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A long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the right bisector of bending angle at a distance $d$ from the bend is $\left(\mu_{0}\right.$ is the permeability of free space) 
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Verified Answer
The correct answer is:
$\frac{\sqrt{3} \mu_{0} I}{2 \pi d}$
We know that
$$
\begin{aligned}
B_{\text {net }} &=2\left[\frac{\mu_{0} i}{4 \pi r}\left(\sin \theta_{1}+\sin \theta_{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{i}{\frac{d \sqrt{3}}{2}} \times\left(\sin 90^{\circ}+\sin 30^{\circ}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times\left(1+\frac{1}{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times \frac{3}{2}\right]=\frac{\sqrt{3} \mu_{0} i}{2 \pi d}
\end{aligned}
$$
$$
\begin{aligned}
B_{\text {net }} &=2\left[\frac{\mu_{0} i}{4 \pi r}\left(\sin \theta_{1}+\sin \theta_{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{i}{\frac{d \sqrt{3}}{2}} \times\left(\sin 90^{\circ}+\sin 30^{\circ}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times\left(1+\frac{1}{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times \frac{3}{2}\right]=\frac{\sqrt{3} \mu_{0} i}{2 \pi d}
\end{aligned}
$$
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