1. Magnetic field at the centre of one circular loop

For a circular loop of radius $a$ carrying current $I$, the magnetic field at its centre is

So, each loop $A$ and $B$ produces a field of magnitude

2. Direction of magnetic field due to each loop

Both currents are anticlockwise as seen in the figure, so by the right-hand rule the magnetic field at the centre due to each loop is perpendicular to the plane of its own loop.

Since the two loops are oriented mutually perpendicular (as implied by the figure in such standard problems), the magnetic fields produced by them at the common centre are also perpendicular to each other.

Thus,

with $B_A \perp B_B$.

3. Resultant magnetic field

For two equal perpendicular vectors, resultant magnitude is

Substituting:

4. Option check

• A: $\dfrac{\sqrt{2}\mu_0 I}{a}$ ❌
• B: $\dfrac{\mu_0 I}{\sqrt{2}a}$ ✅
• C: $\dfrac{\mu_0 I}{2a}$ ❌
• D: $\dfrac{2\mu_0 I}{a}$ ❌

Hence, the correct answer is Option B.