Given: Two identical wires of equal length $l$ carry the same current $I$.

• Wire $A$ is bent into a circle of radius $R$.
• Wire $B$ is bent into a square of side $a$. We need to find $\frac{B_A}{B_B}.$

Magnetic field at the centre of the circular wire

For a circular loop of radius $R$ carrying current $I$, $B_A = \frac{\mu_0 I}{2R}.$

Magnetic field at the centre of the square loop

The magnetic field due to one side of the square at the centre is obtained using the formula for a finite straight wire: $B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2).$

For one side of the square:

• perpendicular distance from centre to side, $r = \frac{a}{2},$
• the angles are $\theta_1 = \theta_2 = 45^\circ.$

So field due to one side is $B_{\text{one side}} = \frac{\mu_0 I}{4\pi (a/2)}(\sin45^\circ + \sin45^\circ).$

Since $\sin45^\circ = \frac{1}{\sqrt{2}}$,

Use equal wire lengths

Since both wires have the same length $l$:

• For the circle: $l = 2\pi R \Rightarrow R = \frac{l}{2\pi}.$

• For the square: $l = 4a \Rightarrow a = \frac{l}{4}.$

Substitute into $B_A$ and $B_B$

For the circle: $B_A = \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2\cdot l/(2\pi)} = \frac{\mu_0 I\pi}{l}.$

For the square:

Take the ratio

Match with options

$\frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}}$ which corresponds to Option B.