Question

Two identical wires $A$ and $B$ , each of length ${}^{\prime }{l}^{\prime }$ , carry the same current $I$ . Wire $A$ is bent into a circle of radius $R$ and wire $B$ is bent to form a square of side ${}^{\prime }{a}^{\prime }$ . If $B_A$ and $B_B$ are the values of the magnetic field at the centres of the circle and square respectively, if the ratio $\frac{B_A}{B_B}$ is $\frac{{\pi }^2}{N\sqrt{2}}$ then the value of $N$ is :

Answer 1

Answer: 8

The wire $A$ which is bent in circular, having a radius, $R$ then the magnitude of the magnetic field,

${\vec{B}}_B=\left(\frac{{\mu }_{0}I}{4\pi \left(\frac{a}{2}\right)}\times \sqrt{2}\right)\times 4$
where, $a=\frac{l}{4}$
it means,
$\frac{{\vec{B}}_A}{{\vec{B}}_B}=\frac{\frac{{\mu }_{0}I}{2\times \frac{I}{2\pi }}}{\left(\frac{{\mu }_{0}I}{4\pi \left(\frac{1}{8}\right)}\times \sqrt{2}\right)\times 4}=\frac{{\pi }^2}{8\sqrt{2}}=\frac{{\pi }^2}{N\sqrt{2}}\Rightarrow N=8$