A conducting square loop initially lies in the XZ plane with its lower edge hinged along the X -axis. Only in the region y ≥ 0 , there is a time dependent magnetic field pointing along the Z -directio

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, $B (t) = B_{0} (cos ω t) \hat{K}$ , 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:

Question diagram: A conducting square loop initially lies in the XZ plane with

Options:

Answer:

A

Solution:

$ϕ = B_{0} cos ω tA sin ω t = \frac{B _{0} A sin 2 ω t}{2} ε = - \frac{d ϕ}{dt} = - B_{0} A cos 2 ω t = (0 \leq t \leq \frac{π}{ω}) ε = 0 (\frac{π}{ω} \leq t \leq \frac{2 π}{ω})$ Ans. Option (A)

TestHub · Accepted Answer

​ ϕ = B 0 ​ cos ω tA sin ω t = 2 B 0 ​ A sin 2 ω t ​ ε = − dt d ϕ ​ = − B 0 ​ A cos 2 ω t = ( 0 ≤ t ≤ ω π ​ ) ε = 0 ( ω π ​ ≤ t ≤ ω 2 π ​ ) ​ Ans. Option (A)

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