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(C) The magnetic field at point $O$ is the vector sum of the magnetic fields due to the four segments of the wire.$1$. For segments $1$ and $3$,the po

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$\frac{\mu_0 I \alpha}{4\pi} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

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(C) The magnetic field at point $O$ is the vector sum of the magnetic fields due to the four segments of the wire.$1$. For segments $1$ and $3$,the point $O$ lies on the axis of the wire. Therefore,the magnetic field due to these segments is $0$.$2$. For the circular arc segments $2$ and $4$,the magnetic field at the center is given by $B = \frac{\mu_0 I 	heta}{4\pi R}$,where $	heta = \alpha$ is the angle subtended at the center.$3$. Using the right-hand rule:- For arc $2$ (radius $R_2$),the current flows in a direction that produces a magnetic field directed into the page (inward).- For arc $4$ (radius $R_1$),the current flows in a direction that produces a magnetic field directed out of the page (outward).$4$. The net magnetic field $B_{net}$ is:$B_{net} = B_4 - B_2 = \frac{\mu_0 I \alpha}{4\pi R_1} - \frac{\mu_0 I \alpha}{4\pi R_2}$$B_{net} = \frac{\mu_0 I \alpha}{4\pi} \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$

The magnetic induction at the centre $O$ of the current-carrying bent wire shown in the adjoining figure is

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