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 Question:
A circular ring of radius \( r \) with a linear charge density \( \lambda \) is rotating with a uniform angular velocity \( \omega \). We need to determine the magnitude of the magnetic field \( B \) produced at its center.

Step 2: Key Formula or Approach:
- Total charge \( q \) on a ring of radius \( r \) with linear charge density \( \lambda \):
\[ q = 2\pi r \lambda \]
- Equivalent current \( I \) due to a charge \( q \) completing one rotation in time period \( T \):
\[ I = \frac{q}{T} = \frac{q \omega}{2\pi} \]
- Magnetic field \( B \) at the center of a circular current loop of radius \( r \):
\[ B = \frac{\mu_0 I}{2r} \]

Step 3: Detailed Explanation:
1. Calculate the total charge \( q \) on the ring:
\[ q = \lambda (2\pi r) \]
2. Determine the equivalent electric current \( I \) produced by the rotation:
\[ I = \frac{q}{T} = \frac{2\pi r \lambda}{\frac{2\pi}{\omega}} = \lambda r \omega \]
3. Calculate the magnetic field \( B \) at the center of the ring:
\[ B = \frac{\mu_0 I}{2r} = \frac{\mu_0 (\lambda r \omega)}{2r} \]
\[ B = \frac{\mu_0 \lambda \omega}{2} \]
We observe that the magnetic field is independent of the radius \( r \) of the ring.

Step 4: Final Answer:
The magnitude of the magnetic field at the center of the rotating ring is \( \frac{\mu_0 \lambda \omega}{2} \).