Question

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 $({\mu }_0$ is the permeability of free space)

• A. $\frac{3{\mu }_{0}I}{2\pi d}$
• B. $\frac{{\mu }_{0}I}{2\pi d}$
• C. $\frac{{\mu }_{0}I}{\sqrt{3}\pi d}$
• D. $\frac{\sqrt{3}{\mu }_{0}I}{2\pi d}$

Answer 1

Answer: (D)

We know that

$B_{net}=2\left[\frac{{\mu }_{0}i}{4\pi r}\left(\sin {\theta }_1+\sin {\theta }_2\right)\right]$
$=2[\frac{{\mu }_0}{4\pi }\times \frac{i}{\frac{d\sqrt{3}}{2}}\times \left(\sin 90^{\circ }+\sin 30^{\circ }\right)$
$=2\left[\frac{{\mu }_0}{4\pi }\times \frac{2i}{d\sqrt{3}}\times \left(1+\frac{1}{2}\right)\right]$
$=2\left[\frac{{\mu }_0}{4\pi }\times \frac{2i}{d\sqrt{3}}\times \frac{3}{2}\right]=\frac{\sqrt{3}{\mu }_{0}i}{2\pi d}$