Question

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

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

Answer 1

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

When a charged ring is rotated, a current will start to flow
$\omega \to$angular velocity, $f\to$ frequency of rotation
$i=$current due to rotation
$q=$charge on ring
$\Rightarrow f=\frac{\omega }{2\pi }$
Also $i=\frac{q}{t}=qf=q\frac{\omega }{2\pi }$
Now,
Electric field at Point P due to ring$P=\frac{1}{4\pi {\epsilon }_0}\frac{qx}{{(x^2+R^2)}^{3/2}}$
$\Rightarrow E_P=\frac{1}{4\pi {\epsilon }_0}\frac{qR}{{\left(2R^2\right)}^{3/2}}=\frac{1}{4\pi {\epsilon }_0}\times \frac{q}{2\sqrt{2}{R}^2}\to 1$
Magnetic field at point P due to rotating Ring; $P=\frac{{\mu }_{0}Ni{R}^2}{2{(R^2+x^2)}^{3/2}}$
$B_P=\frac{{\mu }_0\times 1\times qf{R}^2}{2\times 2\sqrt{2}{R}^3}(N=1)$
$\Rightarrow B_P=\frac{{\mu }_0}{2}\times \frac{q\omega }{2\pi \times 2\sqrt{2}R}$
$\Rightarrow B_P=\frac{{\mu }_0}{4\pi }\frac{qv}{2\sqrt{2}R}\to 2(\because \omega =\frac{v}{R})$
Dividing equation $1$ by equation $2$
$\Rightarrow \frac{E_P}{B_P}=\frac{\frac{1}{4\pi {\epsilon }_0}\times \frac{q}{2\sqrt{2}{R}^2}}{\frac{{\mu }_0}{4\pi }\times \frac{qv}{2\sqrt{2}{R}^2}}$
$\Rightarrow \frac{E_P}{B_P}=\frac{1}{{\mu }_0{\epsilon }_0}\times \frac{1}{v}$
But speed of light$=C=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$
$\Rightarrow C^2=\frac{1}{{\mu }_0{\epsilon }_0}$
$\Rightarrow \frac{E_P}{B_P}=\frac{C^2}{v}$
Answer A $\frac{C^2}{v}$