1. Current division in the two branches

The square loop has two parallel paths between points $A$ and $C$:

• Path $ABC$ has resistance $r$
• Path $ADC$ has resistance $2r$

If total current entering at $A$ and leaving at $C$ is $I$, then currents divide inversely to resistances.

Let current through $ABC$ be $I_1$ and through $ADC$ be $I_2$. Then $I_1:I_2=2r:r=2:1$ and $I_1+I_2=I$ So, $I_1=\frac{2I}{3}, \qquad I_2=\frac{I}{3}$

2. Magnetic field at the centre due to one side of the square

For a finite straight conductor, magnetic field at a point at perpendicular distance $d$ is $B=\frac{\mu_0 i}{4\pi d}(\sin\theta_1+\sin\theta_2)$

At the centre of a square of side $a$, distance from centre to each side is $d=\frac{a}{2}$

For each side, the angles are $\theta_1=\theta_2=45^\circ$

Hence field due to one side carrying current $i$ is $B_{\text{side}}=\frac{\mu_0 i}{4\pi (a/2)}(\sin45^\circ+\sin45^\circ)$ $=\frac{\mu_0 i}{2\pi a}\left(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right)$ $=\frac{\mu_0 i}{2\pi a}(\sqrt2)$ $=\frac{\mu_0 i}{\sqrt2\pi a}$

3. Direction of magnetic field contributions

Current in branch $ABC$ goes from $A \to B \to C$. Current in branch $ADC$ goes from $A \to D \to C$.

At the centre of the square:

• Sides $AB$ and $BC$ produce magnetic field in the same direction.
• Sides $AD$ and $DC$ produce magnetic field in the opposite direction to that of $AB,BC$.

So net field is $B=2B_{AB/BC}-2B_{AD/DC}$ with currents $I_1$ and $I_2$ respectively.

Thus $B=2\left(\frac{\mu_0 I_1}{\sqrt2\pi a}\right)-2\left(\frac{\mu_0 I_2}{\sqrt2\pi a}\right)$ $=\frac{2\mu_0}{\sqrt2\pi a}(I_1-I_2)$

Substitute $I_1=\frac{2I}{3}, \qquad I_2=\frac{I}{3}$

Then $I_1-I_2=\frac{I}{3}$

Therefore $B=\frac{2\mu_0}{\sqrt2\pi a}\cdot \frac{I}{3}$ $=\frac{\sqrt2\mu_0 I}{3\pi a}$

4. Final answer

$\boxed{B=\frac{\sqrt2\mu_0 I}{3\pi a}}$

So the correct option is B.