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## Saturday, December 8, 2007

### AP Physics B & C - Multiple Choice Practice Questions on Electromagnetic Induction

"Live as if you were to die tomorrow. Learn as if you were to live forever."
– Mahatma Gandhi
Let us discuss some multiple choice questions on electromagnetic induction.
Questions (1) and (2) are meant for AP Physics B as well as AP Physics C aspirants where as question (3) is specifically meant for AP Physics C aspirants.
(1) A search coil (a small plane coil of a few turns of insulated copper wire) of area A and negligible resistance has N turns in it. It is kept between the poles of an electromagnet so that the plane of the coil is perpendicular to the magnetic field B produced by the electromagnet. This search coil is connected in series with a resistance R to form a closed circuit. When the current in the electromagnet is reversed, the charge flowing through R is
(a) NAB/R (b) zero (c) NAB (d) 2NAB (e) 2NAB/R
The initial magnetic flux through the search coil is NAB. Since the current is reversed, the final flux is –NAB. Therefore, the change of flux (dФ) is NAB – (–NAB) = 2NAB.
Since the charge, Q = dФ/R, we obtain Q = 2NAB/R.
Suppose the current is not reversed, but instead, the search coil is rotated through 180º. In this case also the flux change will be 2NAB and the charge flowing will be 2NAB/R.
[Reversing the current through the coil of an electromagnet is not a desirable practice. The emf induced in the coil will be very large in the case of large electromagnets].

You may be asked to find the emf induced (instead of the charge) in the coil. In that case, the time in which the search coil is rotated through 180º is to be given. If the time is ‘t’ the induced emf will be 2NAB/t.

(2) Two identical parallel straight conductors P and Q are placed symmetrically on two smooth conducting rails R1 and R2 as shown. A uniform magnetic field exists in the region. The direction of the magnetic field is perpendicular to the plane of the figure and is inwards, as shown (by the arrow tails). If the conductor P is moved along the rails towards Q, keeping it perpendicular to the rails, then the conductor Q will
(a) remain stationary
(b) move towards P
(c) move away from P
(d) rotate in a clockwise direction
(e) rotate in an anticlockwise direction
You can easily find the correct option (c) by applying Lenz’s law. When the conductor P is moved towards Q, the magnetic flux linked with the circuit formed by the rails and the conductors P and Q is decreased since the area of the circuit is decreased. This decrement in magnetic flux is to be opposed in accordance with Lenz’s law. So, the conductor Q has to move away from P, thereby increasing the area of the circuit.
The above explanation is enough for finding the answer to the question. But it is better to see how the movement of the conductor Q happens:
Because of the flux change, an induced current flows in the circuit. The direction of the induced current through the conductor Q results in a magnetic force on Q. Since the force has to be away from P as demanded by Lenz’s law, the current in the circuit has to flow in the clockwise direction (as given by Fleming’s left hand rule). If the direction of the magnetic field is opposite, the correct option will still be (c). But, the current flowing in the circuit will be anticlockwise.
(3) A time varying magnetic field given by B(t) = 2t2 – 4t +2 exists in a region where a plane coil of area 0.2 m2 having 20 turns is placed with its plane perpendicular to the field. The induced emf in the coil at t = 2 s is
(a) zero (b) ) 0.8 V (c) 1.6 V (d) 16 V (e) 160 V
The induced emf is dФ/dt. Here Ф = NAB = 20×0.2(2t2 – 4t +2) = 4(2t2 – 4t +2)
Therefore, dФ/dt = 4(4t – 4) = 16t – 16.
Substituting for t = 2s, dФ/dt = 16 volt.
You will find more questions (with solution) on electromagnetic induction here
as well as here at physicsplus