Question 28: Electromagnetic Induction

Question

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, $\vec{B}(t)=B_0(\cos \omega t) \hat{k}$, 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:

Accepted Answer

Here, we will use the magnetic flux formula and Faraday's law. Faraday's law relates the induced EMF to the rate of change of magnetic flux, which is appropriate for a loop in a changing magnetic field and changing orientation.
magnetic flux formula: $\phi=\vec{B} \cdot \vec{A}$
$$ \Rightarrow \phi=B A \cos 	heta $$
and the induced EMF is given by Faraday's law: $\quad v=\frac{-d \phi}{d t}$
where, $\phi=$ magnetic flux through the loop
$B=$ magnetic field
$A=$ Area of the loop
$	heta=$ Angle between a perpendicular vector to the area and the magnetic field.
Given: $\quad \vec{B}(t)=B_0(\cos \omega t) \hat{k}$
$$ 	heta=\omega t $$
we know, $\quad \phi=\vec{B} \cdot \vec{A}$
$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& =B A \cos (90-	heta) \ \phi & =B A \sin 	heta\end{aligned}$
$$ \begin{aligned} &\begin{aligned} \Rightarrow \phi & =B_0 \cos (\omega t) A \sin (\omega t) \ \Rightarrow \phi & =\frac{B_0 A}{2} \sin (2 \omega t) \ \varepsilon=v & =\frac{-d \phi}{d t}=\frac{-d}{d t}\left[\frac{B_0 A}{2} \sin (2 \omega t)ight] \ & =\frac{-B_0 A}{2}(2 \omega) \cos (2 \omega t) \ \Rightarrow \varepsilon & =V=-B_0 \omega A \cos (2 \omega t) \end{aligned}\ &	ext { (only for half rotation) } \end{aligned} $$
cos function with time period $=\frac{2 \pi}{2 w}$
$$ =\frac{\pi}{w} $$
Hence, option (A) is correct.

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Electromagnetic Induction

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