Q124 / 155 · WBJEE 2026 · Physics
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$\dfrac{\lambda\omega^2}{2 - \mu_0}$
- B$\dfrac{\mu_0\lambda^2\omega}{\sqrt{2}}$
- Correct answerC$\dfrac{\mu_0\lambda\omega}{2}$
- D$\dfrac{\mu_0\lambda}{2\omega^2}$
Solution
Total charge on the ring is $Q = \lambda (2\pi r)$.
Time period of rotation is $T = \dfrac{2\pi}{\omega}$.
Equivalent current due to the rotating charge is $I = \dfrac{Q}{T} = \dfrac{\lambda (2\pi r)}{2\pi / \omega} = \lambda \omega r$.
The magnetic field at the centre of a circular current-carrying loop is $B = \dfrac{\mu_0 I}{2r}$.
Substituting the value of $I$, we get $B = \dfrac{\mu_0 (\lambda \omega r)}{2r} = \dfrac{\mu_0 \lambda \omega}{2}$.
Answer: $\dfrac{\mu_0\lambda\omega}{2}$
Time period of rotation is $T = \dfrac{2\pi}{\omega}$.
Equivalent current due to the rotating charge is $I = \dfrac{Q}{T} = \dfrac{\lambda (2\pi r)}{2\pi / \omega} = \lambda \omega r$.
The magnetic field at the centre of a circular current-carrying loop is $B = \dfrac{\mu_0 I}{2r}$.
Substituting the value of $I$, we get $B = \dfrac{\mu_0 (\lambda \omega r)}{2r} = \dfrac{\mu_0 \lambda \omega}{2}$.
Answer: $\dfrac{\mu_0\lambda\omega}{2}$
Hence, the correct answer is option: 'C'.