The magnetic induction at the centre O in the | vedclass

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(A) The magnetic field at the centre $O$ due to a semi-circular arc of radius $R$ carrying current $i$ is given by $B = \frac{{\mu _0}i}{{4R}}$.In the

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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 $O$ due to a semi-circular arc of radius $R$ carrying current $i$ is given by $B = \frac{{\mu _0}i}{{4R}}$.In the given figure,the current flows through two semi-circular arcs of radii $R_1$ and $R_2$ and two straight radial segments.The magnetic field due to the straight radial segments at the centre $O$ is zero because the position vector of the point $O$ is parallel to the current element $idl$ for these segments.For the inner semi-circular arc of radius $R_1$,the magnetic field $B_1$ at $O$ is directed into the plane of the paper $(\otimes)$ and its magnitude is $B_1 = \frac{{\mu _0}i}{{4R_1}}$.For the outer semi-circular arc of radius $R_2$,the magnetic field $B_2$ at $O$ is directed out of the plane of the paper $(\odot)$ and its magnitude is $B_2 = \frac{{\mu _0}i}{{4R_2}}$.Since $R_1$  $B_2$.The net magnetic field $B_{net}$ is $B_1 - B_2$ directed into the plane of the paper.$B_{net} = \frac{{\mu _0}i}{{4R_1}} - \frac{{\mu _0}i}{{4R_2}} = \frac{{\mu _0}i}{4}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$.

The magnetic induction at the centre $O$ in the figure shown is

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