JEE Main 2020 — Magnetic Effects of Current Question with Solution

Given,

Current in the arc $A$ is $i_1=2 \mathrm{A}$

Current in the arc $B$ is $i_2= 3 \mathrm{A}$

Radius of arc $A$ is $R_1=2 \mathrm{cm}$

Radius of arc $B$ is $R_2=4 \mathrm{cm}$

Magnetic field at the centre of circular current carrying arc is given by

$B_{\mathrm{C}}=\frac{{\mu }_{\mathit{0}}i}{\mathit{4}\pi R}\left(\theta \right) ...\left(1\right)$ 

Where, $\mathrm{\theta }$ is the angle subtended at centre.

Magnetic field at the centre due to arc $A$ and $B$ is given by 

${\mathrm{B}}_A=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_1}{4{\mathrm{\pi R}}_1}\left(2\pi -\frac{\pi }{2}\right) ...\left(2\right)$

${\mathrm{B}}_B=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_2}{4{\mathrm{\pi R}}_2}\left(2\pi -\frac{\pi }{3}\right) ...\left(3\right)$

Dividing equation (2) by (1), we get

$\Rightarrow \frac{B_A}{B_B}=\frac{i_1}{i_2}\times \frac{R_{\mathit{2}}\left(2\mathrm{\pi }-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{R_{\mathit{1}}\left(2\mathrm{\pi }-{\displaystyle \frac{\mathrm{\pi }}{3}}\right)}$

$\Rightarrow \frac{B_A}{B_B}=\frac{2}{3}\times \frac{4}{2}\times \frac{3\mathrm{\pi }}{2}\times \frac{3}{5\mathrm{\pi }}$

$\Rightarrow \frac{B_A}{B_B}=\frac{6}{5}$