Home Physics Electromagnetic Induction Self Induction, Emf, Induced Current The north pole of a bar magnet is moved swif…
Physics Electromagnetic Induction Self Induction, Emf, Induced Current Single Correct MCQ
Published on: September 12, 2026

The north pole of a bar magnet is moved swiftly downward towards a closed coil and then second time it is raised upwards slowly. The magnitude and direction of the induced currents in the two cases will be of

First case

Second case

(a)

Low value clockwise

Higher value anticlockwise

(b)

Low value clockwise

Equal value anticlockwise

(c)

Higher value clockwise

Low value clockwise

(d)

Higher value anticlockwise

Low value clockwise

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Text Solution

Verified by Experts
The correct answer is:
D
Step 1: According to Faraday's Law of Electromagnetic Induction, the induced electromotive force (emf) in a circuit is proportional to the rate of change of magnetic flux through the circuit.
Step 2: In the first case, when the north pole of the magnet is moved swiftly downward towards the coil, the magnetic flux through the coil increases rapidly. According to Lenz's Law, the direction of the induced current will oppose the change in flux; hence, the induced current will flow in a direction that creates a south pole at the end of the coil nearest to the magnet. This current will flow clockwise (CW) when viewing from the north pole of the magnet. As the motion is swift, the magnitude of the induced current will also be higher due to the rapid change in flux.
Step 3: In the second case, when the north pole is raised upwards slowly, the magnetic flux through the coil decreases gradually. Again applying Lenz's Law, the induced current will flow in a direction to oppose the decrease in flux, which means it will exhibit an anticlockwise (ACW) direction when viewed from the same perspective. Since the motion is slow, the induced current will have a lower magnitude compared to the first case.
Therefore, in summary, for the first case we have a higher value of current flowing clockwise, while for the second case we have a lower value of current flowing anticlockwise. Hence the correct option is (d): Higher value anticlockwise for the first case and lower value clockwise for the second case.

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