Question

A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity \(\omega\) with resect to normal axis then the magnetic moment of the loop is :

• A. \(q\,\omega r^2\)
• B. \(\frac{4}{3}\,q\,\omega r^2\)
• C. \(\frac{3}{2}\,q\,\omega r^2\)
• D. \(\frac{1}{2}\,q\,\omega r^2\)

Hint:

A rotating charged ring is equivalent to a current loop with \(I=\frac{q\omega}{2\pi}\).

Answer 1

The Correct Option is D

Time period \(T=\frac{2\pi}{\omega}\), so the equivalent current is \(I=\frac{q}{T}=\frac{q\omega}{2\pi}\).
Magnetic moment \(\mu=IA=\frac{q\omega}{2\pi}\cdot\pi r^2=\frac12q\omega r^2\).
Answer: (4) \(\frac{1}{2}\,q\,\omega r^2\)