question_answer

A long conducting wire carrying a current I is  bent at 120° (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)  \[\frac{3{{\mu }_{0}}l}{2\pi d}\]

B)  \[\frac{{{\mu }_{0}}l}{2\pi d}\]

C)  \[\frac{{{\mu }_{0}}l}{\sqrt{3}\pi d}\]

D)  \[\frac{\sqrt{3}{{\mu }_{0}}l}{2\pi d}\]

Correct Answer: D

Solution :

 We know that \[{{B}_{net}}=2\left[ \frac{{{\mu }_{0}}i}{4\pi r}(sin{{\theta }_{1}}+sin{{\theta }_{2}}) \right]\] \[=2\left[ \frac{{{\mu }_{0}}}{4\pi }\times \frac{i}{\frac{d\sqrt{3}}{2}}\times (sin{{90}^{o}}+sin{{30}^{o}}) \right]\] \[=2\left[ \frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{d\sqrt{3}}\times \left( 1+\frac{1}{2} \right) \right]\] \[=2\left[ \frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{d\sqrt{3}}\times \frac{3}{2} \right]=\frac{\sqrt{3}{{\mu }_{0}}i}{2\pi d}\]