JEE Advanced 2025 · Paper 1 · Physics Q3

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PhysicsSingle correct+3 / -1Question 19 of 48

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). 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:

3D axes: Z up, Y to the right, X out of the page towards the lower left. A square loop hinged along the X-axis is shown tilted from the XZ plane towards +Y by angle ωt (dashed outline marks its initial position in the XZ plane). An upward arrow labelled B(t) shows the field along Z in the region y ≥ 0.

V vs t: from t = 0 to π/ω, V is one full cosine-shaped cycle starting at its negative extreme, rising through zero to a positive peak and back to the negative extreme at π/ω; from π/ω to 2π/ω, V = 0 (flat line on the axis, with vertical jumps at both ends); from 2π/ω to 3π/ω the same cycle repeats.

V vs t: a single sine-like wave over 0 to 2π/ω: V starts at 0, goes negative to a minimum near π/(2ω), crosses zero at π/ω, rises to a positive maximum and returns to 0 at 2π/ω.

V vs t: a single inverted-cosine wave over 0 to 2π/ω: V starts at its negative extreme at t = 0, crosses zero, peaks positive at π/ω, crosses zero again and returns to the negative extreme at 2π/ω.

V vs t: from 0 to π/ω, V is one full sine cycle starting at 0, going negative first, then positive, ending at 0 at π/ω; from π/ω to 2π/ω, V = 0; from 2π/ω to 3π/ω the same sine cycle repeats.

Answer and worked solution

Official answer: A

Where the loop is. The loop starts in the $XZ$ plane and rotates towards $+Y$ . For $0<\omega t<\pi$ it lies in $y\ge0$ ; for $\pi<\omega t<2\pi$ it lies in $y\le0$ , where there is no field.

Flux while in the field. The loop makes angle $\omega t$ with the $Z$ axis, so its area projected normal to $\hat k$ is $A\sin\omega t$ :
$\Phi=B_0\cos\omega t\cdot A\sin\omega t=\frac{B_0A}{2}\sin2\omega t.$

emf.
$V=-\frac{d\Phi}{dt}=-B_0A\omega\cos2\omega t.$

It starts at its negative extreme, goes through a positive peak at $t=\frac{\pi}{2\omega}$ , and is back at the negative extreme at $\frac\pi\omega$ .
It is zero from $\frac\pi\omega$ to $\frac{2\pi}\omega$ .
It then repeats.

This is plot (A).

Topics:Faraday's and Lenz's laws Motional EMF

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