A charge q is uniformly distributed on a nonconducting disc of radius R.It is rotated with an angular speed $ω$ about an axis passing through the centre of mass of the disc and perpendicular to its plane.Find the magnetic moment of the disc.
[Hint: Magnetic moment = $(\frac{q}{2 m})$ (angular momentum)]

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Solution

$\frac{M}{L} = \frac{q}{2 m}$
$M = (\frac{q}{2 m}) L = (\frac{q}{2 m}) (I ω)$ $= (\frac{q}{2 m}) (\frac{1}{2} m R^{2}) (ω) = \frac{q R^{2} ω}{4}$

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A charge q is uniformly distributed on a nonconducting disc of radius R.It is rotated with an angular speed ω about an axis passing through the centre of mass of the disc and perpendicular to its plane.Find the magnetic moment of the disc.[Hint: Magnetic moment =(q2m)(angular momentum)]

ML=q2mM=(q2m)L=(q2m)(Iω)=(q2m)(12mR2)(ω)=qR2ω4

$\frac{M}{L} = \frac{q}{2 m}$
$M = (\frac{q}{2 m}) L = (\frac{q}{2 m}) (I ω)$ $= (\frac{q}{2 m}) (\frac{1}{2} m R^{2}) (ω) = \frac{q R^{2} ω}{4}$