Question

A non-conducting ring carries linear charge density $\lambda$. It is rotating with angular speed $\omega$ about its axis. The magnetic field at its centre is

• A. $\frac{3\mu_{0} \lambda\omega}{2\, \pi}$
• B. $\frac{ \mu_{0} \lambda \omega}{2 }$
• C. $\frac{ \mu_{0} \lambda \omega}{ \pi}$
• D. $\mu_{0} \lambda \omega$

Answer 1

Answer: (B)

$\frac{ \mu_{0} \lambda \omega}{2 }$

Answer 2

Let the radius of the ring is $r$. Charge on the ring, $q = l \lambda=2 \pi r \lambda [\because l =2 \pi r ]$ Current due to the rotation of the ring, $i =\frac{ q }{ t }=\frac{2 \pi r \lambda}{\frac{2 \pi}{\omega}}$ $\left[\because t =\frac{2 \pi}{\omega}\right]$ $=\omega \lambda r$ $i =\omega \lambda r$...(i) $\left[\because t =\frac{2 \pi}{\omega}\right]$ $\therefore$ Magnetic field at the centre of the ring due to the rotation of the charged ring, $B =\frac{\mu_{0}}{2} \cdot \frac{ i }{ r }=\frac{\mu_{0}}{2} \cdot \frac{\omega \lambda r }{ r }$ $=\frac{\mu_{0}}{2} \cdot \omega \lambda$ [using E (i)]