Question 15

Induced current is given by $i = -\frac{1}{R} \frac{d\Phi}{dt} = -\frac{B}{R} \frac{dA}{dt}$, where $A$ is the area of the loop in the region $x > 0$.

1. Loop (P): A semicircle rotating about its center. As it rotates clockwise, the area entering the field increases linearly for $180^\circ$ ($t = 0$ to $T/2$), then decreases linearly for the next $180^\circ$ ($t = T/2$ to $T$). This leads to a constant positive current for $T/2$ followed by a constant negative current for $T/2$. This matches Graph (3).

2. Loop (Q): A $60^\circ$ sector. It enters the field in $T/6$, remains entirely in the $x > 0$ region for $120^\circ$ ($T/3$ duration, where flux is constant), then leaves for $T/6$. This generates pulses of current with intervals of zero current. Given its starting position, it matches Graph (2).

3. Loop (R): A $60^\circ$ sector with a different initial orientation. Similar to (Q), it generates pulses with zero intervals. Based on the provided options, it matches Graph (1).

4. Loop (S): Two opposite $60^\circ$ sectors. By symmetry, for any part of one sector entering the field ($x > 0$), an equivalent part of the other sector is leaving the field. Consequently, the net area in the field remains constant ($dA/dt = 0$) throughout the rotation. Thus, the induced current is zero. This matches Graph (4).

Comparing matches P-3, Q-2, R-1, and S-4 with the given options, we find that option (C) is the correct match.

Correct Option: (C)