*Notes to a video lecture on http://www.unizor.com*

__Problems on Self-Induction__

*Problem A*

Consider a wire of resistance

*bent into a circle of a radius*

**R**_{0}*with its ends connected to a battery producing a voltage*

**r***.*

**U**_{0}There is a switch that can turn the flow of electricity in this circuit ON or OFF during a short time period

*.*

**T**When it turns ON, the resistance of the circuit, as a function of time

*, changes from infinitely large value to the value of*

**R(t)***as*

**R**_{0}

**R(t) = R**_{0}·T/tWhen it turns OFF, the resistance changes from the value of

*to an infinitely large value as*

**R**_{0}

**R(t) = R**_{0}·T/(T−t)What is the EMF of self-induction

*during the current switching ON or OFF?*

**U**_{i}*Solution*

**Switching ON**According to the Ohm's Law, the electric current in the circuit is

**I(t) = U**_{0}/R(t) = U_{0}·t /(R_{0}·T)The magnetic field intensity inside a loop, as we discussed in the lecture on magnetism of a current in a loop, is

**B(t) = μ**_{0}·I(t) /(2r) = μ_{0}·U_{0}·t /(2r·R_{0}·T)The magnetic flux going through the wire loop of the radius

*is a product of the intensity of the magnetic field by the area of a loop*

**r**

**Φ(t) = B(t)·πr² = μ**_{0}·U_{0}·t·πr² /(2r·R_{0}·T) = πμ_{0}·U_{0}·t·r /(2R_{0}·T)The EMF of self-induction is the first derivative of the magnetic flux

by time with a minus sign (minus sign because the induced EMF works to

prevent the increase in the current)

**U**d_{i}= −**Φ/**d**t = −πμ**_{0}·U_{0}·r /(2R_{0}·T)It's negative, so it reduces the overall voltage in the circuit and,

therefore, reduces the electric current in it. At the end of switching

ON (at

*) the current will not reach its maximum value but will continue to rise after the switch is completely ON until it reaches it.*

**t=T**

**Switching OFF**According to the Ohm's Law, the electric current in the circuit is

**I(t) = U**_{0}/R(t) = U_{0}·(T−t) /(R_{0}·T)The magnetic field intensity inside a loop, as we discussed in the lecture on magnetism of a current in a loop, is

**B(t) = μ**_{0}·I(t) /(2r) = μ_{0}·U_{0}·(T−t) /(2r·R_{0}·T)The magnetic flux going through the wire loop of the radius

*is a product of the intensity of the magnetic field by the area of a loop*

**r**

**Φ(t) = B(t)·πr² = μ**_{0}·U_{0}·(T−t)·πr² /(2r·R_{0}·T) = πμ_{0}·U_{0}·(T−t)·r /(2R_{0}·T)The EMF of self-induction is the first derivative of the magnetic flux

by time with a minus sign (minus sign because the induced EMF works to

prevent the increase in the current)

**U**d_{i}= −**Φ/**d**t = πμ**_{0}·U_{0}·r /(2R_{0}·T)It's positive, so it adds to the overall voltage in the circuit and,

therefore, increases the electric current in it, which, potentially, can

be harmful for electronic devices on the circuit.

*Problem B*

In the same setting as in

*Problem A*calculate minimum time

*of switching a circuit OFF in order for the current in a circuit grow no more then 10% from the theoretical value*

**T**_{min}*,*

**I**_{0}= U_{0}/R_{0}*.*

**I**_{0}= U_{0}/R_{0}*Solution*

Induced EMF equals to (from

*Problem A*)

**U**_{i}= πμ_{0}·U_{0}·r /(2R_{0}·T)It's a constant and does not depend on time

*.*

**t**The resistance of a circuit is growing with time from

*at*

**R**_{0}*to infinity at*

**t=0***. Therefore, the maximum electric current will be observed at*

**t=T***, when the resistance is minimal, and will equal to*

**t=0**

**I**_{max}= (U_{0}+U_{i}) /R_{0}= I_{0}+ (U_{i}/R_{0})In order for this current not to exceed

*by 10% the following inequality must be satisfied*

**I**_{0}

**U**_{i}/R_{0}≤ 0.1·U_{0}/R_{0}The resistance cancels and the remaining inequality is

**U**_{i}≤ 0.1·U_{0}Using the value of

*calculated in*

**U**_{i}*Problem A*, we obtain the following inequality that should be resolved for time interval

**T**

**πμ**_{0}·U_{0}·r /(2R_{0}·T) ≤ 0.1·U_{0}Voltage

*cancels out and the remaining inequality for time interval*

**U**_{0}*is*

**T**

**T ≥ πμ**_{0}·r /(0.2R_{0})**Practical case**

Assume the following conditions

**μ**N/A²_{0}= 4π·10^{−7}

**r = 0.1**m

**R**Ω_{0}= 200Then the time interval of switching OFF must be at least

**T = πμ**sec_{0}·r /(0.2R_{0}) ≅ 10^{−8}That is a very short interval of time. Mechanical switches usually are

much slower, which means the electric current under normal usage in

homes will not significantly rise when we switch the electricity OFF.

**IMPORTANT:**

Check formula units

Check formula units

*[N·m → W → V·A·sec]*

**(N/A²)·m/Ω = N·m/(A²·Ω) =**

*[Ω → V/A]*

= (V·A·sec)/(A²·Ω) =

= (V·A·sec)/(A²·Ω) =

= (V·A·sec·A)/(A²·V) = sec= (V·A·sec·A)/(A²·V) = sec

as it is supposed to be.

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