JEE Advanced 2025 Paper 1 Q03 Physics Electrostatics & Circuits EM Induction Medium

JEE Advanced 2025 Paper 1 · Q03 · EM Induction

A conducting square loop initially lies in the $XZ$ plane with its lower edge hinged along the $X$-axis. Only in the region $y \ge 0$, there is a time dependent magnetic field pointing along the $Z$-direction, $\vec{B}(t) = B_0(\cos\omega t)\hat{k}$, where $B_0$ is a constant. The magnetic field is zero everywhere else. At time $t = 0$, the loop starts rotating with constant angular speed $\omega$ about the $X$-axis in the clockwise direction as viewed from the $+X$ axis (as shown in the figure above). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. ($V$) in the loop as a function of time?

  1. A. See option A in figure
  2. B. See option B in figure
  3. C. See option C in figure
  4. D. See option D in figure
Reveal answer + step-by-step solution

Correct answer:A

Solution

Let the loop side be $a$. Define $\theta(t)=\omega t$ as the angle of the loop with the $XZ$ plane. The loop is in the region $y\ge 0$ (where $\vec B$ exists) only during $0\le \omega t\le \pi$; on the second half of each rotation the loop is in $y<0$ where $\vec B = 0$, so $\Phi = 0$ and $V=0$.

While the loop is inside the field region, the projected area normal to $\hat k$ is $A_\perp = a^2\sin(\omega t)$, so $$\Phi(t) = B_0\cos(\omega t)\,a^2\sin(\omega t) = \tfrac{a^2 B_0}{2}\sin(2\omega t).$$ Hence $V = -d\Phi/dt = -a^2 B_0\,\omega\cos(2\omega t)$ for $0<\omega t<\pi$, and $V=0$ for $\pi<\omega t<2\pi$. The plot is a full $\cos$-like oscillation with period $\pi/\omega$ during the first half, then flat zero for the second half, repeating with period $2\pi/\omega$. This matches plot (A).

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →