A nonconducting ring of radius r and mass m carries a total charge q that is uniformly distributed over it. The ring is rotated about its axis with an angular speed ω . If l represents the angular momentum of the ring

Question

Quantrex Academy · Accepted Answer

Correct answer: A. μ = (q)/(2m) l

The equivalent current due to the rotating charge is given by

I = \dfrac{q}{T} = \dfrac{q \omega}{2\pi}

.

The magnetic moment of the ring is

\mu = I A = \left(\dfrac{q \omega}{2\pi}\right) (\pi r^2) = \dfrac{1}{2} q \omega r^2

.

The angular momentum of the ring about its axis of rotation is

l = m r^2 \omega

.

Taking the ratio of magnetic moment to angular momentum, we get:

\dfrac{\mu}{l} = \dfrac{\dfrac{1}{2} q \omega r^2}{m r^2 \omega} = \dfrac{q}{2m}

Therefore,

\mu = \dfrac{q}{2m} l

.

A nonconducting ring of radius r and mass m carries a total charge...

Options

Correct answer

Step-by-step solution

More from Magnetic Field