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(C) The magnetic field at the centre of a circular loop of radius '$a$' carrying current '$I$' is given by $B = \frac{\mu_0 I}{2 a}$.Since the two loo

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$\frac{\mu_0 I}{\sqrt{2} a}$

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(C) The magnetic field at the centre of a circular loop of radius '$a$' carrying current '$I$' is given by $B = \frac{\mu_0 I}{2 a}$.Since the two loops $A$ and $B$ are perpendicular to each other,the magnetic fields produced by them at the centre $O$ will also be perpendicular.Let $B_A$ be the magnetic field due to loop $A$ and $B_B$ be the magnetic field due to loop $B$. Then $B_A = B_B = \frac{\mu_0 I}{2 a}$.The net magnetic field $B_{net}$ at the centre is the vector sum of $B_A$ and $B_B$:$B_{net} = \sqrt{B_A^2 + B_B^2} = \sqrt{\left(\frac{\mu_0 I}{2 a}\right)^2 + \left(\frac{\mu_0 I}{2 a}\right)^2}$$B_{net} = \sqrt{2 \left(\frac{\mu_0 I}{2 a}\right)^2} = \sqrt{2} \cdot \frac{\mu_0 I}{2 a} = \frac{\mu_0 I}{\sqrt{2} a}$.

Two insulated circular loops $A$ and $B$ of radius '$a$' carry a current of '$I$' in the directions as shown in the figure. The magnitude of the magnetic induction at the centre $O$ will be:

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