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: Understanding the Question:
A thin ring of radius $R$ carries a uniformly distributed static charge $q$. When the ring spins about its center at a constant rotation frequency $f$, this moving charge creates a circular current. We need to find the resulting magnetic induction ($B$) at the geometric center of the spinning ring.

Step 2: Key Formula or Approach:
1. A rotating charge generates an effective electrical current loop given by the charge passing a point per unit time:
$$I = \frac{q}{T} = qf$$ where $T = \frac{1}{f}$ is the period of rotation. 2. The magnetic field induction $B$ at the center of a circular loop of radius $R$ carrying an electric current $I$ is given by Biot-Savart's law:
$$B = \frac{\mu_0 I}{2R}$$

Step 3: Detailed Explanation:
First, let's find the effective electric current $I$ created by the spinning ring. The entire charge $q$ completes one full revolution in a time period $T$. Since frequency $f = \frac{1}{T}$, the current can be expressed as:
$$I = qf$$ Now, substitute this current expression into the standard formula for the magnetic field at the center of a circular loop:
$$B = \frac{\mu_0 (qf)}{2R} = \frac{\mu_0 qf}{2R}$$ This algebraic substitution matches the configuration given in option (A).

Step 4: Final Answer:
The magnetic induction at the center of the ring is $\frac{\mu_0 qf}{2R}$, which corresponds to option (A).