Question

A non-conducting ring of radius $R$ has charge $Q$ distributed unevenly over it. If it rotates with an angular velocity $\omega$, the equivalent current will be

• A. $0$
• B. $Q\omega$
• C. $\frac{Q\omega }{2\pi }$
• D. $\frac{Q\omega }{2\pi R}$

Answer 1

The correct option is C $\frac{Q\omega }{2\pi }$

With each rotation, charge $Q$ crosses any fixed point $\text{P}$ on the rim of the ring.

And, number of rotations per second $=\omega /2\pi$.

Therefore, by definition, charge crossing $\text{P}$ per second is current $I$

$\therefore I=\frac{dq}{dt}=\frac{Q}{T}$

$=\frac{Q}{\frac{2\pi }{\omega }}=\frac{Q\omega }{2\pi }$

Hence, option $\left(C\right)$ is the correct answer.

$\begin{array}{c}\text{Why this Question?}\\ \text{Time rate of flow of charge through a}\\ \text{cross-sectional area is called current.}\end{array}$