Question

A conductor of length l has shape of a semicylinder of radius R(<<l). Crossection of the conductor is shown in figure. Thickness of conductor is t(<<R) and conductivity of its material is $\delta ={\delta }_{0}cos\theta$. If an ideal battery of emf V is connected across its end faces, then magnetic field at point O on the axis of semicylinder is $n\frac{{\mu }_0{\delta }_{0}Vt}{l}$. Here, n is

Answer 1

$B_{net}$ at O $={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}dBcos\theta$
$dB=\frac{{\mu }_0}{2\pi }\frac{dI}{R}$
$dI=\frac{V}{dR}=\frac{VRtd\theta }{pl}=\frac{VRt{\delta }_{0}cos\theta }{l}$
$B_{net}={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{{\mu }_0}{2\pi R}\frac{VRt{\delta }_{0}co{s}^2\theta d\theta }{l}=\frac{{\mu }_{0}Vt{\delta }_0}{2\pi l}{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}co{s}^2\theta d\theta =\frac{{\mu }_{0}Vt{\delta }_0}{4\pi l}$