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

Answer 1

The correct option is D $\frac{\pi B{r}^2}{t}(1-\frac{\pi }{4})$

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