Question

A long conducting wire carrying a current $l$ is bent at ${120}^o$(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. $\frac{\sqrt{3}{\mu }_{0}l}{2\pi d}$

Answer 1

The correct option is D $\frac{\sqrt{3}{\mu }_{0}l}{2\pi d}$

Magnetic field at point P due to each segment is

$\frac{{\mu }_0}{4\pi }\times \frac{I(sin{90}^{\circ }+sin{30}^{\circ })}{d\frac{\sqrt{3}}{2}}$

$=\frac{{\mu }_0}{4\pi }\frac{2I}{d\sqrt{3}}\frac{3}{2}$

Hence total magnetic field at P=$2\times \frac{\sqrt{3}{\mu }_{0}I}{4\pi d}$

$=\frac{\sqrt{3}{\mu }_{0}I}{2\pi d}$