Question

A non-conducting ring with radius of $10cm$ is uniformly charged with total positive charge of $10\mu C$. The ring rotates at constant angular speed of $20rad/s$ about an axis passing through its centre perpendicular to plane of the ring. What is magnitude of the magnetic field on the axis of the ring at $5cm$ from the centre?

• A. $143pT$
• B. $12pT$
• C. $9pT$
• D. $2pT$

Answer 1

The correct option is A $143pT$

Given : radius of ring,$r=10cm={10}^{-1}m$

charge on ring,$q=10\ast {10}^{-6}$

$\omega =20rad/s$

distance from centre,$a=5cm=5\ast {10}^{-2}m$

To find: magnitude of magnetic field at a,$B=?$

Solution: As we know that,

magnetic field at distance a from centre of coil is,

$==>B=\frac{{\mu }_{0}I{r}^2}{2(r^2+a^2{)}^{3/2}}$

As, $I=q/t=\frac{q\omega }{2\pi }$

On equating above both equations,we get

$==>B=\frac{{\mu }_{0}q\omega r^2}{4\pi (r^2+a^2{)}^{3/2}}$

Substitute all the required values,we get

$==>B=\frac{{10}^{-7}\ast 10\ast {10}^{-6}\ast 20\ast ({10}^{-1}{)}^2}{\left(\right({10}^{-1}{)}^2+(5\ast {10}^{-2}{)}^2{)}^{3/2}}$

$==>B=\frac{20\ast {10}^{-14}}{(1.953125{)}^{1/2}\ast {10}^{-3}}$

$==>B=\frac{20\ast {10}^{-11}}{1.39754248594}$

$==>B=14.31\ast {10}^{-11}$

$==>B=143.1\ast {10}^{-12}T$

$==>B\simeq 143pT$

hence,

The correct opt : A