Question

A square frame of side l m carries a current I, it produces a magnetic field $B_1$ at its centre. Same current is passed through a circular coil having the same perimeter as the square. Magnetic field at the centre of circular coil is $B_2$. Then, the ratio $\frac{B_1}{B_2}$ is

• A. $\frac{2\sqrt{2}}{3\pi }$
• B. $\frac{8\sqrt{2}}{\pi }$
• C. $\frac{5\sqrt{3}}{\sqrt{7}\pi }$
• D. $1$

Answer 1

Answer: (B)

$\frac{8\sqrt{2}}{\pi }$

For circle
$I=2\pi R$
$\Rightarrow R=\frac{l}{2\pi }$
For square $l=4a,a=\frac{l}{4}$
Magnetic field at the centre of circular coil
$B_2=\frac{{\mu }_{0}i}{2R}=\frac{{\mu }_{0}i}{2\cdot \frac{l}{2\pi }}=\frac{{\mu }_0\pi i}{l}$
Magnetic field at the cnetre of square from.
$B_2=4[\frac{{\mu }_0}{4\pi }\cdot \frac{i}{a}(sin{45}^o+sin{45}^o\left)\right]$
$=\frac{2\sqrt{2}{\mu }_{o}i}{\pi a}=\frac{8\sqrt{2}{\mu }_{o}i}{\pi l}$
$\Rightarrow \frac{B_1}{B_2}=\frac{8\sqrt{2}}{\pi }$