Topic: Electromagnetics
A ring of radius R carries a linear charge density $\lambda$. It is rotating with angular speed $\omega$. The magnetic field at its center is
- A. $\frac{3\mu_{0}\lambda\omega}{2}$
- B. $\frac{\mu_{0}\lambda\omega}{2}$
- C. $\frac{\mu_{0}\lambda\omega}{\pi}$
- D. $\mu_{0}\lambda\omega$
Answer: B. $\frac{\mu_{0}\lambda\omega}{2}$
Explanation: Total charge on the ring is $q = 2\pi R \cdot \lambda$. The current equivalent to the rotating charge is $I = \frac{q}{T} = \frac{2\pi R \lambda}{2\pi/\omega} = \lambda R \omega$. The magnetic field at the center of a circular current loop is $B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (\lambda R \omega)}{2R} = \frac{\mu_0 \lambda \omega}{2}$.