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(C) The wire consists of three parts: two semi-infinite straight wires and a circular arc of $270^\circ$ ($3\pi/2$ radians).Let the magnetic field due

Vedclass · Accepted Answer

$\frac{\mu_0}{4 \pi} \frac{I}{a}\left[\frac{3 \pi}{2}+2\right]$

Vedclass · Answer

(C) The wire consists of three parts: two semi-infinite straight wires and a circular arc of $270^\circ$ ($3\pi/2$ radians).Let the magnetic field due to the straight wire $(1)$ at $O$ be $B_1$. Since $O$ lies on the line of the wire, $B_1 = \frac{\mu_0 I}{4 \pi a}$.Let the magnetic field due to the circular arc $(2)$ at $O$ be $B_2$. The angle subtended is $	heta = 3\pi/2$. Thus, $B_2 = \frac{\mu_0 I}{4 \pi a} 	heta = \frac{\mu_0 I}{4 \pi a} \left(\frac{3 \pi}{2}ight)$.Let the magnetic field due to the straight wire $(3)$ at $O$ be $B_3$. Since $O$ lies on the line of the wire, $B_3 = \frac{\mu_0 I}{4 \pi a}$.All three fields are directed into the page $(\otimes)$.The total magnetic field $B = B_1 + B_2 + B_3 = \frac{\mu_0 I}{4 \pi a} + \frac{\mu_0 I}{4 \pi a} \left(\frac{3 \pi}{2}ight) + \frac{\mu_0 I}{4 \pi a} = \frac{\mu_0 I}{4 \pi a} \left[1 + \frac{3 \pi}{2} + 1ight] = \frac{\mu_0 I}{4 \pi a} \left[\frac{3 \pi}{2} + 2ight]$.

An infinite wire has a circular bend of radius $a$, and carries a current $I$ as shown in the figure. The magnitude of the magnetic field at the origin $O$ is given by:

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