</b></i><br /><br /><i><b>P<sub>1</sub> = 0.2²·880 = 35.2W</b></i><br /><br />Consumed power of <b>35.2W</b> is less than <b>55W</b>, it will produce less light than it is manufactured for.<br /><br /><i><b>P<sub>2</sub> = 0.2²·220 = 8.8W</b></i><br /><br />Consumed power of <b>8.8W</b> is significantly less than <b>55W</b>, it will be hardly lit at all.<br /><br /><br /><br /><i>Problem F</i><br /><br />Two incandescent lamps are connected in <b><u>parallel</u></b> to a source of electricity with voltage <b>220V</b>.<br /><br />The first lamp has characteristic <b>220V</b> and <b>55W</b>.<br /><br />The second lamp has characteristic <b>110V</b> and <b>55W</b>.<br /><br />What electric current will be going through a common part of a circuit?<br /><br />Will the amount of heat produced by each lamb be greater or less than the one it's designed to produce?<br /><br /><br /><br /><i>Solution</i><br /><br /><i><b>P = U·I = I²·R = U²/R</b></i><br /><br /><i><b>R = U²/P</b></i><br /><br /><i><b>R<sub>1</sub> = 220²/55 = 880Ω</b></i><br /><br /><i><b>R<sub>2</sub> = 110²/55 = 220Ω</b></i><br /><br /><i><b>R<sub>total</sub> = 1 <span style="font-size: medium;">/</span> (1/R<sub>1</sub> + 1/R<sub>2</sub>) = 176Ω</b></i><br /><br /><i><b>I<sub>common</sub> = U/R<sub>total</sub> = 220/176 = 1.25A</b></i><br /><br /><i><b>P<sub>1</sub> = 220²/880 = 55W</b></i><br /><br />Consumed power of <b>55W</b> is exactly the same as normal, it will produce exactly the same amount of heat and light as it is manufactured for<br /><br /><i><b>P<sub>2</sub> = 220²/220 = 220W</b></i><br /><br />Consumed power of <b>220W</b> is 4 times greater than normal <b>55W</b>, it will produced a lot of heat and light, but will burn soon.<br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-78413332746538910772020-04-19T10:03:00.001-07:002020-04-19T10:03:12.475-07:00Unizor - Physics4Teens - Electromagnetism - Electric Heat<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Direct Current - Electric Heat</u><br /><br /><br /><br />Consider a circuit with a source of electricity that produces a difference in <i>electric potential </i> (<i>voltage</i>) <i><b>U</b>(volts)</i> between its terminals. The only other component in this simple circuit is a <i>resistor</i> with <i>resistance <b>R</b>(ohms)</i>.<br /><br /><img src="http://www.unizor.com/Pictures/SimpleCircuit.png" style="height: 200px; width: 200px;" /><br /><br />According to the Ohm's Law, the electric current going through a resistor is<br /><br /><i><b>I = U/R</b>(amperes)</i><br /><br /><br /><br />Recall that a difference in <i>electric potential </i> (<i>voltage</i>) of <i><b>1V</b>(one volt)</i> between two terminals of a source of electricity means that it requires one <i>joule</i> of work (<i><b>1J</b></i>) to move one <i>coulomb</i> of positive electric charge (<i><b>+1C</b></i>) between these terminals.<br /><br />Therefore,<br /><br /><i><b>1V·1C = 1J</b></i> and <i><b>1V = 1J/1C</b></i><br /><br /><br /><br />Further recall that the electric current of <i><b>1A</b>(one ampere)</i> is the flow of electricity, when <i><b>1C</b>(one coulomb)</i> of electricity is moving across the wire within <i><b>1sec</b>(one second)</i>.<br />Therefore,<br /><br /><i><b>1A = 1C/1sec</b></i> and <i><b>1A·1sec = 1C</b></i><br /><br /><br /><br />In case of our circuit with the voltage between the terminals of a source of electricity <i><b>U</b></i>, electric current <i><b>I</b></i> and resistor <i><b>R</b></i> we have <i><b>I</b></i> <i>coulombs</i> of electric charge going through a circuit each <i>second</i>.<br /><br />So, it <i><b>t</b></i> <i>seconds</i> the source of electricity moves <i><b>I·t</b></i> <i>coulombs</i> of electricity.<br /><br /><br /><br />Moving one coulomb (<i><b>1C</b></i>) of electricity between terminals with one volt (<i><b>1V</b></i>) of a difference in electric potential requires one joule (<i><b>1J</b></i>) of work.<br /><br />Therefore, moving this one coulomb (<i><b>1C</b></i>) of electricity between terminals with a difference of electric potential (<i>voltage</i>) <i><b>U</b> volts</i> requires <i><b>U</b> joules</i> of work.<br /><br />Continuing this logic, moving <i><b>I·t</b></i> <i>coulombs</i> of electricity between the terminals with the voltage <i><b>U</b> volts</i> requires <i><b>U·I·t</b> joules</i> of work.<br /><br /><br /><br />So, the work performed by a source of electricity with voltage <i><b>U</b> (volts)</i> that moves <i><b>I</b></i> coulombs electricity every second (<i>amperage</i>) through a resistor with resistance <i><b>R</b> ohms</i> during time <i><b>t</b> sec</i> is (in <i>joules</i>)<br /><br /><i><b>W = U·I·t</b></i><br /><br /><br /><br />From the Ohm's Law <i><b>U=I·R</b></i> immediately follows<br /><br /><i><b>W = I²·R·t</b></i> and <i><b>W = U²·t / R</b></i><br /><br /><br /><br />In most cases we will assume that wires have zero resistance and only <br />resistors have the property of resistance. So, in our simple circuit the<br /> only resistance to the moving electrons is in a resistor <i><b>R</b></i>.<br /><br />Since the source of electricity performs work, this work is supposed to <br />do something. According to the Law of Energy Conservation, the work is <br />just a transformation of one form of energy into another.<br /><br />In our case the only result of the work performed by a source of <br />electricity is moving electrons in a circuit that, meeting resistance of<br /> the atoms inside a resistor, push them around, thus increasing their <br />chaotic movement inside a resistor. This chaotic movement is <i>heat</i> that can be measured, for example, by measuring a temperature of a resistor.<br /><br /><br /><br />The bottom line is that a source of electricity that has <i>voltage</i> <i><b>V</b></i>, producing an electric current with <i>amperage</i> <i><b>I</b></i> going through a resistor with <i>resistance</i> <i><b>R</b></i> during <i>time</i> <i><b>t</b></i> performs work<br /><br /><i><b>W = U·I·t = I²·R·t = U²·t / R</b></i><br /><br />which is converted to <i>heat</i> that increases the temperature of a resistor.<br /><br /><br /><br />This process is used in, for example, in incandescent lamps. The rising <br />temperature of a tungsten spiral inside such a lamp results in heat and <br />light radiation that lights and heats up the space around a lamp.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-8356448146759109442020-04-11T20:30:00.001-07:002020-04-11T20:30:43.457-07:00Unizor - Physics4Teens - Electromagnetism - DC Ohm's Law - Problems 3<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/X9yGunGRKhA" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current - Ohm's Law -<br />Problems 3</u><br /><br /><i>Problem A</i><br />Consider the following circuit with the voltage <i><b>U</b></i> and all resistances given.<br /><img src="http://www.unizor.com/Pictures/Ohms_Problem3A.jpg" style="height: 180px; width: 200px;" /><br />Ignore for now the arrows on the picture, they will be explained in the solution.<br />The task is to determine all the electric currents <i><b>I<sub>1</sub></b></i>, <i><b>I<sub>2</sub></b></i>, <i><b>I<sub>3</sub></b></i>, <i><b>I<sub>4</sub></b></i>, <i><b>I<sub>5</sub></b></i> going through each resistor and the current <i><b>I</b></i> in the common segment of the circuit.<br /><br /><i>Solution</i><br />This problem differs from seemingly analogous one presented before. There is no symmetry here, no points with the same electric potential that can be connected or disconnected for convenience.<br />We will solve it using the most general brute force approach.<br /><br />First of all, we will randomly choose the direction of the unknown electric currents in each segment of a circuit. Randomly, because we don't really know the direction, except in the common segment with current <i><b>I</b></i>. In this segment the direction of current is from a positive terminal of the battery to a negative one (as you recall, by definition, it's opposite to the flow of electrons). In other cases we cannot determine the direction, and we choose any particular one. After the calculations it might be positive (if we correctly choose the direction) or negative (if in reality the direction is opposite).<br /><br />We have six unknown values of the electric current in this problem. These values are related to each other because they have to obey certain laws of Physics, which we will use to have sufficient number of equations to find algebraically these unknown values.<br /><br />Next, let's use the principle of balance of the current at every connection. Amount of electricity going into this connection per unit of time must be equal to amount electricity going out of it.<br /><i><b>I = I<sub>1</sub> + I<sub>2</sub></b></i><br /><i><b>I<sub>1</sub> = I<sub>3</sub> + I<sub>4</sub></b></i><br /><i><b>I<sub>2</sub> + I<sub>3</sub> = I<sub>5</sub></b></i><br /><i><b>I<sub>4</sub> + I<sub>5</sub> = I</b></i><br /><br />To add more equations (we need six), let's consider a voltage drop on any part of a circuit with the same chosen direction of the current.<br /><br />Going along the top of the circuit through resistors with resistances <i><b>R</b></i> and <i><b>3R</b></i>, the combined voltage drop should be equal to a voltage between terminals of a battery<br /><i><b>I<sub>1</sub>·R + I<sub>4</sub>·(3R) = U</b></i><br /><br />Going along the bottom of the circuit through resistors with resistances <i><b>3R</b></i> and <i><b>R</b></i>, the combined voltage drop should also be equal to a voltage between terminals of a battery<br /><i><b>I<sub>2</sub>·(3R) + I<sub>5</sub>·R = U</b></i><br /><br />Going along the top of the circuit through resistor with resistances <i><b>R</b></i>, then down through resistor <i><b>2R</b></i> and right through resistor <i><b>R</b></i>, the combined voltage drop should also be equal to a voltage between terminals of a battery<br /><i><b>I<sub>2</sub>·R + I<sub>3</sub>·(2R) + I<sub>5</sub>·R = U</b></i><br /><br />Going along the bottom of the circuit through resistor with resistances <i><b>3R</b></i>, then up through resistor <i><b>2R</b></i> (against the current, so we should subtract the voltage drop) and right through resistor <i><b>3R</b></i>, the combined voltage drop should also be equal to a voltage between terminals of a battery<br /><i><b>I<sub>2</sub>·(3R) − I<sub>3</sub>·(2R) + I<sub>4</sub>·(3R) = U</b></i><br /><br />We have got eight linear equations, with only six unknowns, which means that our system contains dependent equations, but it's OK, it can be algebraically solved.<br /><br /><i>Solution by substitution</i><br />(long by straight forward):<br /><br /><i>I=x,<br />I<sub>1</sub>=y,<br />I<sub>2</sub>=z,<br />I<sub>3</sub>=u,<br />I<sub>4</sub>=v,<br />I<sub>5</sub>=w,<br />A=U/R<br /><br />x=y+z<br />y=u+v<br />z+u=w<br />v+w=x<br />y+3v=A<br />3z+w=A<br />y+2u+w=A<br />3z-2u+3v=A<br />______________<br /><b>w=z+u</b><br />x=y+z<br />y=u+v<br />v+(z+u)=x<br />y+3v=A<br />3z+(z+u)=A<br />y+2u+(z+u)=A<br />3z-2u+3v=A<br />______________<br />x=y+z<br />y=u+v<br />v+z+u=x<br />y+3v=A<br />4z+u=A<br />y+3u+z=A<br />3z-2u+3v=A<br />______________<br /><b>v=y-u</b><br />x=y+z<br />v+z+u=x<br />y+3v=A<br />4z+u=A<br />y+3u+z=A<br />3z-2u+3v=A<br />______________<br />x=y+z<br />(y-u)+z+u=x<br />y+3(y-u)=A<br />4z+u=A<br />y+3u+z=A<br />3z-2u+3(y-u)=A<br />______________<br /><b>u=A-4z</b><br />x=y+z<br />y+z=x </i>same as above<i><br />y+3(y-u)=A<br />y+3u+z=A<br />3z-2u+3(y-u)=A<br />______________<br />x=y+z<br />4y-3(A-4z)=A<br />y+3(A-4z)+z=A<br />3z-5(A-4z)+3y=A<br />______________<br /><b>z=x-y</b><br />4y+12z=4A<br />y-11z=-2A<br />3y+23z=6A<br />______________<br />3x-2y=A<br />-11x+12y=-2A<br />23x-20y=6A<br />______________<br /><b>2y=3x-A</b><br />-11x+12y=-2A<br />23x-20y=6A<br />______________<br />-11x+6(3x-A)=-2A<br />23x-10(3x-A)=6A<br />______________<br />7x=4A<br />-7x=-4A </i>same as above<i><br />______________<br />x=4A/7<br />y=(12A/7 - A)/2 = 5A/14<br />z=3A/14 u=A-12A/14 = A/7<br />v=5A/14 - A/7 = 3A/14<br />w=3A/14 + A/7 = 5A/14<br />______________<br />I = 4U/(7R)<br />I<sub>1</sub> = 5U/(14R)<br />I<sub>2</sub> = 3U/(14R)<br />I<sub>3</sub> = U/(7R)<br />I<sub>4</sub> = 3U/(14R)<br />I<sub>5</sub> = 5U/(14R)</i><br />Notice the anti-symmetry in resistors (top left is the same as bottom right <i>R</i>, top right is the same as bottom left <i>3R</i>).<br />It results in equality of the corresponding currents:<br /><i>I<sub>1</sub> = I<sub>5</sub></i>,<br /><i>I<sub>2</sub> = I<sub>4</sub></i>.<br />______________<br /><br />Knowing the current<br /><i><b>I=4U/(7R)</b></i>,<br />we can calculate the total resistance of the circuit using the Ohm's Law:<br /><i><b>R<sub>circuit</sub> = U/I = (7/4)·R</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-11079963767014585662020-04-05T19:19:00.001-07:002020-04-05T19:19:11.063-07:00Unizor - Physics4Teens - Electromagnetism - DC Ohm's Law - Problems 2<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/iRhSQV-Y-eQ" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current - Ohm's Law -<br />Problems 2</u><br /><br /><i>Problem A</i><br />Two resistors <i><b>R<sub>1</sub></b></i> and <i><b>R<sub>2</sub></b></i> are connected parallel to each other in a circuit.<br />Prove that the currents flowing through them are inversely proportional to their resistances, that is <i><b>I<sub>1</sub>/I<sub>2</sub>=R<sub>2</sub>/R<sub>1</sub></b></i>.<br /><br /><i>Hint</i><br />It follows directly from the Ohm's Law.<br /><br /><i>Problem B</i><br />Given a square <i><b>ABCD</b></i> with diagonal <i><b>AC</b></i>.<br />Each side of this square and diagonal <i><b>AC</b></i> are resistors with resistance <i><b>R</b></i> each.<br />Points <i><b>A</b></i> and <i><b>D</b></i> are connected to a source of electricity with voltage <i><b>U</b></i>.<br />Determine resistance of an entire circuit <i><b>R<sub>ABCD</sub></b></i> and current <i><b>I<sub>AC</sub></b></i> along the diagonal <i><b>AC</b></i>.<br /><br /><i>Solution</i><br />Redraw the circuit as<br /><img src="http://www.unizor.com/Pictures/Ohms_Problem2B.jpg" style="height: 180px; width: 200px;" /><br />and start with calculating the resistance of elements in a thin rectangle.<br /><i><b>1/R<sub>ABC</sub> = 1/R + 1/(2R) = 3/(2R)</b></i><br /><i><b>R<sub>ABC</sub> = 2R/3</b></i><br />Now the total resistance of a circuit is calculated as a parallel connection of the resitors with resistance <i><b>(2R/3)+R=5R/3</b></i> at the top branch and another <i><b>R</b></i> at the bottom of a drawing<br /><i><b>1/R<sub>ABCD</sub> = [1/(5R/3)] + 1/R =<br />= 3/(5R) + 1/R = 8/(5R)</b></i><br />Total resistance of a circuit is<br /><i><b>R<sub>ABCD</sub> = 5R/8</b></i><br />Current from the source of electricity in the common part of a circuit is, therefore,<br /><i><b>I = 8U/(5R)</b></i><br />To find the current along diagonal <i><b>AC</b></i>, we need to know the voltage between points <i><b>A</b></i> and <i><b>C</b></i>.<br />The voltage between points <i><b>A</b></i> and <i><b>D</b></i> is <i><b>U</b></i>.<br />The current in a common wire to the top branch with resistance <i><b>5R/3</b></i> is, therefore,<br /><i><b>I<sub>top</sub> = 3U/(5R)</b></i><br />The voltage in that branch drops from point <i><b>A</b></i> to point <i><b>C</b></i> by<br /><i><b>U<sub>ABC</sub> = I<sub>top</sub>·R<sub>ABC</sub> =<br />= 3U/(5R) · 2R/3 = 2U/5</b></i><br />Therefore, the current in the diagonal <i><b>AC</b></i> is<br /><i><b>I<sub>AC</sub> =</b></i><br /><i><b>R<sub>ABCD</sub> = 5R/8</b></i><br /><i><b>I<sub>AC</sub> = 2U/(5R)</b></i><br /><br /><i>Problem C</i><br />Given a cube <i><b>ABCDEFGH</b></i> made with identical resistors <i><b>R</b></i> on each edge, 12 resistors altogether.<br />It is connected to the source of electricity at two vertices on its main diagonal <i><b>AG</b></i>.<br /><img src="http://www.unizor.com/Pictures/Ohms_Problem2C.jpg" style="height: 180px; width: 200px;" /><br />What is the total resistance <i><b>R<sub>cube</sub></b></i> of this cube?<br /><br /><i>Solution</i><br />The cube is a symmetrical figure relatively to its main diagonal <i><b>AG</b></i> and, since all resistors are the same, one on each edge, we can use this consideration to state that the main current from the source of electricity to point <b><i>A</i></b> is divided into three identical currents along edges <i><b>AB</b></i>, <i><b>AD</b></i> and <i><b>AE</b></i>.<br />Therefore, a voltage drop at points <i><b>B</b></i>, <i><b>D</b></i> and <i><b>E</b></i> is exactly the same and there is no difference in electric potential between these three vertices.<br />If there is no difference in electric potential between points <i><b>B</b></i>, <i><b>D</b></i> and <i><b>E</b></i>, there will be no movement of electrons between these three points if we merge these three points into one point <i><b>X</b></i>. The flow of electricity will not change by this merging. Analogously, the main current from the source of electricity to point <b><i>G</i></b> is divided into three identical currents along edges <i><b>GC</b></i>, <i><b>GF</b></i> and <i><b>GH</b></i>.<br />Therefore, a voltage drop at points <i><b>C</b></i>, <i><b>F</b></i> and <i><b>H</b></i> is exactly the same and there is no difference in electric potential between these three vertices.<br />If there is no difference in electric potential between points <i><b>C</b></i>, <i><b>F</b></i> and <i><b>H</b></i>, there will be no movement of electrons between these three points if we merge these three points into one point <i><b>Y</b></i>. The flow of electricity will not change by this merging.<br />The following picture is the result of this transformation.<br /><img src="http://www.unizor.com/Pictures/Ohms_Problem2Cmodified.jpg" style="height: 180px; width: 200px;" /><br />Regardless of seemingly complex picture, it actually represents a simple circuit.<br />Between points <i><b>A</b></i> and <i><b>X</b></i> there are three identical parallel resistors. Their combined resistance is <i><b>R<sub>AX</sub></b></i>.<br /><i><b>R<sub>AX</sub> = R/3</b></i><br />Between points <i><b>X</b></i> and <i><b>Y</b></i> there are six identical parallel resistors. Their combined resistance is <i><b>R<sub>XY</sub></b></i>.<br /><i><b>R<sub>XY</sub> = R/6</b></i><br />Between points <i><b>Y</b></i> and <i><b>G</b></i> there are three identical parallel resistors. Their combined resistance is <i><b>R<sub>YG</sub></b></i>.<br /><i><b>R<sub>YG</sub> = R/3</b></i><br />These three groups are connected in a series with combined resistance<br /><i><b>R<sub>cube</sub> = R/3 + R/6 R/3 = 5R/6</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-30236835826341273242020-04-04T20:58:00.001-07:002020-04-04T20:58:01.771-07:00Unizor - Physics4Teens - Electromagnetism - DC Ohm's Law - Problems 1<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/i6AhoCWziXY" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current - Ohm's Law -<br />Problems 1</u><br /><br /><i>Problem A</i><br />Determine currents <i><b>I</b></i>, <i><b>I<sub>1</sub></b></i> and <i><b>I<sub>2</sub></b></i>, if voltage at the source of electric power <i><b>U</b></i> and resistances <i><b>R<sub>1</sub></b></i>, <i><b>R<sub>2</sub></b></i> and <i><b>R<sub>3</sub></b></i> are given. <img src="http://www.unizor.com/Pictures/Ohms_Problem1A.jpg" style="height: 140px; width: 200px;" /><br /><i>Solution</i><br /><i><b>R<sub>23</sub> = R<sub>2</sub> + R<sub>3</sub></b></i><br /><i><b>1/R = 1/R<sub>1</sub> + 1/R<sub>23</sub></b></i><br /><i><b>I = U/R</b></i><br /><i><b>I<sub>1</sub> = U/R<sub>1</sub></b></i><br /><i><b>I<sub>2</sub> = U/R<sub>23</sub></b></i><br /><br /><i>Problem B</i><br />Consider a circuit that contains the source of electric field with difference of electric potential on its terminals (<i>voltage</i>) <i><b>U</b></i> and a resistor with resistance <i><b>R</b></i>. The electric field forces the electrons to flow through the circuit, which means the field performs some work.<br />Express the work performed by an electric field during time <i><b>t</b></i> and its power in terms of voltage <i><b>U</b></i>, current <i><b>I</b></i> and resistance <i><b>R</b></i>.<br /><br /><i>Solution</i><br />Assume that <i><b>q</b></i> coulombs of electricity flows through a circuit during time <i><b>t</b></i>.<br />Using the definition of the difference of potentials (<i>voltage</i>) <i><b>U</b></i> and the definition of the electric current <b><i>I=q/t</i></b>, the work performed by a field <i><b>A</b></i> equals to<br /><i><b>A = U·q = U·I·t</b></i><br />and from the Ohm's Law <b><i>I=U/R</i></b> or <b><i>U=I·R</i></b>, the latter can be written as<br /><i><b>A = I²·R·t = U²·t/R</b></i><br />From the definition of power as work performed in a unit of time <i><b>W=A/t</b></i> follows<br /><i><b>W = I²·R = U²/R</b></i><br /><br /><i>Problem C</i><br />The Energy Conservation Law dictates that the work performed by an electric field to move electrons around a circuit defined by <i>voltage <b>U</b></i>, <i>amperage <b>I</b></i> and <i>resistance <b>R</b></i> should result in something tangible.<br />This tangible result is the heat. The molecules inside the resistor are pushed around by electrons circulated in the circuit and, as a result, the temperature of the resistor increases.<br />The amount of heat produced in the resistor equals to amount of work performed by the electric field (<b>Joule-Lenz Law</b>).<br />Consider the circuit with a resistor being a regular incandescent lamp with a spiral made of tungsten with mass <i><b>m=0.001kg</b></i>.<br />The spiral resistance is <i><b>R=200Ω</b></i>.<br />The voltage produced by a source of electric field is <i><b>110V</b></i>.<br />Tungsten specific heat capacity is <i><b>c=134J/(kg·K)</b></i>.<br />We switch on the electricity in a circuit for the period of <i><b>t=2sec</b></i>.<br />What will be the rise in temperature of the tungsten spiral at the end?<br /><br /><i>Solution</i><br />Work performed by an electric field:<br /><i><b>A = U²·t/R</b></i><br />(same as amount of heat inside the resistor)<br />The amount of heat is<br /><i><b>Q = 110²·2/200 = 121 joules</b></i><br />Heat <i><b>Q</b></i>, mass <i><b>m</b></i>, change of temperature Δ<i><b>T</b></i> and specific heat <i><b>c</b></i> are connected by the following law:<br /><i><b>Q = c·m·</b></i>Δ<i><b>T</b></i><br />Therefore,<br />Δ<i><b>T = Q <span style="font-size: medium;">/</span> (c·m)</b></i><br />Using the values given,<br />Δ<i><b>T = 121/(134·0.001) = 903°K</b></i><br />This will be the rise of temperature after 2 sec of working electricity.<br /><br /><i>Problem D</i><br />Determine currents <i><b>I</b></i>, <i><b>I<sub>1</sub></b></i> and <i><b>I<sub>2</sub></b></i>, if voltage at the source of electric power <i><b>U</b></i> and resistance <i><b>R</b></i> are given. <img src="http://www.unizor.com/Pictures/Ohms_Problem1D.jpg" style="height: 180px; width: 200px;" /><br /><br /><i>Solution</i><br />The problem with this circuit is that it cannot be considered as consisting from parallel and series resistors. The resistor <i><b>3R</b></i>, connecting top and bottom branches of a circuit, prevents it.<br />However, notice that the circuit is symmetrical, its top part (resistors <i><b>R</b></i> and <i><b>2R</b></i>) is the same as the bottom part. From this symmetry we conclude that<br /><i><b>I<sub>1</sub> = I<sub>2</sub> = I/2</b></i><br />and the voltage drops on resistors with resistance <i><b>R</b></i> on the top and on the bottom are the same<br />Δ<i><b>U = (I/2)·R</b></i><br />Therefore, the difference in electric potentials at the ends of resistor <i><b>3R</b></i> is zero and <b>there is no electric current going through it, and it can be removed and excluded from all the calculations</b>.<br />What's left after we exclude resistor <i><b>3R</b></i> from consideration is parallel connection of two pairs of resistors connected in a series.<br />Total resistance of each branch (top or bottom) is a sum of two resistors <i><b>R</b></i> and <i><b>2R</b></i> connected in a series<br /><i><b>R<sub>branch</sub> = R + 2R = 3R</b></i><br />Total resistance of two branches together is calculated from a formula for parallel connection of two resistors with resistance <i><b>3R</b></i> each<br /><i><b>1/R<sub>total</sub> = 1/(3R) + 1/(3R)</b></i><br /><i><b>R<sub>total</sub> = 3R/2</b></i><br />Now we can calculated the electric current <i><b>I</b></i> from the Ohm's Law<br /><i><b>I = U/R<sub>total</sub> = 2U/(3R)</b></i><br /><i><b>I<sub>1</sub> = I<sub>2</sub> = I/2 = U/(3R)</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-3480164561343523102020-04-04T20:48:00.001-07:002020-04-04T20:48:27.913-07:00Unizor - Physics4Teens - Ohm's Law - Batteries, Generators<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/wXGpQDPTGd8" width="480"></iframe><br /><br /><br /><br /><a class="style-scope ytcp-video-metadata-info" href="https://youtu.be/KuCvcUHi-iM" style="background-color: #f4f4f4; font-family: Roboto, Noto, sans-serif; font-size: 15px; text-decoration-line: none; white-space: nowrap;" target="_blank">https://youtu.be/KuCvcUHi-iM</a><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current -<br />Batteries, Generators</u><br /><br />In talking about electricity we will often use examples with heat distribution.<br />Consider a house with a heating system.<br /><img src="http://www.unizor.com/Pictures/Boiler.png" style="height: 300px; width: 200px;" /><br />A radiator, connected to a boiler by pipes, is installed in a room. The boiler heats up the water that goes from its outgoing pipe to an intake of a radiator, where it releases its heat, goes out of a radiator and comes back to a boiler intake pipe. There it is heated again and a circle continues.<br /><br />The boiler produces energy (heat), pipes transport it to a radiator, where it's released to a room.<br /><br />With an electric circuit that contains some source of electric energy (battery, generator etc.) and a resistor the process is very similar.<br />Inside a source of energy some electrons are separated from their nuclei and sent to one terminal, which becomes negatively charged, while those atoms that lost some of their electrons are sent to the other terminal, which becomes positively charged.<br />This process generates potential energy, since positive and negative charges are attracted to each other, and a battery separates them and does not let them to connect back within it.<br /><br />If we have a circuit from one terminal to one end of a resistor and from another end of a resistor to another terminal of the source of electric energy, the flow of electrons resembles the flow of water through a pipe from the boiler to a radiator, heating the room and flowing back to the boiler. The water in the radiator, heating the room, is like electrons making their way through a resistor, hitting its atoms that resist the movement of electrons. The water pipes are like wiring that connects a resistor to the source of electric energy. The boiler is like a source of electric energy (a battery or a generator). Full analogy.<br /><br />The potential energy of positive and negative charges in presence of a circuit is transformed into kinetic energy of electrons moving along a circuit, partially spending its energy on dragging through atoms of resistor and ending up on the positive terminal of the battery.<br />A simple mechanical equivalent is the process of a child climbing up a ladder to the top of a slide, thus generating a potential energy and, instead of going back down the stairs, slides down a slope, partially spending kinetic energy to friction.<br /><br />Using the analogy of heater and radiator, let's analyze what happens if we have more then one source of electricity. In particular, what happens if we connect batteries in a series or parallel to each other.<br /><br /><i>Series Connection</i><br /><br />Consider first the situation with two boilers connected in a series. That is, outgoing pipe with hot water from one boiler goes to another boiler, which adds more heat to the water before it's sent from it to a radiator, from which the water goes back to the intake of the first radiator.<br /><br />Obviously, the water, being heated twice, will be more hot than heated only once. So, the amount of water going to a radiator will be the same as with one boiler, but its temperature will be greater.<br /><br />Similar process occurs when two batteries are connected in a series as follows: <img src="http://www.unizor.com/Pictures/BatteriesSeries.png" style="height: 100px; width: 200px;" /><br />Battery <i>A</i> is capable of separating certain number of electrons from their nuclei, so does battery <i>B</i>.<br />When connected in a series, battery <i>B</i> separates its own electrons plus is given the already separated electrons form battery <i>A</i>, thus increasing the number of electrons separated from their nuclei. That creates a larger difference in potential between positive terminal of <i>A</i> and negative terminal of <i>B</i>.<br /><br />Therefore, batteries in a series increase the difference of potential (<i>voltage</i>) between their free ends.<br />For example, if a single AA battery provides 1.5V on its terminals, two such batteries connected in a series provide 3V.<br /><br />This type of connection in a series is used when we connect 2 standard AA batteries to power TV remote control that requires 3V to work properly. Or we connect 6 such batteries in a series to get 9V difference in potentials to power a toy car.<br /><br />So, all batteries connected in a series work at their usual capacity, last exactly as if they work separately, just the voltage is increased.<br /><br /><u>Conclusion:</u> <b>The <i>voltage</i> of an assembly of several sources of electric power connected in <i>series</i> is a sum of all individual voltages of its components</b>.<br /><br /><br /><i>Parallel Connection</i><br /><br />Again, let's start with boiler analogy. Consider two identical boilers connected parallel to each other, that is outgoing pipes with hot water from both of them are joined and go to a radiator, where the hot water releases its heat, cools down, and the cold water on its way out is split into both intakes of two boilers.<br /><br />Under these conditions the temperature of the hot water is the same as if only one boiler worked. At the same time, each boiler can work with half of its power because the total amount of hot water needed to heat the room is now split into two boilers.<br /><br />Now consider two identical batteries connected parallel to each other.<br /><img src="http://www.unizor.com/Pictures/BatteriesParallel.png" style="height: 200px; width: 160px;" /><br />The difference in electric potential (<i>voltage</i>) on the free ends of this assembly is the same as the voltage on the ends of each battery. So, <b>the <i>voltage</i> in parallel connection remains the same</b>.<br /><br />At the same time the electrons needed by electric power consumer (a lamp, a refrigerator etc.) connected into a circuit with these batteries will consume half of needed electrons from one battery and another half from another.<br />So, <b>the batteries in parallel connection share the burden and will work longer</b>. Actually, about twice as long in this configuration than if only one battery supplied the power to a device that consumes the electricity.<br /><br />For example, if a single AA battery provides 1.5V on its terminals and works for 1 hour supplying electricity to a small lamp, two such batteries connected in parallel provide the same 1.5V for this lamp, but will last about 2 hours.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-27235475065224806412020-03-29T08:10:00.001-07:002020-03-29T08:10:22.839-07:00Unizor - Physics4Teens - Electromagnetism - Votage Drop<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ueLvyFe-Y24" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current - Voltage Drop</u><br /><br />Imagine a waterfall going down the rocks. As it hits each rock, it loses some of its potential energy.<br />Analogously, consider several radiators in a series connected to a boiler that supplies hot water for them. As the water goes from one radiator to another, it cools down, losing its heat energy.<br />The same happens with electrons, as they go along the circuit and meet one resistance after another. Going through each of them, they lose their energy.<br /><br />Consider the following circuit<br /><img src="http://www.unizor.com/Pictures/ResistorSeries.jpg" style="height: 120px; width: 200px;" /><br />Assume, we know how to measure the difference in potential (<i>voltage</i>) between any two points on this circuit. Measuring the voltage between the terminals of the battery gives some value <i><b>U</b></i>.<br />Measuring the voltage between the positive terminal of the battery (the longer thin line with a plus sign) and the point in-between the resistors gives some other value <i><b>U<sub>1</sub></b></i>.<br />Measuring the voltage between the negative terminal of the battery (the shorter thick line with a minus sign) and the point in-between the resistors gives yet other value <i><b>U<sub>2</sub></b></i>.<br /><br />Notice that the amount of electricity going through this circuit per unit of time (<i>electric current</i> or <i>amperage</i>) is the same everywhere since it's a closed loop and equals <i><b>I</b></i>.<br /><br />Now let's apply the Ohm's Law to an entire circuit, keeping in mind that the voltage between the terminals of the battery is <i><b>U</b></i> and the total resistance of an entire circuit is <i><b>R=R<sub>1</sub>+R<sub>2</sub></b></i>:<br /><i><b>I = U <span style="font-size: medium;">/</span> R = U <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>)</b></i><br /><br />Applying the Ohm's Law to a segment of a circuit that includes only resistor <i><b>R<sub>1</sub></b></i> and knowing the voltage on its two ends <i><b>U<sub>1</sub></b></i>, we obtain<br /><i><b>I = U<sub>1</sub> <span style="font-size: medium;">/</span> R<sub>1</sub></b></i><br /><br />Equating two different expressions for the electric current <i><b>I</b></i>, we can find the voltage <i><b>U<sub>1</sub></b></i>:<br /><i><b>I = U <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) = U<sub>1</sub> <span style="font-size: medium;">/</span> R<sub>1</sub></b></i><br />from which the value of <i><b>U<sub>1</sub></b></i> is<br /><i><b>U<sub>1</sub> = U·R<sub>1</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>)</b></i><br /><br />Let's do the same calculation for the second resistor.<br /><i><b>I = U <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) = U<sub>2</sub> <span style="font-size: medium;">/</span> R<sub>2</sub></b></i><br />from which the value of <i><b>U<sub>2</sub></b></i> is<br /><i><b>U<sub>2</sub> = U·R<sub>2</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>)</b></i><br /><br />Notice that <nobr><i><b>U<sub>1</sub> + U<sub>2</sub> = U</b></i>.</nobr><br />Indeed,<br /><i><b>U·R<sub>1</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) +<br />+ U·R<sub>2</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) =<br />= U·(R<sub>1</sub>+R<sub>2</sub>) <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) = U</b></i><br /><br />Completely analogous calculations can be provided with three or more resistors connected in a <i>series</i>.<br />So, the difference in electric potential or <b>voltage drop</b> between the terminals of a battery <i><b>U</b></i> is split between the voltage drops on each resistor in a <i>series</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-10918279177176819342020-03-28T12:52:00.001-07:002020-03-28T12:52:18.732-07:00Unizor - Physics4Teens - Electromagnetism - Ohm's Law<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/B9xIOeHul_0" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current - Ohm's Law</u><br /><br />Simply speaking, experiment shows that the current in a conductor (<i>amperage</i>) is proportional to a difference in potentials (<i>voltage</i>) on its ends. This is the <b>Ohm's Law</b>.<br /><br />Obviously, some physical explanation of this law is appropriate, and this is what this lecture is about.<br /><br />Electric current is the flow of electrons inside a conductor. Just as a reminder, the traditionally defined direction of the current is opposite to the direction of the flow of electrons.<br /><br />Even without any electric field electrons inside a conductor are not exactly spin each around its own nucleus, staying on the orbit forever. Some of them, especially those on the outer orbit, can tear off their orbit, jump to another nucleus' orbit, then going somewhere else etc., thus creating certain chaotic environment.<br />When an electric field is present, the chaotic movement of electrons becomes directional to a degree, depending on the strength of the field, thus creating a flow of electrons - an <i>electric current</i>.<br /><br />This flow of electrons occurs when there is an excess of electrons on one end and deficiency (or less of an excess) of electrons on the other end, which creates an electric field inside a conductor that forces some light electrons to leave their atoms and move, while heavy nuclei with remaining electrons stay put.<br />Electrons are repelled from the end, where there is excess of them and attracted (or repelled less) from the other end.<br /><br />On their way from one end of a conductor to another these electrons must go through a maze of atoms, many of which have lost some electrons because of the electric field and, therefore, are positively charged, attracting negative electrons in an attempt to compensate lost electrons. Some succeed by capturing free electrons, some not, some lose more electrons in their chaotic but directional movement.<br /><br />The flow of electrons inside a conductor can be compared with a waterfall from the high level, where potential energy relative to gravitational field of the Earth is greater to a lower level with lesser potential energy.<br /><br />As water falling down the waterfall, hitting all the stones on its way, electrons hit atoms and lose some of their energy, some of them get absorbed by atoms and don't reach the other end of a conductor, thus diminishing the flow.<br /><br />The very important aspect in this movement of electrons is the properties of the material a conductor is made of.<br /><br />Different atoms of different materials have different characteristics of capturing and releasing electrons, when subjected to the forces of an electric field. However, certain common laws of conductivity can be logically explained and used in practical applications.<br /><br /><i>Ohm's Law</i><br /><br />Stronger electric field produced by a greater difference in electric potential (<i>voltage <b>U</b></i>) between the ends of a conductor should cause more intense movement of electrons, greater number of electrons going from one end of a conductor and reaching the other end per unit of time (<i>amperage <b>I</b></i>). This qualitative property is perfectly understood. Mathematically speaking, <i><b>I</b></i> is a <i>monotonically increasing</i> function of <i><b>U</b></i>.<br /><br />It was an experimental fact that this monotonic function to a very high precision is <i>linear</i>:<br /><i><b>I = σ·U</b></i><br />where <i><b>σ</b></i> is called a <b>conductivity</b> of a conductor.<br />In more common cases this law is written not in terms of <b>conductivity</b>, but in terms of <b>resistivity</b>:<br /><i><b>I = U <span style="font-size: medium;">/</span> R</b></i><br />where <i><b>R</b></i> is called a <b>resistance</b> of a conductor.<br /><br />In most cases we will use the Ohm's Law in that last form, using a <b>resistance</b> as the characteristic of a conductor.<br /><br /><i>Resistance</i><br /><br />In practical life we use some conductors with a very low resistance, like copper wire, to connect some electrical appliance, like a regular incandescent lamp bulb, to a source of electric power. In this arrangement we usually consider wiring as having no resistance and concentrate on the properties of an appliance as the one with the electrical resistance <i><b>R</b></i>.<br /><br />Schematically, resistor is pictured as a rectangle or a zigzag line connected by straight lines of wires to a source of electric power<br /><img src="http://www.unizor.com/Pictures/Resistor.jpg" style="height: 120px; width: 200px;" /><br />Arrows on this picture show the <u>traditionally</u> defined direction of electricity from positive to negative terminal of the source of electric power - opposite to a direction of electrons' movement.<br /><br />Consider an extreme case when the resistor is non-conductive, like there is only vacuum in between its two ends. Symbolically, it's equivalent to <i><b>R=∞</b></i>. The Ohm's Law in this case results in <i><b>I=U/R=0</b></i>, which means that the circuit is broken and there is no current in it.<br /><br />In an opposite extreme case with <i><b>R=0</b></i> we have <i><b>I=U/R=∞</b></i>, which is the so called "short".<br />In practical life it happens when you detach two wires coming to a lamp and connect them. Without the bulb, which has some significant resistance, the electric current in a circuit would almost instantaneously grow very high and you might get electrocuted.<br /><br />Let's use an incandescent lamp as an example of a resistor. More precisely, the resistor is a tangent spiral (filament) inside the lamp.<br /><br />Consider what happens with the resistance of this tangent spiral if we will make it longer.<br />Obviously, the electrons will have to go through more obstacles on their way from one end of a spiral to another, which will slow their movement more than when a spiral was shorter. It's logical to expect that doubling the length of a resistor will double its resistance, and it is confirmed by experiments.<br />In general, for linearly-shaped resistors their resistance should be proportional to the length.<br /><br />The immediate consequence of this consideration is that <b>two resistors, sequentially connected one after another in a <i>series</i>, will have a combined resistance equal to a sum of resistance of each one</b>.<br /><img src="http://www.unizor.com/Pictures/ResistorSeries.jpg" style="height: 120px; width: 200px;" /><br /><br />Now, instead of making a spiral longer, let's make it thicker. Electrons will have more space to move, more freedom of direction and more electrons can travel across the spiral per unit of time. Doubling the thickness of a spiral is similar to doubling the width of a highway with more cars moving on it per unit of time. So, the resistance of a thicker spiral will decrease by the factor of increased cross-section area of a tangent spiral.<br />In general, for linearly-shaped resistors their resistance should be inversely proportional to the cross-section area.<br /><br />Increasing the thickness of a spiral is logically equivalent to using two spirals connected parallel to each other, as on this circuit diagram. <img src="http://www.unizor.com/Pictures/ResistorParallel.jpg" style="height: 120px; width: 200px;" /><br />The number of electrons per unit of time (electric current or <i>amperage</i>) coming from the common wire is split into two parallel flows, and all the electrons passing through a common part of a circuit per unit of time should be equal to a sum of numbers of electrons passing through each of the parallel segments.<br />The immediate consequence of this consideration is that <b>two resistors, connected parallel to each other, will result in the electric current in a common wire to be a sum of electric currents in each one of them</b>.<br /><i><b>I = I<sub>1</sub> + I<sub>2</sub></b></i><br /><br />Now let's apply the Ohm's Law to an entire circuit, and each of two parallel branches, keeping in mind that the voltage between the terminals of the battery is <i><b>U</b></i>, the same as the voltage between the ends of each resistor, and the resistance of each branch is known as <i><b>R<sub>1</sub></b></i> and <i><b>R<sub>2</sub></b></i>.<br /><br />Assuming the total resistance of two parallel branches is <i><b>R</b></i>, the Ohm's Law gives<br /><i><b>I = U <span style="font-size: medium;">/</span> R</b></i><br /><br />Applying the same principle to one branch with resistor <i><b>R<sub>1</sub></b></i>, we have<br /><i><b>I<sub>1</sub> = U <span style="font-size: medium;">/</span> R<sub>1</sub></b></i><br /><br />The same for another branch:<br /><i><b>I<sub>2</sub> = U <span style="font-size: medium;">/</span> R<sub>2</sub></b></i><br /><br />Now the original equation about a current in the common segment of a wire equaled to a sum of currents in two branches looks like<br /><i><b>U <span style="font-size: medium;">/</span> R = U <span style="font-size: medium;">/</span> R<sub>1</sub> + U <span style="font-size: medium;">/</span> R<sub>2</sub></b></i><br />or<br /><i><b>1<span style="font-size: medium;">/</span>R = 1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub></b></i><br /><br />From the last equation follows the expression for a combined resistance of two resistors installed parallel to each other<br /><i><b>R = 1 <span style="font-size: medium;">/</span> (1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub>)</b></i><br />This formula looks more cumbersome and the one for <i><b>1<span style="font-size: medium;">/</span>R</b></i> above is more commonly occurs.<br /><br />Instead of resistance, it can be expressed in terms of <i>conductivity</i> of an entire assembly <i>σ=1/R</i> of two parallel resistors with conductivity <i>σ<sub>1</sub>=1/R<sub>1</sub></i> and <i>σ<sub>2</sub>=1/R<sub>2</sub></i>:<br /><i><b>σ = σ<sub>1</sub> + σ<sub>2</sub></b></i><br /><br />For two identical resistors of resistance <i><b>r</b></i> each the combined resistance <i><b>R</b></i> would be<br /><i><b>1/R = 1/r + 1/r = 2/r</b></i><br /><i><b>R = r/2</b></i><br /><br />More generally, we can derive the resistance of <i><b>n</b></i> identical resistors of resistance <i><b>r</b></i> each connected parallel to each other. In this case the electric current <i><b>I</b></i> in the common wire in split into <i><b>n</b></i> identical flows, each having an amperage of <i><b>I/n</b></i>. From the Ohm's Law the voltage at the ends of each resistor should be <i><b>U=I·r/n</b></i>, from which follows that <i><b>R=r/n</b></i> is the resistance of an entire assembly of <i><b>n</b></i> identical resistors with resistance <i><b>r</b></i> each.<br /><br /><i>Example</i><br /><br />Consider a circuit with resistors <i><b>R<sub>1</sub></b></i> and <i><b>R<sub>2</sub></b></i> connected <i>parallel</i> to each other and resistor <i><b>R<sub>3</sub></b></i> installed after both of them in a <i>series</i>.<br />What would be the resistance <i><b>R</b></i> of a combined assembly of these three resistors?<br /><br />First, determine the resistance of two <i>parallel</i> resistors as a unit.<br /><i><b>R<sub>12</sub> = 1 <span style="font-size: medium;">/</span> (1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub>)</b></i><br />This this unit of two parallel resistors with resistance <i><b>R<sub>12</sub></b></i> is in a <i>series</i> with resistor <i><b>R<sub>3</sub></b></i>, their combined resistance is the sum of their individual resistances<br /><i><b>R = R<sub>12</sub> + R<sub>3</sub> = </b></i>[<i><b>1 <span style="font-size: medium;">/</span> (1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub>)</b></i>]<i><b> + R<sub>3</sub></b></i><br /><br /><br /><i>Resistance Units<br />of Measurement</i><br /><br />The Ohm's Law allows to easily establish the units of measurement for the resistance of any component in an electric circuit - <b>ohms Ω</b>.<br />From the original form of the Ohm's Law<br /><i><b>I = U <span style="font-size: medium;">/</span> R</b></i><br />follows that<br /><i><b>R = U <span style="font-size: medium;">/</span> I</b></i><br /><br />This allows to establish a unit of measurement for resistance of any component of an electric circuit.<br />If the difference in potential (<i>voltage</i>) between one end of such a component and another is <i><b>1V </b>(volt)</i> and the electric current going through it is <i><b>1A </b>(ampere)</i>, this component by definition has a resistance of <i><b>1Ω </b>(ohm)</i>.<br /><br />Consequently, the resistance of a component in an electric circuit with the current going through it <i><b>I</b> (ampere)</i> with voltage on its end <i><b>U </b>(volt)</i> equals to <i><b>U <span style="font-size: medium;">/</span> I</b> (ohm)</i>.<br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-62565975928265431572020-03-11T09:41:00.001-07:002020-03-11T09:41:49.722-07:00Unizor - Physics4Teens - Electromagnetism - Electric Current - Speed of ...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/36iXMPKENRs" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Speed of Electrons</u><br /><br /><i>Electric current</i> is a movement of electrons. We know from experience that, when we turn on the switch, the lights in the room are lit practically immediately. Does it mean that electrons from one terminal of a switch go to the light fixture and back to another terminal of a switch that fast?<br />No.<br /><br />Let's calculate the real speed of electrons, first, theoretically and then in some practical case.<br /><br />Assume, the <i>amperage</i> of the electric current going through a wire, that is the number of <i>coulombs</i> of electric charge going through a wire per second, is <i><b>I</b></i>, and the wire has cross-section area <i><b>A</b></i>.<br />Assume further that we know all the physical characteristics of a material our wire is made of, which will be introduced as needed.<br /><br />Based on this information, our plan is to determine the number of electrons going through the wire per unit of time and, knowing the density of electrons per linear unit of length in the wire, determine the linear speed of these electrons.<br /><br />Obviously, to determine the linear density of electrons, we will need physical characteristics of a wire.<br /><br />The number of electrons going through a wire per unit of time is easily determined from the <i>amperage</i> <i><b>I</b></i>. Since <i><b>I</b></i> represents the number of <i>coulombs</i> of electric charge going through a wire per second, we just have to divide this by the charge of a single electron in <i>coulombs</i> <i><b>e=−1.60217646·10<sup>−19</sup></b>C</i>.<br />So, the number of electrons going through a wire per second is<br /><i><b>N<sub>e</sub> = I <span style="font-size: medium;">/</span> e</b></i>.<br /><br />Now we will determine the linear density of electrons in the wire.<br /><br />First of all, we have to know how many active electrons in an atom of material our wire is made of participate in the transfer of electric charge, because not all electrons of each atom are freely moving in the electric field, but only those on the outer orbit. Let's assume, this number is <i><b>n<sub>e</sub></b></i>.<br />Using this number, we convert the number of electrons participating in the transfer of electric charge <i><b>N<sub>e</sub></b></i> into the number of atoms <i><b>N<sub>atoms</sub></b></i> in that part of a wire occupied by all electrons transferring the given charge per second <i><b>I</b></i>.<br /><i><b>N<sub>atoms</sub> = N<sub>e</sub> <span style="font-size: medium;">/</span> n<sub>e</sub> = I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e)</b></i>.<br /><br />Next, from the number of atoms we will find their mass and, using the density of wire material, the volume.<br />Dividing the volume by a cross-section of a wire, we will get the length of a segment of wire occupied by those electrons transferring charge per second, which is the <b>speed of electrons</b> or <b>drift</b>.<br /><br />Knowing the number of atoms, to get to their mass, we will use the <i>Avogadro number</i><br /><i><b>N<sub>A</sub>=6.02214076·10<sup>23</sup></b></i><br />that represents the number of particles in one <i>mole</i> of a substance. One <i>mole</i> of material our wire is made of is the number of grams equal to its <i>atomic mass</i> <i><b>m<sub>a</sub></b></i>, known for any material used for a wire.<br />If <i><b>N<sub>A</sub></b></i> atoms have mass of <i><b>m<sub>a</sub></b></i> gram, <i><b>N<sub>atoms</sub></b></i> have total mass<br /><i><b>M<sub>atoms</sub> = m<sub>a</sub> · N<sub>atoms</sub> <span style="font-size: medium;">/</span> N<sub>A</sub> =<br />= m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>)</b></i>.<br /><br />Knowing the total mass <i><b>M<sub>atoms</sub></b></i> of all atoms that contain all electrons traveling through a wire per second, we calculate the volume <i><b>V<sub>atoms</sub></b></i> by using the <i>density</i> <i><b>ρ</b></i> of a material the wire is made of.<br /><i><b>V<sub>atoms</sub> = M<sub>atoms</sub> <span style="font-size: medium;">/</span> ρ =<br />= m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>·ρ)</b></i><br /><br />Dividing the volume <i><b>V<sub>atoms</sub></b></i> by the cross-section area of a wire, we will get the length of the wire <i><b>L</b></i> occupied by electrons traveling through it in one second<br /><i><b>L = V<sub>atoms</sub> <span style="font-size: medium;">/</span> A =<br />= m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>·ρ·A)</b></i><br />where<br /><i><b>m<sub>a</sub></b></i> - <i>atomic mass</i> of wire's material (assuming it's one atom molecules, like copper)<br /><i><b>I</b></i> - <i>electric current - amperage</i> in the wire<br /><i><b>n<sub>e</sub></b></i> - <i>number of active electrons in each atom</i> of wire's material that participate in the transfer of electric charge<br /><i><b>e</b></i> - <i>electric charge of one electron</i><br /><i><b>N<sub>A</sub></b></i> - <i>number of atoms in 1 mol of conducting material - Avogadro Number</i><br /><i><b>ρ</b></i> - <i>density</i> of wire's material<br /><i><b>A</b></i> - <i>cross-section area</i> of a wire<br /><br />The above formula represents the length of a wire occupied by all active electrons traveling through it during one second, <b>which is the speed of movement of electrons making up an electric current</b>, called <b>drift</b>.<br /><br />Let get to practical examples.<br /><br />Assume, the <i>voltage</i> or the difference of <i>electric potential <b>E</b></i> between two ends of a copper wire is maintained at <b>110V</b> (standard voltage for apartments in the USA).<br />This wire connects a lamp that consumes <b>120W</b> of electric power <i><b>P</b></i> (or <i>wattage</i>).<br />Let the cross-section area of a wire be <b>3 mm²</b>.<br /><br />First of all, let's calculate the amount of electricity moving through the wire per unit of time - <i>amperage <b>I</b></i>.<br />As we know, the <i>amperage <b>I</b></i>, multiplied by <i>voltage <b>E</b></i>, is the electric power <i><b>P</b></i> (<i>wattage</i>).<br />Therefore,<br /><i><b>E = 110</b>V</i><br /><i><b>P = 120</b>W</i><br /><i><b>I = P/E = 60W/110V ≅ 1.09</b>A</i><br /><br />The <i>atomic number</i> of copper is <i><b>Z=29</b></i>. It means, the atom of copper has 29 protons in the nucleus and 29 electrons orbiting a nucleus. These 29 electrons are in four orbits: 2+8+18+1.<br />The outer orbit has only one electron that participates in the movement of electric charge, so the number of active electrons in an atom of copper is<br /><i><b>n<sub>e</sub>=1</b></i>.<br /><br />The nucleus of an atom of copper has 29 protons and 34 or 36 neutrons, its <i>atomic mass</i> is<br /><i><b>m<sub>a</sub> = 63.546</b>g/mol</i><br /><br />The electric charge of one electron is<br /><i><b>e = −1.60217646·10<sup>−19</sup></b>C</i>.<br /><br />The Avogadro number is<br /><i><b>N<sub>A</sub>=6.02214076·10<sup>23</sup></b></i><br /><br />Density of copper is<br /><i><b>ρ = 8.96</b> g/cm³ <b>= 0.00896</b> g/mm³</i><br /><br />Cross-section area of a wire is<br /><i><b>A = 3</b> mm²</i><br /><br />Using the formula above with values listed, we obtain<br /><i><b>L = m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>·ρ·A)</b></i><br />we get<br /><i><b>L ≅ 0.0267</b> mm/sec</i><br />This is the speed of electrons traveling along a copper wire in this case.<br />Pretty slow!Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-67528497808565416752020-02-24T20:42:00.001-08:002020-02-24T20:42:02.809-08:00Unizor - Physics4Teens - Electromagnetism - Electric Current<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/IygOGBg68mg" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Electric Current</u><br /><br /><i>Conductors and Dielectrics</i><br /><br /><i>Electric current</i> is a flow of electric charge. Since the actual carrier of electric charge is excess or deficiency of electrons, we need certain material where electrons can travel. So, vacuum cannot be a <i>conductor</i> of electricity because there is no electrons in it, but many metals, like copper, can. But we know that the electrons are orbiting the nuclei of the atoms. So, why do they travel?<br /><br />The answer is: <b>the force of an external electric field</b> pushes or pulls electrons off their orbits and, as a result, they move inside the material where electric field is present towards or away from the source of the electric field, depending on whether the source is positively (deficiency of electrons) or negatively (excess of electrons) charged.<br /><br />Consider a copper wire. It contains atoms of copper with 29 electrons in each atom, orbiting on different orbits around corresponding nuclei with 29 protons and from 34 to 36 neutrons in each.<br />Electrons stay on their orbit until some outside electric field comes into play. When it does, if its intensity is sufficient to push or pull light electrons off their orbits, while heavy nuclei stay in place, these electrons move in one or another direction as a result of different forces acting on them, the major of which is the intensity of the outside electric field. General direction of electrons is defined by the vector of intensity of the electric field. That makes copper a good <i>conductor</i> of electricity.<br /><br />On the other hand, there are materials, like glass, where electrons are connected stronger to their nuclei, which makes more difficult to push them off their orbits. these materials do not conduct electricity, they are called <i>insulators</i> or <i>dielectrics</i>.<br /><br />Ideal <i>conductor</i>, connected to an electrically charged object, makes an extension of this object. Since electrons are freely moving between the original object and an attached conductor, both constitute a new object with an electric charge evenly distributed between its parts.<br />Ideal <i>dielectric</i>, attached to an electrically charged object, does not share its electrons with this object, so the object remains the only one charged.<br /><br />In practical cases there are no ideal conductors (except under certain conditions of superconductivity under temperatures close to absolute zero) and no ideal dielectrics (except absolute vacuum that has no electrons at all).<br /><br />Metals are usually good conductors because their nuclei are relatively not easily moved from their places, while electrons are easily pushed off their orbits.<br />We use this property of <i>conductors</i> to direct the electrons to perform some work, like lighting the bulbs or moving electrical cars.<br /><br />IMPORTANT NOTICE:<br /><i>Conductivity</i> is related to movement of electrons and is a measure of how easily electrons are pushed form their orbits by outside electric field.<br />This should not be mixed with <i>permittivity</i> defined for electric fields and is a measure of propagation of electric field inside some substance.<br /><br /><i>Electric Current</i><br /><br />If the source of the field is a positive charge located near one end of a copper wire, electrons inside the wire would go towards that end. If the negative charge is the source of the field, electrons will move towards the opposite end.<br /><br />If there is nothing on the opposite end of a copper wire, electrons, after being pushed towards one of the edges, will stop. If, however, there is an opposite charge on the other end of a wire, electrons will move from the negatively charged end to the positively charged one until both charges neutralize each other and whatever end was missing electrons (positively charged) will be compensated by electrons that are in excess on the negatively charged end.<br /><br />Imagine now that we manage to keep one end of the wire constantly charged positively, while another end constantly charged negatively. Then electrons from the negatively charged end will flow to the positively charged end as long as we can keep these constant opposite charges on both ends. We will have a constant flow of electricity, which is called <i>electric current</i> (or simply <i>current</i> in the context of electricity).<br /><br />This process of maintaining constant flow of electricity is analogous to maintaining constant flow of water down the water slide using a pump that constantly pumps the water from a pool to the top of a slide, from which it flows down because of the difference in heights and gravity.<br /><br />While the presence of the electric field is felt almost instantaneously (actually, with a speed close to a speed of light), the electrons that carry electrical charge are not moving from a negatively charged end of a copper wire to the positively charged end with this speed.<br /><br />A good analogy is the pipe filled with water and a pump connected to one of its ends. As soon as the pump starts working, the water it pumps starts its trip along the pipe and pushes the neighboring molecule of water. Those, in turn, push the next ones etc. So, the water will come from another end of a pipe almost instantaneously (actually, after a time interval needed for the sound waves in the water to cover the length of a pipe), but it's the "old" water already present in the pipe before the pump started working. "New" water that is physically pushed into a pipe by a pump will eventually reach the other end, but not that fast.<br /><br />Finally, let's talk about measurement of the <i>electric current</i>.<br />The natural way of measurement of the flow of water in the pipe, as exemplified above, would be amount of water flowing out of a pipe per unit of time.<br />In our case of electric charge we can do the same - measure the flow by amount of electricity (in <i>coulombs</i>) traveling from one source of electric field to another (with opposite charge) per unit of time.<br /><br />The unit of measurement of the <i>electric current</i> is <i>ampere</i>, where <i>1 ampere</i> is the flow of electricity, when <i>1 coulomb</i> of electricity is moving across the wire within <i>1 second</i>.<br /><b><i>1 A = 1 C / 1 sec</i></b>.<br /><br />Recall the definition of a unit <i>volt</i> as a difference in electric potential between points <i>A</i> and <i>B</i> such that moving one <i>coulomb</i> of electric charge between these points requires one <i>joule</i> of work. Therefore,<br /><i><b>1 J = 1 V · 1 C</b></i><br />From the definition of <i>ampere</i> above<br /><b><i>1 C = 1 A · 1 sec</i></b>.<br />Therefore,<br /><i><b>1 J = 1 V · 1 A · 1 sec</b></i><br /><i><b>1 V · 1 A = 1 J / 1 sec</b></i><br />As we know,<br /><i><b>1 J / 1 sec = 1 W (watt)</b></i><br />So, electric current of <i>1 ampere</i> between points with difference of potential <i>1 volt</i> performs work of <i>1 watt</i>, that is <i>1 joule per second</i>.<br /><br />There is a direct analogy between electricity and mechanics with <i>force</i> analogous to <i>voltage</i> and <i>speed</i> analogous to <i>amperage</i><br /><i><b>Force · Distance = Work</b></i><br /><i><b>Force · Distance / Time =<br />= Work / Time = Power</b></i><br /><i><b>Force · (Distance / Time) =<br />= Force · Speed = Power</b></i><br /><i><b>Voltage · Amperage = Power</b></i><br /><br />Let's consider a slightly more complicated example of the <i>electric current</i>.<br />Assume that at one end of a copper wire we have a source of electric field with negative charge and at another end of this wire we have another source of electric field also with negative charge. Both ends will repel electrons inside a wire. However, if the charges are not equal, the larger one will push stronger, and electrons will move away from it towards the other end of a wire.<br /><br />The situation with two unequal negative charges is analogous to a water pipe with two pumps of different power pumping water into it from both ends. The stronger pump will overcome the weaker and the water will move from a stronger pump to the weaker.<br /><br />So, the most important factor in determining the direction of electrons in the wire is the <i>intensity of electric field</i> produced by electric charges. For multiple sources of electric field their vectors of intensity are added. From a general viewpoint, if there is a difference in intensity of electric fields, electrons will travel in the direction defined by a stronger force. In practical situation, when two sources of electricity are applied to two ends of a wire, one positive and one negative, one end attracts electrons and another pushes them away, the flow of electrons will be always from negative to positive charge.<br /><br />Assume, the intensity of electric field at the end <i>A</i> of a wire is <i><b>E<sub>A</sub></b></i> and intensity at the other end <i>B</i> is <i><b>E<sub>B</sub></b></i>. If both charges at points <i>A</i> and <i>B</i> are positive or both negative, the vectors <i><b>E<sub>A</sub></b></i> and <i><b>E<sub>B</sub></b></i> inside a wire are oppositely directed. If the charges are of different sings (which is a typical situation in practical applications of electricity), these vectors are directed the same way.<br /><br />The force acting on each coulomb of electricity inside a wire is a vector sum of both intensities:<br /><i><b>E = E<sub>A</sub> + E<sub>B</sub></b></i><br />The work needed to move one coulomb of electricity is, therefore,<br /><i><b>W = E·L</b></i>,<br />where <i><b>L</b></i> is the length of a wire.<br />This value <i><b>W</b></i> represents the difference of <i>electric potentials</i> of the electric field between points A and B, that is the <i>voltage</i> <i><b>V<sub>AB</sub></b></i> between them.<br />The difference in <i>intensity of an electric field</i> corresponds to the non-zero <i>voltage</i> between these points.<br /><br />If we can maintain the difference in electric field's potential between the two ends of a wire (non-zero <i>voltage</i> between them), the intensity of an electric field will push electrons from one end of a wire to another. This is how <b>direct electric current</b> is maintained.<br /><br />As electrons move from one end to another, they leave "holes" - spots where they used to be, which are "moving" in the opposite direction. Since we conditionally associate "negative" charge with electrons and "positive" charge with the absence of electrons ("holes"), we can say that the direction of positive charges is opposite to that of negative.<br /><br />For historical reasons, because electrons were not discovered yet, the direction of positive charges (that is, "holes" that are left, when electrons leave their places) was defined as a direction of the electric current.<br /><br />The word <i>direct</i> means that the direction of the flow of electrons does not change with time and goes from the end with negative charge to the end with positive charge, which implies that the direction of the electric current (the direction of "holes" left by electrons) is opposite, from positive to negative end. For practical reasons we will not consider the case of the same sign of charges on both ends.<br /><br />In most practical cases there is a device that separates the electrons from the neutral atoms within some object, thereby producing negative and positive charges on its terminals. If there is some conductor of electricity between these terminals, electrons will move from one terminal to another along this conductor, which constitutes a <b>direct electric current</b> in it.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-14102784939056414392020-02-20T14:11:00.001-08:002020-02-20T14:11:09.732-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Capacitors<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/obgGC5itTsk" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Capacitors</u><br /><br /><i>Electric Field of<br />a Uniformly Charged Disk</i><br /><br />Please refer to Problems 4 of "Electric Field", as we will use its results.<br />The problem there derives a formula of intensity of the electric field produced by an infinitely thin disk of radius <i><b>R</b></i> charged with surface density of electricity <i><b>σ</b></i> at point <i><b>P</b></i> positioned at height <i><b>h</b></i> above the center of this disk in case the space around this disk is filled with material with <i>dielectric constant</i> (also known as <i>relative permittivity</i>) <i><b>ε<sub>r</sub></b></i>.<br /><br />The direction of the vector of electric field intensity is along the perpendicular from point <i><b>P</b></i> to a charged disk and its magnitude equals to<br /><i><b>E(h) =<br />= </b></i>[<i><b>σ/(2ε<sub>r</sub>·ε<sub>0</sub>)</b></i>]<i><b>·</b></i>[<i><b>1−1/√<span style="text-decoration-line: overline;">1+R²/h² </span></b></i>]<br /><br />From the above formula for any media filling the space around a charged disk we see that the greater <i>dielectric constant</i> <i><b>ε<sub>r</sub></b></i> for the media that fills the space around the charged disk - the smaller intensity of the field around it.<br /><br />Assume now that our goal is to generate electric charge (excess or deficiency of electrons) and store it somehow. Since electrons are not easily produced from nothing, we should take some electrically neutral object <i>X</i>, separate part of its electrons from the atoms and place them into a different object <i>X<sub>−</sub></i>, which becomes negatively charged, leaving old object positively charged, which we can label now <i>X<sub>+</sub></i>.<br /><br />For practical reasons objects <i>X<sub>−</sub></i> and <i>X<sub>+</sub></i> should be near each other and we should prevent any kind of exchange of electric charge, like a spark, between them, which would negate our efforts to separate negative and positive charges. Considering the proximity of <i>X<sub>−</sub></i> and <i>X<sub>+</sub></i>, it would be beneficial for avoiding any exchange of the charge between them to have a uniform distribution of charge in each object.<br /><br />The best configuration of objects <i>X<sub>−</sub></i> and <i>X<sub>+</sub></i> that allows approximately uniform distribution of charge in the presence of opposite charge nearby is when both objects are thin flat plates positioned parallel to each other. Other configurations will cause the concentration of charges in places close to the opposite charge and higher intensity of the electric field between them, which might result in a discharging spark. Another configuration might be of two concentric spheres, but it's not very practical.<br /><br />Consider now two relatively large infinitely thin disks of radius <i><b>R</b></i> with opposite charges <i><b>+Q</b></i> and <i><b>−Q</b></i> positioned parallel to each other, perpendicular to a line connecting their centers and at distance <i><b>d</b></i> from each other. Let's measure an intensity of the electric field at any point between these disks on a center line at distance <i><b>h</b></i> from positively charged disk <nobr>(<i><b>0 ≤ h ≤ d</b></i>).</nobr><br /><br />The intensity of a combined field of two disks is a vector sum of intensities of all components of this field. The direction of the intensity of both fields is along the center line between them. The probe charge is <i><b>+1C</b></i>, so the positively charged disk will repel it, while negatively charged one will attract it. So, we can calculate the magnitude of each field and add them together to get the magnitude of the combined field.<br /><br />The density of the electric charge for these disks equals to a total charge <i><b>Q</b></i> divided by a surface area <i><b>A</b></i>. There are two opposite surfaces of each disk, but in case we have two close to each other parallel disks with opposite charges the electric charge of each disk (excess or deficiency of electrons) concentrates on a surface that is close to another disk with an opposite charge. So for each disk the absolute value of charge density is constant that equals to<br /><i><b>σ = Q/A = Q/(πR²)</b></i><br /><br />The magnitude of the field intensity from a positively charged disk at distance <i><b>h</b></i> from its surface is<br /><i><b>E(h) =<br />= </b></i>[<i><b>σ/(2ε<sub>r</sub>·ε<sub>0</sub>)</b></i>]<i><b>·</b></i>[<i><b>1−1/√<span style="text-decoration-line: overline;">1+R²/h² </span></b></i>]<br /><br />The magnitude of the field intensity from a negatively charged disk at distance <i><b>d−h</b></i> from its surface is <i><b>E(d−h)</b></i>.<br /><br />Just as an observation, let's notice that, in case the disks are very large and the distance between them very small, both <i><b>E(h)</b></i> and <i><b>E(d−h)</b></i> are approximately equal to<br /><i><b>E = σ/(2ε<sub>r</sub>·ε<sub>0</sub>)</b></i><br /><br />Since both vectors of intensity are directed from positive to negative disk (positively charged disk repels the probe charge of <i><b>+1C</b></i> positioned in-between the disks, and the negatively charged disk attracts it), the magnitude of the intensity of the combined electric field produced by both disks equals to<br /><i><b><span style="text-decoration-line: overline;">E</span>(h) = E(h)+E(d−h)</b></i><br />Expressed in all its details, the formula is quite large and difficult to analyze.<br /><br />At this point we will do what physicists usually do with a cumbersome formula - assume that in practice certain really small values can be assumed as infinitely small and certain large values to be infinitely large.<br />Indeed, if <i><b>h→0</b></i> or <i><b>h→d</b></i> the assumption above is correct. Representing graphically intensity <i><b>E</b></i> as a function of <i><b>h</b></i> for <i><b>R=10</b></i> and <i><b>d=0.1</b></i> on a segment <nobr><i><b>0 ≤ h ≤ d</b></i></nobr> shows hardly visible rise in the middle of a segment with <i><b>h=d/2</b></i>.<br /><br />So, for practical reasons we will assume that the distance between the disks <i><b>d</b></i> is very small, while the radius of disks <i><b>R</b></i> is very large. Since the point we measure the intensity is between the disks at the distance <i><b>h</b></i> from one of them, variables <i><b>h</b></i> and <i><b>d−h</b></i> can also be assumed as very small.<br />This assumption leads to consider the values <i><b>R²/h²</b></i> and <i><b>R²/(d−h)²</b></i> as infinitely large, which result in the following approximate formula for the intensity of a combined field between the disks:<br /><i><b><span style="text-decoration-line: overline;">E</span> ≅ σ/(2ε<sub>r</sub>·ε<sub>0</sub>) + σ/(2ε<sub>r</sub>·ε<sub>0</sub>) =<br />= σ/(ε<sub>r</sub>·ε<sub>0</sub>)</b></i><br />So, approximately, the electric field between the large parallel disks on a small distance from each other is uniform and depends only on the density of electric charge on the disks <i><b>σ</b></i> and the dielectric constant of the media between them <i><b>ε<sub>r</sub></b></i>.<br /><br />Let's analyze a process of discharging of electricity between two disks.<br />Since our purpose is to store the charges and to avoid discharge, we should know how much electricity <i><b>Q</b></i> we can store in these two disks before electrons jump from a negatively charged disk to a positively charged because of the force of attraction.<br /><br />The discharge will be more difficult if the force of attraction between the disks is less. The force of attraction is characterized by the intensity of the field between the disks. Knowing intensity of the field between the disks <i><b><span style="text-decoration-line: overline;">E</span></b></i>, which is the force acting on a unit charge, we can calculate the work needed by a charge of <i><b>+1C</b></i> to overcome a distance <i><b>d</b></i> between the disks by multiplying the intensity (force) by the distance.<br /><br />The work needed to discharge <i><b>+1C</b></i>, that is the work needed to move <i><b>+1C</b></i> of charge from one disk to another (the <i>voltage</i> between the disks) is<br /><i><b>V = <span style="text-decoration-line: overline;">E</span>·d = σ·d/(ε<sub>r</sub>·ε<sub>0</sub>)</b></i><br /><br />Let's recall that the purpose of our work is to store as much electric charge in these two disks as possible. But with growing charge <i><b>Q</b></i> proportionally grows the density of electricity <i><b>σ</b></i> and proportionally grows the <i>intensity</i> of the field and the <i>voltage</i> between the disks.<br /><br />Let's introduce the new characteristic that defines the ability of our two disk construction, called <b>capacitor</b>, to hold electric charge - <i>capacity</i> <i><b>C = Q/V</b></i>.<br />Defined as such, the <i>capacity</i> of a <b>capacitor</b> described above equals to<br /><i><b>C = Q/V = σ·A/</b></i>[<i><b>σ·d/(ε<sub>r</sub>·ε<sub>0</sub>)</b></i>]<i><b> =<br />= A·(ε<sub>r</sub>·ε<sub>0</sub>)/d</b></i><br /><br />As we see, the <i>capacity</i> of our <b>capacitor</b> depends on three major factors:<br />(a) the area of the disks (<i>capacity</i> is proportional to this area)<br />(b) the distance between disks (<i>capacity</i> is inversely proportional to this area)<br />(c) the dielectric constant of the media between the disks (<i>capacity</i> is proportional to a dielectric constant of the media).<br /><br />To satisfy the purpose of storing as much electricity in the <b>capacitor</b> as possible, we should increase the area of the disks, decrease the distance between the disks and isolate disks from each other with the media with high dielectric constant.<br /><br />The disk shape of a capacitor is not really important, as long as it contains two flat parallel shapes on a small distance from each other. The easiest variation from the practical standpoint is two rectangles. Although different shape leads to slightly different results in the value of <i>capacity</i>, the difference is negligible, and it disappears completely when the distance between planes becomes an infinitesimal value. So, the definition of <i>capacity</i><br /><i><b>C = A·(ε<sub>r</sub>·ε<sub>0</sub>)/d</b></i><br />is quite universal and is widely used as a characteristic of a <b>capacitor</b>.<br /><br />Finally, if we adapt a rectangular form of a capacitor, it's easy to significantly increase the area <i><b>A</b></i> by folding or rolling both surfaces together. This technique is actually utilized in manufacturing capacitors of large capacity and it's called <i>parallel connection of capacitors</i>.<br />In such a way of connecting capacitors the total capacity of an assembly equals to a sum of capacities of individual capacitors<br /><i><b>C = C<sub>1</sub>+C<sub>2</sub>+C<sub>3</sub>+C<sub>4</sub>+...</b></i><br />This picture depicts an idea to stack smaller individual capacitors to enlarge the area, thus allowing to store more electric charge in a smaller volume.<br /><img src="http://www.unizor.com/Pictures/Capacitor.png" style="height: 200px; width: 200px;" />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0