A conducting square loop initially lies in the XZ plane with its lower edge hinged along the X-axis. Only in the region , there is a time dependent magnetic field pointing along the Z-direction, where is a constant. The magnetic field is zero everywher

Question

A conducting square loop initially lies in the XZ 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, $\vec{B} (t) = B_{0} (\cos ω t) \hat{k},$

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? [2025]

Smart Achievers · Accepted Answer

(1)

Magnetic flux

$ϕ = B A \cos (90 ° - ω t) = (B_{0} \cos ω t) . A . \sin ω t$

$= \frac{B_{0} A}{2} \sin 2 ω t$

Induced emf, $e = - \frac{d ϕ}{d t}$

$= - \frac{B_{0} A}{2} (2 ω) \cos 2 ω t \propto - \cos 2 ω t,$ $ω_{eq} = 2 ω$

so, $T_{eq} = \frac{2 π}{ω_{eq}} = \frac{2 π}{2 ω} = \frac{π}{ω}$

Therefore, during the half revolution, there will be one cycle of emf. During the second half revolution, the loop will be out of the magnetic field, so the emf will remain zero.

Solution

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