“There is no democracy in physics. We can't
say that some second-rate guy has as much right to an opinion as Fermi.”

–Luis Walter Alvarez

Today we shall
discuss a few questions involving electromagnetic induction. Questions in this
section are generally interesting and we have discussed many typical questions
on various occasions on this site.

(1) A single turn
plane circular conducting loop of area

*A*and resistance*R*is placed in a uniform magnetic field of flux density*B*which has a time rate of change. The plane of the loop is perpendicular to the magnetic field. If the emf induced in the loop is*V*, the time rate of change of the magnetic flux density is
(a)

*V/A*
(b)

*V/RA*
(c)

*R**V/A*
(d)

*A/V*
(e)

*AR/V*
The induced emf

*V*is given by*V =*

*∆*

*Φ/*

*∆*

*t*where

*∆*

*Φ*is the change of magnetic flux occurring in a small time

*∆*

*t*.

Since

*Φ = BA*we have*V =*

*∆*

*BA/*

*∆*

*t*

The time rate of
change of the magnetic flux density is

*∆**B/**∆**t*=*V/A*
(2) A long straight
power line carries a current

*I*which*decreases*with time at a uniform rate. A plane circular conducting loop is arranged below the power line as shown in the figure. Which one among the following statements is true?
(b) A uniformly decreasing current is induced in
the loop.

(c) A uniformly increasing current is induced in
the loop.

(d) A steady current is induced in the loop and
it flows in the anticlockwise direction

(e) A steady current is induced in the loop and
it flows in the clockwise direction

Since the current in
the power line is changing, the magnetic flux linked with the circular loop is
changing. Therefore there must be an induced current in the loop. The induced
current in the loop must be steady since the rate of decrease of magnetic flux
is steady (because of the uniform decrease of current in the power line).

The magnetic field
lines produced by the current in the power line are directed normally into the
plane of the loop. Since the current in the power line

*decreases*with time, the induced current in the loop must supply magnetic flux lines in the*same*direction, in accordance with Lenz’s law (for opposing the reduction of the flux). Therefore the induced current in the circular loop must flow in the*clockwise*direction [Option (e)].
The following
questions are meant for AP Physics C aspirants:

(3) Two horizontal
conducting rails AB and CD of negligible resistance are connected by a
conductor BC of resistance R. Another conducting rod PQ of length L and
negligible resistance can slide without friction along the rails (Fig.). The
plane ABCD is horizontal and a constant magnetic field B tesla acts
perpendicular to the plane ABCD. A small constant horizontal force F is applied
on the slider PQ perpendicular to its length so that it slides with a constant
velocity ‘v’. What is the value of the velocity v?

(a) FR/BL

(b) FR/B

^{2}L^{2}
(c) FR/B

^{2}L
(d) FR/BL

^{2}
(e) FR

^{2}/B^{2}L^{2}
On applying the force F, the rod PQ starts to
move from rest with an acceleration. When the rod moves the magnetic flux
linked with the circuit PBCQ changes and an emf is induced in the circuit.
Obviously this is the motional emf BLv.

[Note that when a conductor of length L moves
with velocity ‘v’ at right angles to a magnetic field of flux density B, the
motional voltage generated between its ends is BLv].

Since we have a

*closed*circuit PBCQ, the emf BLv drives a current ‘I’ through it. PQ is therefore a current carrying conductor moving at right angles to a magnetic field. A magnetic force ILB acts opposite to the direction of motion of the conductor (in accordance with Lenz’s law). The opposing magnetic force goes on increasing with the increase in velocity of the conductor until the magnitude of the magnetic force becomes equal to that of the applied force F. The conductor thereafter continues to move with the terminal velocity acquired by it. The velocity of the rod after the initial accelerated motion is now constant.
Equating the
magnitudes of the applied force F and the magnetic force ILB we have

But I = BLv/R

Therefore F = B

^{2}L^{2}v/R
This gives v = FR/B

^{2}L^{2}
(4) An inductance L
and a resistance R are connected in series with a battery and switch S as shown
in the figure. The switch is closed at time

*t =*0. Which one among the following graphs gives the variation of the voltage*V*across the inductance as a function of time_{L}*t*?
There will be a
voltage drop across the inductance
only if the current in it changes. When the switch is closed the current in the
series LR circuit will rise rapidly initially and will finally settle at the
final maximum value.

[The final maximum
current

*I*_{0}in the LR circuit is*V*_{0}/R where*V*_{0}is the emf of the battery. The current*I*in the LR circuit during the growth at any instant*t*is given by*I = I*

_{0}[1 – e

^{–Rt/L}] where e is the base of natural logarithms]

Since the

*rate of**variation*of current_{ }is maximum initially, the voltage induced in the inductance is maximum initially. The*rate of variation*of current is non-linear and finally becomes zero. Therefore, the emf induced in the inductance finally becomes zero and the variation of the voltage*V*across the inductance as a function of time_{L}*t*is correctly represented by graph (b).
[Note that graph (d)
is incorrect since the variation of the voltage
is linear].

You can access all
posts on electromagnetic induction on this
site by clicking on the label ‘electromagnetic induction’ below this
post.

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