Question

A long conducting wire carrying a current $I$ is bent at ${120}^0$ (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

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

Let the magnetic field due to wire (1) and (2) at point P be $B_1$ and $B_2$ respectively.

From geometry, we get $x=dcos{30}^o=\frac{\sqrt{3}d}{2}$

Given : ${\theta }_1={30}^o$ ${\theta }_2={90}^o$

For wire (1) : $B_1=\frac{{\mu }_{o}I}{4\pi x}(sin{\theta }_1+sin{\theta }_2)$

$B_1=\frac{{\mu }_{o}I}{4\pi \times \frac{\sqrt{3}d}{2}}(sin{30}^o+sin{90}^o)$

$B_1=\frac{{\mu }_{o}I}{4\pi \times \frac{\sqrt{3}d}{2}}(0.5+1)$ $⟹B_1=\frac{{\mu }_o\sqrt{3}I}{4\pi d}$

Similarly $B_2=\frac{{\mu }_o\sqrt{3}I}{4\pi d}$

Total magnetic field at point P $B_T=B_1+B_2=\frac{{\mu }_o\sqrt{3}I}{2\pi d}$