Question

Two identical wires $A$ and $B$ , each of length $'𝑙'$ , 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 $'a'$ . 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

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 $