1. Physical situation

A bar magnet moves along the axis of a conducting loop and just fits through it. As the magnet approaches, enters, passes the center, and moves away, the magnetic flux through the loop changes with time.

By Faraday’s law,

So the induced e.m.f. depends on how the flux through the loop changes.

2. Nature of flux change

• When the magnet is far away, flux through the loop is very small and changes very little. $\Rightarrow \mathcal{E} \approx 0$

• As the magnet approaches the loop, the magnetic flux through the loop increases. Hence, $\frac{d\Phi_B}{dt} > 0 \quad \Rightarrow \quad \mathcal{E} < 0$ (taking the sign convention from Faraday’s law)

• When the magnet is symmetrically placed at the center of the loop, flux is maximum, so its rate of change is zero. $\frac{d\Phi_B}{dt} = 0 \Rightarrow \mathcal{E}=0$

• As the magnet moves out of the loop, the flux decreases. $\frac{d\Phi_B}{dt} < 0 \Rightarrow \quad \mathcal{E} > 0$

• When the magnet is again far away, flux change becomes negligible. $\Rightarrow \mathcal{E} \approx 0$

So the e.m.f. graph must have:

• one pulse of one sign while entering,
• zero at the middle,
• one pulse of opposite sign while leaving.

3. Relative magnitudes of the two pulses

Since the magnet just passes through the loop, the flux changes very rapidly near entry and exit. The graph is approximately two opposite peaks of similar shape, separated by a zero crossing at the center.

Thus the correct graph is the one showing:

• zero initially,
• a peak of one sign,
• crossing through zero,
• a peak of opposite sign,
• then back to zero.

4. Matching with options

This standard induced e.m.f. vs time graph corresponds to Option C.

5. Comparison with stored answer

Derived answer: C
Stored correct answer: C

They match.