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(A) The magnetic field at the centre of a semicircular wire of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{4R}$.For the larger sem

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$\frac{\mu_0 I}{4}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

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(A) The magnetic field at the centre of a semicircular wire of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{4R}$.For the larger semicircle of radius $R_1$,the magnetic field at $O$ is $B_1 = \frac{\mu_0 I}{4R_1}$ (directed outwards,using the right-hand rule).For the smaller semicircle of radius $R_2$,the magnetic field at $O$ is $B_2 = \frac{\mu_0 I}{4R_2}$ (directed inwards,using the right-hand rule).The straight segments $PQ$ and $SR$ do not contribute to the magnetic field at $O$ because the point $O$ lies on their axis.The net magnetic field at $O$ is $B_{net} = B_1 - B_2 = \frac{\mu_0 I}{4R_1} - \frac{\mu_0 I}{4R_2} = \frac{\mu_0 I}{4} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.

The wire loop $PQRSP$ formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown. The magnitude of the magnetic field at the centre '$O$' is

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