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(D) According to the Biot-Savart Law,the magnetic field $dB$ due to a current element $idl$ is given by $dB = \frac{\mu_0}{4\pi} \frac{idl \sin 	heta

Vedclass · Accepted Answer

$\frac{\mu_0}{4\pi} \cdot \frac{\pi i}{r} 	ext{ tesla}$

Vedclass · Answer

(D) According to the Biot-Savart Law,the magnetic field $dB$ due to a current element $idl$ is given by $dB = \frac{\mu_0}{4\pi} \frac{idl \sin 	heta}{r^2}$.For the straight segments of the wire,the angle $	heta$ between the current element and the position vector to the centre $O$ is either $0^{\circ}$ or $180^{\circ}$. Since $\sin(0^{\circ}) = 0$ and $\sin(180^{\circ}) = 0$,the straight parts contribute zero magnetic field at the centre $O$.For the semicircular portion,the magnetic field at the centre is half of that produced by a full circular loop of radius $r$. The magnetic field at the centre of a full circular loop is $B = \frac{\mu_0 i}{2r}$.Therefore,the magnetic field due to the semicircle is $B_{semi} = \frac{1}{2} \left( \frac{\mu_0 i}{2r} ight) = \frac{\mu_0 i}{4r}$.This can be rewritten as $\frac{\mu_0}{4\pi} \cdot \frac{\pi i}{r} 	ext{ tesla}$.

$A$ portion of a conductive wire is bent in the form of a semicircle of radius $r$ as shown in the figure. At the centre $O$ of the semicircle,the magnetic induction will be:

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