A current I flows in a wire bent in the shape | vedclass

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(B) The magnetic field at the center of a complete circular loop carrying current $I$ is given by $B_{total} = \frac{\mu_0 I}{2 R}$.Since the given wi

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$\frac{\mu_0 I}{8 R}$

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(B) The magnetic field at the center of a complete circular loop carrying current $I$ is given by $B_{total} = \frac{\mu_0 I}{2 R}$.Since the given wire is a quarter of a circular loop, the angle subtended at the center is $90^\circ$ or $\frac{\pi}{2}$ radians.The magnetic field due to an arc subtending an angle $	heta$ at the center is $B = \frac{\mu_0 I 	heta}{4 \pi R}$.For a quarter ring, $	heta = \frac{\pi}{2}$.Substituting the value of $	heta$, we get $B = \frac{\mu_0 I (\pi/2)}{4 \pi R} = \frac{\mu_0 I}{8 R}$.Thus, the magnetic field induction at point $O$ is $\frac{\mu_0 I}{8 R}$.

$A$ current $I$ flows in a wire bent in the shape of a quarter ring of radius $R$ as shown in the figure. The magnetic field induction $B$ at point $O$ is

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