Question

A wire in the form of a circular loop of radius $ r $ lies with its plane normal to a magnetic field $ B. $ If the wire is pulled to take a square shape in the same plane in time $ t $ , the emf induced in the loop is given by

• A. $ \frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{10} \right) $
• B. $ \frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{8} \right) $
• C. $ \frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{6} \right) $
• D. $ \frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{4} \right) $

Answer 1

Answer: (D)

Induced $ emf(e) $
$ =\frac{\text{magnetic field $\times$ change in area}}{\text{time}}$
$=\frac{B\Delta A}{t} $
Since, the circumference of the circular loop $ =2\pi r $ ,
The side of the square loop $ =\frac{2\pi r}{4}=\frac{\pi r}{2} $
Therefore, $ \Delta A=\pi r^{2}-{{\left( \frac{\pi \,r}{2} \right)}^{2}}=\pi {{r}^{2}}\left( 1-\frac{\pi }{4} \right) $ $ \therefore e=\frac{B(\pi r^{2})}{t}\left( 1-\frac{\pi }{4} \right) $