The magnetic induction at O due to the whole | vedclass

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(C) The conductor consists of three parts: two straight segments $AB$ and $CD$,and a semicircular arc $BC$ of radius $r$.For the straight segments $AB

Vedclass · Accepted Answer

$\frac{\mu_0 i}{4r}$

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(C) The conductor consists of three parts: two straight segments $AB$ and $CD$,and a semicircular arc $BC$ of radius $r$.For the straight segments $AB$ and $CD$,the point $O$ lies on the axis of the current-carrying wires. According to the Biot-Savart law,the magnetic field $dB = \frac{\mu_0}{4\pi} \frac{i dl \sin 	heta}{r^2}$. For these segments,the angle $	heta$ between the current element $dl$ and the position vector is $0^\circ$ or $180^\circ$,so $\sin 	heta = 0$. Thus,the magnetic induction due to $AB$ and $CD$ is zero.For the semicircular arc $BC$,the magnetic field at the center $O$ is given by $B = \frac{1}{2} \left( \frac{\mu_0 i}{2r} ight) = \frac{\mu_0 i}{4r}$.Therefore,the total magnetic induction at $O$ is $\frac{\mu_0 i}{4r}$.

The magnetic induction at $O$ due to the whole length of the conductor is:

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