JEE Advanced 2025 — Paper 1 — PHY-I solution

JEE Advanced2025Paper 1PHY-I

A conducting square loop initially lies in the $X Z$ plane with its lower edge hinged along the $X$ -axis. Only in the region $y \geq 0$ , there is a time dependent magnetic field pointing along the $Z$ -direction, $B (t) = B_{0} (cos ω t) \hat{K}$ , 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 $ω$ about the $X$ axis in the clockwise direction as viewed from the $+ X$ axis (as shown in the figure). 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:

Solution

While the loop is in the region $y \geq 0$ (during half the rotation period), the area vector of the loop makes angle $ω t$ with $\hat{Z}$ , so the flux through it is:

$ϕ (t) = B_{0} cos (ω t) \cdot A sin (ω t) = \frac{B _{0} A}{2} sin (2 ω t)$ , valid for $0 \leq t \leq π / ω$ .

The induced EMF in this interval is:

$ε = - \frac{d ϕ}{d t} = - B_{0} A ω cos (2 ω t)$

During $π / ω \leq t \leq 2 π / ω$ the loop is in the region $y < 0$ where $B = 0$ , so $ε = 0$ .

The pattern then repeats. This matches option (A): one full cosine cycle of EMF during each half-rotation, separated by flat zero intervals.

← Back to full paper

Practice more PHY-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →