A bar magnet is pushed into a closed coil at a steady speed. Each time it starts from the same position and stops at the same position inside the coil. The motion takes 0.4 s the first time and 0.1 s the second time. The ratio of the heat produced in the coil the second time to the heat produced the first time is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4
The induced current heats the coil. The magnet starts and stops at the same places, so the change of flux is the same both times. Only the time taken is different.
Let \( \Delta\Phi \) be the change of flux through each turn and \( R \) the resistance of the coil. In time \( t \), the emf and the current are
\( \begin{aligned} \varepsilon &= \frac{N\Delta\Phi}{t} \\ I &= \frac{N\Delta\Phi}{Rt} \end{aligned} \)
The heat produced is
\( \begin{aligned} H &= I^{2}Rt \\ &= \left(\frac{N\Delta\Phi}{Rt}\right)^{2}Rt \\ &= \frac{(N\Delta\Phi)^{2}}{Rt} \end{aligned} \)
So \( H \) is inversely proportional to \( t \).
\( \begin{aligned} \frac{H_2}{H_1} &= \frac{t_1}{t_2} \\ &= \frac{0.4}{0.1} \\ &= 4 \end{aligned} \)
Therefore, the faster push produces more heat. The person also has to do more work to push the magnet in.
NCERT (2026-27):
Faraday’s Law of Induction – “The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.”
Lenz’s Law and Conservation of Energy – “… In this situation, the bar magnet experiences a repulsive force due to the induced current. Therefore, a person has to do work in moving the magnet. … This energy is dissipated by Joule heating produced by the induced current.”