A non-conducting thin disc of radius $R$ charged uniformly over one side rotates with constant angular velocity $ω$ about an axis passing through its periphery point $P$ and perpendicular to its plane. The electrostatic potential at the point $P$ due to charges on the disc is $V_{0}$ before the rotation. The magnetic field at $P$ due to the moving charges upon rotation is $k μ_{0} \in_{0} ω V_{0} .$ Find the value of $k .$

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Detailed Solution

$d q = \frac{1}{4 π \in_{0}} \frac{d q}{r}$
Integrating $\frac{1}{4 π \in_{0}} \int \frac{d q}{r} = V_{0} (is given)$
Current due to $d q = \frac{d q ω}{2 π}$
dB due to $d q = \frac{μ_{0} (\frac{d p ω}{2 π})}{2 π}$
Integrating, we get $B = \frac{μ_{0} ω}{4 π} \int \frac{d q}{r}$
$= \frac{μ_{0} ω}{4 π} (4 π \in_{0} V_{0}) = μ_{0} \in_{0} ω V_{0}$
The result is true for any non-conducting lamina.