Question

A uniformly charged ring of radius $R$ is rotated about its axis with constant linear speed $v$ of each of it's particles. The ratio of electric field to magnetic field at a point $P$ on the axis of the ring at a distant $x=R$ from the centre of the ring is :
$(c$ is speed of light$)$

• A. $\frac{c^2}{v}$
• B. $\frac{v^2}{c}$
• C. $\frac{c}{v}$
• D. $\frac{v}{c}$

Answer 1

The correct option is A $\frac{c^2}{v}$

Let $Q$ be the charge on the ring. The electric field at point $P$ is,
$\therefore E=\frac{1}{4\pi {ϵ}_0}\frac{Qx}{(x^2+R^2{)}^{3/2}}=\frac{1}{4\pi {ϵ}_0}\frac{QR}{(2R^2{)}^{3/2}}$
The rotating charged (Q) ring is equivalent to a ring in which current flows, such that
$I=\frac{Qv}{2\pi R}$
The magnetic field at point $P$ due to moving charge is
$B=\frac{{\mu }_0}{4\pi }\frac{2\pi I{R}^2}{(x^2+R^2{)}^{3/2}}=\frac{{\mu }_0}{4\pi }\frac{QvR}{(2R^2{)}^{3/2}}$
$\therefore \frac{E}{B}=\frac{1}{{\mu }_0{ϵ}_{0}v}=\frac{c^2}{v}$