Question

A square frame of side $1m$ 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 $B/B^{\prime }$ 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,
$\tan 45^{\circ }=\frac{CF}{OF},OF=CF=\frac{1}{2}m$
The magnetic field at the centre O of the square frame is
$B=\frac{{\mu }_0}{4\pi }\frac{I}{(1/2)}\left[\sin 45^{\circ }+\sin 45^{\circ }\right]\times 4=\frac{{\mu }_0}{4\pi }\frac{16I}{\sqrt{2}}$
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
$B^{\prime }=\frac{{\mu }_0}{4\pi }\frac{2\pi l}{r}=\frac{{\mu }_0}{4\pi }\frac{2\pi I}{(2/\pi )}=\frac{{\mu }_0}{4\pi }{\pi }^{2}I$
$\therefore \frac{B}{B^{\prime }}=\frac{16}{\sqrt{2}{\pi }^2}$
$=\frac{8\sqrt{2}}{{\pi }^2}$