uniformly charged disc of radius R and total charge Q is rotating about its axis passing through the centre of diae and perpendicular to the plane of dise, with an angular velocity `omega`. Calculate its magnetic dipole moment. Also find the ratio of angular momentum to that with the calculated magnetic moment of the system.

Explanation

Consider a disc of radius R having centre at O.
Given that, Q is the total charge on disc.
Consider a small element of thickness dx at a distance x from the centre.
dq = Charge on this small element ` = Q/(pi R^(2)) xx 2pi x dx `

Magnetic moment `(d mu)` of mall element ` = 1/2 d q omegax^(2)`
` = 1/2 (Q/(piR^(2)) * 2pi xdx) omegax^(2)`
` dmu = (Qomega)/R^(2) x^(3) dx `
The total magnetic moment can be obtained by integrating between the limits 0 to R.
`mu = intdmu = underset(0)overset(R)int (Qomega)/R^(2) x^(3) dx `
` = (Qomega)/R^(2) [x^(4)/4]_(0)^(R) `
` = 1/4 (Qomega)/R^(2) R^(4)`
` = 1/4 Q omegaR^(2)`
Angular momentum , ` L = I omega = (MR^(2))/(2) omega`
` "Magnetic moment"/"Angular momentum" = 1/4 (QomegaR^(2))/(MR^(2) omega) xx 2 `
` = Q/(2M)`