Question

A wire frame of area $3.92\times {10}^{-4}{\text{m}}^2$ and resistance $20\Omega$ is suspended freely from a $0.392\text{m}$ long thread. There is a uniform magnetic field of $0.784\text{T}$ and the plane of wire frame is perpendicular to the magnetic field. The frame is made to oscillate under gravity by displacing it through $2\times {10}^{-2}\text{m}$ from its initial position along the direction of magnetic field. The plane of the frame is always along the direction of thread and does not rotate about it.

• A. Induced emf as a function of time is $3\times {10}^{-6}\sin 10t$.
• B. Induced emf as a function of time is $2\times {10}^{-6}\sin 10t$.
• C. Maximum current in the frame is ${10}^{-7}\text{A}$.
• D. Maximum current in the frame is ${10}^{-9}\text{A}$.

Answer 1

Answer: (B), (C)

Maximum current in the frame is ${10}^{-7}\text{A}$.

At the instant, when the thread makes an angle $\theta$ with the vertical, the magnetic flux through the frame,

$\varphi =BA\cos \theta$


The induced emf,

$E=-\frac{d\varphi }{dt}=BA\sin \theta \frac{d\theta }{dt}.......\left(1\right)$

The restoring torque on the frame,

$\tau =-mg\left(l\sin \theta \right)$

For small $\theta ,\sin \theta \approx \theta$

$\frac{d^2\theta }{d{t}^2}=\frac{mgl}{I}(-\theta )(\because \tau =I\alpha =I\frac{d^2\theta }{d{t}^2})$

$\frac{d^2\theta }{d{t}^2}=\frac{mgl}{m{l}^2}(-\theta )=\frac{g}{l}(-\theta )$

Putting, $\frac{g}{l}={\omega }^2$, we get

$\frac{d^2\theta }{d{t}^2}=-{\omega }^2\theta ..........\left(2\right)$

This represents $\text{SHM}$, and so assuming variation in $\theta$ as,

$\theta ={\theta }_0\sin \omega t$

Substituting the value of $\theta$ in equation $\left(1\right)$, we get

$E=BA\left({\theta }_0\sin \omega t\right)\frac{d}{dt}\left({\theta }_0\sin \omega t\right)(\because \sin \theta \approx \theta )$

$=BA{\theta }_0\sin \omega t\left({\theta }_0\omega \cos \omega t\right)$

$\therefore E=\frac{1}{2}BA\omega {\theta }_0^2\sin 2\omega t$

Here,

$\omega =\sqrt{\frac{g}{l}}=\sqrt{\frac{9.8}{0.392}}=5\text{rad/s}$

${\theta }_0=\frac{x_0}{l}=\frac{2\times {10}^{-2}}{0.392}=0.05\text{rad}$

So,

$E=\frac{1}{2}\times \left(0.784\right)\times (3.92\times {10}^{-4})\times 5\times {0.05}^2\times \sin (2\times 5\times t)$

$E=2\times {10}^{-6}\sin 10t$

Therefore, option $\left(B\right)$ is correct.

Now, maximum induced emf,

$E_{max}=2\times {10}^{-6}\text{V}$

So, the maximum induced current is,

$i_{max}=\frac{E_{max}}{R}=\frac{2\times {10}^{-6}}{20}={10}^{-7}\text{A}$

Hence, $\left(C\right)$ is also correct.

Why this question? This question is given to use the concept of $SHM$ in electro magnetic induction.