Current I is flowing in a conductor shaped as | vedclass

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(A) The conductor consists of three parts: two semi-infinite straight wires ($A$ and $C$) and a three-quarter circular arc $(B)$.$1$. For the straight

Vedclass · Accepted Answer

$\frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} + 1 \right)$

Vedclass · Answer

(A) The conductor consists of three parts: two semi-infinite straight wires ($A$ and $C$) and a three-quarter circular arc $(B)$.$1$. For the straight wire $A$,the point $O$ lies on its axis,so the magnetic field $B_A = 0$.$2$. For the straight wire $C$,the distance from $O$ is $r$. The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is $B_C = \frac{\mu_0 I}{4\pi r}$. Using the right-hand rule,the direction is into the page $(\otimes)$.$3$. For the three-quarter circular arc $B$,the angle subtended is $	heta = \frac{3\pi}{2}$. The magnetic field is $B_B = \frac{\mu_0 I 	heta}{4\pi r} = \frac{\mu_0 I (3\pi/2)}{4\pi r}$. The direction is also into the page $(\otimes)$.$4$. The net magnetic field at $O$ is $B_{net} = B_A + B_B + B_C = 0 + \frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} ight) + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} + 1 ight)$.

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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