Question

The magnetic field at $O$ due to current in the wire segment BC of the infinite wire forming a loop as shown in figure is

• A. $\frac{{\mu }_{0}I}{4\pi d}(\cos {\varphi }_1+\cos {\varphi }_2)$
• B. $\frac{{\mu }_0}{4\pi }\frac{2I}{d}$
• C. $\frac{{\mu }_0}{4\pi }\frac{I}{d}(\sin {\varphi }_1+\sin {\varphi }_2)$
• D. $\frac{{\mu }_0}{4\pi }\frac{I}{d}$

Answer 1

The correct option is A $\frac{{\mu }_{0}I}{4\pi d}(\cos {\varphi }_1+\cos {\varphi }_2)$

Using $B=\frac{{\mu }_{0}I}{4\pi a}[\sin {\theta }_1+\sin {\theta }_2]$

But ${\theta }_1+{\varphi }_1={90}^{o}or{\theta }_1={90}^o-{\varphi }_1$

${\varphi }_1=\sin ({90}^o-{\varphi }_1)=\cos {\varphi }_1$

Similarly
$\sin {\theta }_2=\cos {\varphi }_2$

$B_{net}=\frac{{\mu }_{0}I}{4\pi d}(\cos {\varphi }_1+\cos {\varphi }_2)$