Question

A thin ring of radius $R$ meter has charge $q$ coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of $f\ \text{rev s}^{-1}$. The magnetic induction at the center of the ring is

• A. $\frac{\mu_0 qf}{2R}$
• B. $\frac{\mu_0 q}{2fR}$
• C. $\frac{\mu_0 qf}{2\pi R}$
• D. $\frac{\mu_0 q}{2\pi fR}$

Hint:

Whenever a problem connects a rotating charge $q$ to a magnetic field, always convert the rotation speed to a standard current branch using $I = qf$ or $I = \frac{qv}{2\pi R}$. This allows you to immediately reuse all standard current-based magnetic field formulas.

Answer 1

The Correct Option is A

Step 1: Turn the rotating charge into a current.
As the ring spins, the whole charge $q$ sweeps past any point once per revolution. With $f$ revolutions per second, the equivalent current is $$I = \frac{q}{T} = qf,$$ since the period $T = 1/f$.
Step 2: Recall the field at the centre of a loop.
A circular loop of radius $R$ carrying current $I$ produces a field at its centre of $$B = \frac{\mu_0 I}{2R}.$$
Step 3: Substitute the effective current.
Replacing $I$ with $qf$, $$B = \frac{\mu_0\,(qf)}{2R}.$$
Step 4: Tidy the expression.
$$B = \frac{\mu_0 q f}{2R}.$$
Step 5: Check the dependences.
The field rises with more charge $q$ and faster spin $f$, and falls for a larger ring $R$ - all physically reasonable.
Step 6: State the result.
The magnetic induction at the centre is $\dfrac{\mu_0 q f}{2R}.$
\[ \boxed{B = \frac{\mu_0 q f}{2R}} \]