Match Rotating Conducting Loops with Induced Current Variation

To determine the correct matching between List-I and List-II, we apply Faraday's Law of Electromagnetic Induction and Lenz's Law to analyze the rate of change of magnetic flux through each rotating loop.

General Mathematical Principle:

The uniform magnetic field $\vec{B} = B\hat{k}$ exists in the region $x > 0$. Each loop rotates clockwise in the $XY$ plane about the origin $O$ with constant angular speed $\omega = \frac{2\pi}{T}$.

The magnetic flux $\Phi(t)$ linked with a conducting loop (or segment) inside $x > 0$ at time $t$ is: $\Phi(t) = B A(t)$

where $A(t)$ is the area of the loop currently inside the magnetic field region $x > 0$.

By Faraday's Law, the induced electromotive force (EMF) is: $\mathcal{E}(t) = -\frac{d\Phi}{dt} = -B \frac{dA}{dt}$

The induced current $i(t) = \frac{\mathcal{E}(t)}{R_{\text{loop}}}$ is directly proportional to the rate at which area enters or leaves the magnetic field region: $i(t) \propto -\frac{dA}{dt}$

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Step-by-Step Matching:

1. Loop (P):

   • The loop is a semicircle of radius $R$ centered at $O$.
   • For $0 < t < T/2$, as the semicircle enters $x > 0$, area increases at a constant rate: $\frac{dA}{dt} = \frac{1}{2} R^2 \omega = \text{constant} > 0 \implies i(t) = \text{constant} > 0$
   • For $T/2 < t < T$, as the semicircle exits $x > 0$, area decreases at a constant rate: $\frac{dA}{dt} = -\frac{1}{2} R^2 \omega = \text{constant} < 0 \implies i(t) = \text{constant} < 0$
   • This step-function behavior matches Graph (3).
   • $\text{P} \rightarrow 3$

2. Loop (Q):

   • The loop consists of two $60^\circ$ sectors separated by a $60^\circ$ angular gap.
   • During the first half-period ($0 < t < T/2$):
     • The first $60^\circ$ sector enters $x > 0$ during $0 < t < T/6$, so $\frac{dA}{dt} > 0 \implies i(t) > 0$.
     • The $60^\circ$ gap passes through the boundary during $T/6 < t < 2T/6$, so $\frac{dA}{dt} = 0 \implies i(t) = 0$.
     • The second $60^\circ$ sector enters during $2T/6 < t < 3T/6$, so $\frac{dA}{dt} > 0 \implies i(t) > 0$.
   • During the second half-period ($T/2 < t < T$), an identical pattern of negative current pulses occurs as the sectors exit $x > 0$.
   • This pulse pattern matches Graph (2).
   • $\text{Q} \rightarrow 2$

3. Loop (R):

   • The loop is a single sector of angle $60^\circ$.
   • For $0 < t < T/6$: The sector enters $x > 0$, so $i(t) = \text{constant} > 0$.
   • For $T/6 < t < T/2$: The entire sector is inside $x > 0$, so flux is constant and $i(t) = 0$.
   • For $T/2 < t < T/2 + T/6$: The sector exits $x > 0$, so $i(t) = \text{constant} < 0$.
   • For $T/2 + T/6 < t < T$: The sector is entirely outside $x > 0$, so $i(t) = 0$.
   • This matches Graph (1).
   • $\text{R} \rightarrow 1$

4. Loop (S):

   • The loop is a figure-eight ("bowtie") configuration consisting of two identical $60^\circ$ sectors joined symmetrically at $O$.
   • Due to the crossover at $O$, the current flows in opposite directions in the two lobes.
   • As one sector enters the region $x > 0$, the opposite sector leaves $x > 0$ at the exact same rate.
   • Consequently, the net rate of change of magnetic flux through the total loop is zero at all times: $\frac{d\Phi}{dt} = 0 \implies i(t) = 0 \quad \text{for all } t \in [0, T]$
   • This matches Graph (4).
   • $\text{S} \rightarrow 4$

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Conclusion:

$\text{P} \rightarrow 3, \quad \text{Q} \rightarrow 2, \quad \text{R} \rightarrow 1, \quad \text{S} \rightarrow 4$

Thus, the correct option is C.