Question

A current loop having two circular arc is as shown in fig and carries a current of $10A$ . The magnetic field at point $O$ is $\frac{N \pi }{14}\times 10^{- 7}T$ . Then find N..

Answer 1

$\overset{ \rightarrow }{B}_{0}$ $=\overset{ \rightarrow }{B}_{P Q}+\overset{ \rightarrow }{B}_{Q R}+\overset{ \rightarrow }{B}_{R S}+\overset{ \rightarrow }{B}_{S P}$
$=\overset{ \rightarrow }{0}+\frac{\left(\mu \right)_{0} I}{4 \pi r_{1}}\left(\frac{\pi }{3}\right)\hat{k}+\overset{ \rightarrow }{0}+\frac{\left(\mu \right)_{0} I}{4 \pi r_{2}}\left(\frac{\pi }{3}\right)\left(\right.-\hat{k}\left.\right)$
$=\frac{\mu _{0}}{12}\left[\frac{1}{r_{1}} - \frac{1}{r_{2}}\right]\hat{k}$
$=\frac{4 \pi \times 10^{- 7} \times 10}{12}\left[\frac{1}{4 \times 10^{- 2}} - \frac{1}{7 \times 10^{- 2}}\right]$
$=\frac{4 \pi \times 10^{- 7} \times 10}{12 \times 10^{- 2}}\left[\frac{7 - 4}{4 \times 7}\right]=\frac{5 \pi }{14}\times 10^{- 7}T$