Question

A charge q is uniformly distributed on a non-conducting disc of radius R. It is rotated with an angular speed co about an axis passing through the centre of mass of the disc and perpendicular to its plane. Find the magnetic moment of the disc.

Answer 1

The magnetic moment of the disc is :- qωR²/4

• Let us assume a circular strip at a distance x from the centre of width dx, then change dq = 2πxdx*q/(πR²)

• As current is rate of flow of charge so dI = dq/dt = 2πxdx*q*ω/(πR²*2π)

Magnetic moment = IA ⇒ dm = dI*dA , upon integrating we get

M = qωR²/4

Answer 2

The magnetic moment of the disc is (qR²ω)/4

Explanation:

The magnetic dipole moment is given as:

M/L = q/2m

⇒ M = (q/2m)L

Angular momentum = L = ω (given)

Now, the magnetic dipole moment  becomes,

M = (q/2m) × ω

The angular momentum is given by the formula,

ω = 1/2 mR²

On substituting the angular momentum, we get,

M = (q/2m) × (1/2 mR²)

∴ M = 1/4 × (qR²ω)