Question

A closed loop of wire consists of a pair of equal semicircles, of radius $3.7\text{cm}$, lying in mutually perpendicular planes. The loop was formed by folding a plane circular loop along a diameter until the two halves became perpendicular. A uniform magnetic field $\overrightarrow{B}$ of magnitude $76\text{mT}$ is directed perpendicular to the fold diameter and makes equal angles ${45}^{\circ }$ with the planes of the semicircles as shown in fig. The magnetic field is reduced to zero at a uniform rate during a time interval of $4.5\text{ms}$. Determine the magnitude of the induced emf in the loop during this interval.

• A. $51\text{mV}$
• B. $71\text{mV}$
• C. $81\text{mV}$
• D. $91\text{mV}$

Answer 1

The correct option is A $51\text{mV}$

The total magnetic flux through the loop is given by,

$\varphi =2\times \left(\frac{\pi r^2}{2}\right)B\cos {45}^{\circ }$

By Faraday's law, the magnitude of induced emf is,

$E=|-\frac{\Delta \varphi }{\Delta t}|=\pi r^2\cos {45}^{\circ }\frac{\Delta B}{\Delta t}$

$=\pi (3.7\times {10}^{-2}{)}^2\times \frac{1}{\sqrt{2}}\times \frac{76\times {10}^{-3}}{4.5\times {10}^{-3}}$

$\approx 51\times {10}^{-3}\text{V}=51\text{mV}$

Hence, $\left(A\right)$ is the correct answer.

Why this question? This question has been asked to test your understanding of change in magnetic flux along with the concept of geometry applied to it.