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 $emf(e)$
$=\frac{\text{magnetic field × change in area}}{\text{time}}$
$=\frac{B\Delta A}{t}$
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, $\Delta A=\pi r^2-{\left(\frac{\pi r}{2}\right)}^2=\pi r^2\left(1-\frac{\pi }{4}\right)$ $\therefore e=\frac{B(\pi r^2)}{t}\left(1-\frac{\pi }{4}\right)$