Question

A square frame of side $1 \,m$ carries a current $I$, produces a magnetic field $B$ at its centre. The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is $B$' The ratio $B / B'$ is

• A. $\frac{8}{\pi^{2}}$
• B. $\frac{8 \sqrt{2}}{\pi^{2}}$
• C. $\frac{16}{\pi^{2}}$
• D. $\frac{16 \sqrt{2}}{\pi^{2}}$

Answer 1

Answer: (B)

Refer figure,
$\tan 45^{\circ}=\frac{C F}{O F}, \quad O F=C F=\frac{1}{2} \,m$
The magnetic field at the centre O of the square frame is
$B=\frac{\mu_{0}}{4 \pi} \frac{I}{(1 / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right] \times 4=\frac{\mu_{0}}{4 \pi} \frac{16 I}{\sqrt{2}}$
When the frame is taken as circular coil of radius $r$, then
$2 \pi r=4 $ or $r=\frac{2}{\pi}$
The magnetic field at the centre of the circular coil carrying same current $I$ is
$B'=\frac{\mu_{0}}{4 \pi} \frac{2 \pi l}{r}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi I}{(2 / \pi)}=\frac{\mu_{0}}{4 \pi} \pi^{2} I $
$\therefore \frac{B}{B'}=\frac{16}{\sqrt{2} \pi^{2}}$
$=\frac{8 \sqrt{2}}{\pi^{2}}$