## Thursday, July 16, 2020

### Problems on Self-Induction: UNIZOR.COM - Physics4Teens - Electromagnetis...

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

Problems on Self-Induction

Problem A

Consider a wire of resistance R0 bent into a circle of a radius r with its ends connected to a battery producing a voltage U0.

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 R(t), changes from infinitely large value to the value of R0 as

R(t) = R0·T/t

When it turns OFF, the resistance changes from the value of R0 to an infinitely large value as

R(t) = R0·T/(T−t)

What is the EMF of self-induction Ui during the current switching ON or OFF?

Solution

Switching ON

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

I(t) = U0 /R(t) = U0·t /(R0·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) = μ0·I(t) /(2r) = μ0·U0·t /(2r·R0·T)

The magnetic flux going through the wire loop of the radius r is a product of the intensity of the magnetic field by the area of a loop

Φ(t) = B(t)·πr² = μ0·U0·t·πr² /(2r·R0·T) = πμ0·U0·t·r /(2R0·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)

Ui = −dΦ/dt = −πμ0·U0·r /(2R0·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 t=T) the current will not reach its maximum value but will continue to rise after the switch is completely ON until it reaches it.

Switching OFF

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

I(t) = U0 /R(t) = U0·(T−t) /(R0·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) = μ0·I(t) /(2r) = μ0·U0·(T−t) /(2r·R0·T)

The magnetic flux going through the wire loop of the radius r is a product of the intensity of the magnetic field by the area of a loop

Φ(t) = B(t)·πr² = μ0·U0·(T−t)·πr² /(2r·R0·T) = πμ0·U0·(T−t)·r /(2R0·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)

Ui = −dΦ/dt = πμ0·U0·r /(2R0·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 Tmin of switching a circuit OFF in order for the current in a circuit grow no more then 10% from the theoretical value

I0 = U0 /R0,I0 = U0 /R0.

Solution

Induced EMF equals to (from Problem A)

Ui = πμ0·U0·r /(2R0·T)

It's a constant and does not depend on time t.

The resistance of a circuit is growing with time from R0 at t=0 to infinity at t=T. Therefore, the maximum electric current will be observed at t=0, when the resistance is minimal, and will equal to

Imax = (U0+Ui) /R0 = I0 + (Ui /R0)

In order for this current not to exceed I0 by 10% the following inequality must be satisfied

Ui /R0 ≤ 0.1·U0 /R0

The resistance cancels and the remaining inequality is

Ui ≤ 0.1·U0

Using the value of Ui calculated in Problem A, we obtain the following inequality that should be resolved for time interval T

πμ0·U0·r /(2R0·T) ≤ 0.1·U0

Voltage U0 cancels out and the remaining inequality for time interval T is

T ≥ πμ0·r /(0.2R0)

Practical case

Assume the following conditions

μ0 = 4π·10−7 N/A²

r = 0.1 m

R0 = 200 Ω

Then the time interval of switching OFF must be at least

T = πμ0·r /(0.2R0) ≅ 10−8 sec

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

(N/A²)·m/Ω = N·m/(A²·Ω) =

[N·m → W → V·A·sec]

= (V·A·sec)/(A²·Ω) =

[Ω → V/A]

= (V·A·sec·A)/(A²·V) = sec

as it is supposed to be.