Question

There is a ring of radius \( r \) having linear charge density \( \lambda \) and rotating with a uniform angular velocity \( \omega \). The magnitude of the magnetic field produced by this ring at its own centre would be (\( \mu_0 \)= permeability of air)

• A. \( \frac{\lambda \omega^2}{2 - \mu_0} \)
• B. \( \frac{\mu_0 \lambda^2 \omega}{\sqrt{2}} \)
• C. \( \frac{\mu_0 \lambda \omega}{2} \)
• D. \( \frac{\mu_0 \lambda}{2\omega^2} \)

Hint:

Whenever a charge \( q \) is distributed uniformly over a ring of radius \( r \) rotating with angular velocity \( \omega \), the equivalent current is always \( I = \frac{q\omega}{2\pi} \). Since \( q = 2\pi r \lambda \), the radius cancels out during the magnetic field calculation at the center.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A rotating charged body creates a circular current.
Current is defined as the rate of flow of charge. For a rotating ring, the charge passing through a point in one period is the total charge on the ring.
A circular current loop produces a magnetic field at its center, which can be calculated using the Biot-Savart Law.
Step 2: Key Formula or Approach:
Total charge on the ring: \(Q = \text{linear density} \times \text{circumference} = \lambda (2\pi r)\).
Equivalent current: \(I = \frac{Q}{T} = Q \cdot f = Q \cdot \frac{\omega}{2\pi}\).
Magnetic field at the center of a circular loop: \(B = \frac{\mu_0 I}{2r}\).
Step 3: Detailed Explanation:
Step A: Find the total charge \(Q\) on the ring.
The ring has a radius \(r\) and linear charge density \(\lambda\).
\[ Q = \lambda \cdot (2\pi r) \]
Step B: Find the equivalent current \(I\) produced by the rotation.
The ring completes one rotation in time \(T = \frac{2\pi}{\omega}\).
During this time \(T\), the entire charge \(Q\) passes through any fixed cross-section.
\[ I = \frac{Q}{T} = \frac{\lambda(2\pi r)}{2\pi / \omega} \]
\[ I = \lambda r \omega \]
Step C: Calculate the magnetic field \(B\) at the center.
The formula for the magnetic field at the center of a circular current-carrying loop is:
\[ B = \frac{\mu_0 I}{2r} \]
Substituting the expression for \(I\):
\[ B = \frac{\mu_0 (\lambda r \omega)}{2r} \]
The radius \(r\) cancels out from the numerator and denominator:
\[ B = \frac{\mu_0 \lambda \omega}{2} \]
This matches option (C).
Step 4: Final Answer:
The magnetic field is independent of the radius of the ring when expressed in terms of linear charge density and angular velocity. The final expression is \(B = \frac{\mu_0 \lambda \omega}{2}\).