Question

An infinite wire is bent in the form of $L,$ carries a current $I$. The magnetic field at the point $O$ is :

• A. $zero$
• B. $\frac{{\mu }_0}{4\pi }\times \frac{I}{d}$
• C. $\frac{{\mu }_0}{2\pi }\times \frac{I}{d}\{\frac{1}{\sqrt{2}}+1\}$
• D. $\frac{{\mu }_0}{4\pi }\times \frac{I}{d}\{\frac{1}{\sqrt{2}}+1\}$

Answer 1

The correct option is C $\frac{{\mu }_0}{2\pi }\times \frac{I}{d}\{\frac{1}{\sqrt{2}}+1\}$

Magnetic field at point $P$ due to a finite wire $B=\frac{{\mu }_0}{4\pi }\frac{I}{r}(\sin {\varphi }_1+\sin {\varphi }_2)$
For point O,
${\varphi }_1={90}^o$; ${\varphi }_2={45}^o$ for each of the arm of the $L$ shaped wire.

$B=2\times \frac{{\mu }_0}{4\pi }\frac{I}{d}(\sin {\varphi }_1+\sin {\varphi }_2)$

$=\frac{{\mu }_{0}I}{2\pi d}(\sin {90}^o+\sin {45}^o)$

$=\frac{{\mu }_{0}i}{2\pi d}(1+\frac{1}{\sqrt{2}})$