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(A) The conductor consists of three parts: two semi-infinite straight wires and a circular arc of $270^\circ$ (or $\frac{3\pi}{2}$ radians).$1$. For t

Vedclass · Accepted Answer

$\frac{\mu_{0} I}{4 \pi r}\left(\frac{3 \pi}{2}+1\right)$

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(A) The conductor consists of three parts: two semi-infinite straight wires and a circular arc of $270^\circ$ (or $\frac{3\pi}{2}$ radians).$1$. For the straight wire $(A)$,the point $O$ lies on its axis,so the magnetic field $B_A = 0$.$2$. For the circular arc $(B)$,the magnetic field at the centre is $B_B = \frac{\mu_0 I 	heta}{4 \pi r} = \frac{\mu_0 I (3\pi/2)}{4 \pi r} = \frac{3 \mu_0 I}{8 r}$. The direction is inwards (using the right-hand rule).$3$. For the straight wire $(C)$,the point $O$ is at a perpendicular distance $r$ from the wire. The magnetic field due to a semi-infinite wire is $B_C = \frac{\mu_0 I}{4 \pi r}$. The direction is also inwards.$4$. The total magnetic field $B = B_A + B_B + B_C = 0 + \frac{3 \mu_0 I}{8 r} + \frac{\mu_0 I}{4 \pi r}$.Factoring out $\frac{\mu_0 I}{4 \pi r}$,we get $B = \frac{\mu_0 I}{4 \pi r} \left( \frac{3\pi}{2} + 1 ight)$.

Current $I$ is flowing in a conductor shaped as shown in the figure. The radius of the curved part is $r$ and the length of the straight portions is very large. The value of the magnetic field at the centre $O$ will be

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