A wire A , bent in the shape of an arc of a circle, carrying a...

Given,

Current in the arc $A$ is ${i}_{1} = 2A$

Current in the arc $B$ is ${i}_{2} = 3A$

Radius of arc $A$ is ${R}_{1} = 2cm$

Radius of arc $B$ is ${R}_{2} = 4cm$

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

${B}_{C} = \dfrac{{\mu}_{0}i}{4\piR}\left(\theta\right) . . . ( 1 )$

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

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

${B}_{A} = \dfrac{{\mu}_{0}{i}_{1}}{4{\pi R}_{1}}\left(2\pi - \dfrac{\pi}{2}\right) . . . \left(2\right)$

${B}_{B} = \dfrac{{\mu}_{0}{i}_{2}}{4{\pi R}_{2}}\left(2\pi - \dfrac{\pi}{3}\right) . . . \left(3\right)$

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

$⇒ \dfrac{{B}_{A}}{{B}_{B}} = \dfrac{{i}_{1}}{{i}_{2}} \times \dfrac{{R}_{2}\left(2\pi - \dfrac{\pi}{2}\right)}{{R}_{1}\left(2\pi - \dfrac{\pi}{3}\right)}$

$⇒ \dfrac{{B}_{A}}{{B}_{B}} = \dfrac{2}{3} \times \dfrac{4}{2} \times \dfrac{3\pi}{2} \times \dfrac{3}{5\pi}$

$⇒ \dfrac{{B}_{A}}{{B}_{B}} = \dfrac{6}{5}$