Question

A thin ring of radius $r$ carries a uniformly distributed charge. The ring rotates at a constant angular speed of $n$ revolutions per second about an axis passing through centre and perpendicular to its plane. If $B$ is the magnitude of magnetic field at the centre of ring, then the charge carried by the ring is

• A. $\frac{2rB}{{\mu }_{0}n}$
• B. $\frac{{\mu }_{0}n}{2rB}$
• C. $\frac{{\mu }_{0}n}{r}$
• D. $\frac{2r}{Bq}$

Answer 1

The correct option is A $\frac{2rB}{{\mu }_{0}n}$

Magnitude of magnetic field at the centre of circular current carrying wire is given by,
$B=\frac{{\mu }_{0}I}{2r}...\left(i\right)$
$\Rightarrow$Frequency of revolution ($n\text{rev/sec})$ of charge $\left(q\right)$ (uniformly distributed on ring) is:
$f=\frac{1}{T}=n...\left(ii\right)$
$\Rightarrow$current in wire due to revolving charge
$I=\frac{q}{T}=qf...\left(iii\right)$
$\therefore I=qn$
From Eq. $\left(i\right)$,
$B=\frac{{\mu }_0\left(qn\right)}{2r}$

$\therefore q=\frac{2rB}{{\mu }_{0}n}$