Question

A ring of radius $r$ is uniformly charged with charge $q$. If the ring is rotated about it's axis with angular frequency $\omega$, then the magnetic induction at its centre will be :

• A. $\frac{{\mu }_{0}q\omega }{4\pi r}$
• B. ${10}^{-7}\times \frac{9}{\omega r}$
• C. ${10}^{-7}\times \frac{r}{q\omega }$
• D. ${10}^{-7}\times \frac{9\omega }{r}$

Answer 1

The correct option is A $\frac{{\mu }_{0}q\omega }{4\pi r}$

Since the total charge is $q$ and we know that current $=\frac{charge}{time}$

$I=\frac{q}{t}$

The angular velocity is equal to $\omega =\frac{2\pi }{t}$

Time period, $t=\frac{2\pi }{\omega }$

$I=\frac{q\omega }{2\pi }$

We know that magnetic field at the center of a current carrying conductor.

$B=\frac{{\mu }_{0}I}{2r}$

Putting the value of $I$ from the above equation, we get

$\begin{array}{c}B=\left(\frac{{\mu }_0}{2r}\right)\times \frac{q\omega }{2\pi }\\ \Rightarrow B=\frac{{\mu }_{0}q\omega }{4\pi r}\end{array}$

Hence the total magnetic induction at the center of the ring will be $\frac{{\mu }_{0}q\omega }{4\pi r}$