Question

A charge ‘q’ is given to a solid conducting sphere of mass ‘m’ and is rotating about it’s diameter with angular velocity ω. Then ratio of it’s magnetic moment to its angular moment
​

Answer 1

Given:

A charge ‘q’ is given to a solid conducting sphere of mass ‘m’ and is rotating about it’s diameter with angular velocity ω.

To find:

Ratio of magnetic moment to its angular momentum.

Calculation:

We know that angular velocity is given by the ratio of angular displacement and time:

$\boxed{ \omega = \dfrac{ \theta}{t} = \dfrac{2\pi}{t} }$

Let magnetic moment be denoted as M:

$\sf{ \therefore \: M = i \times area}$

$\sf{ = > M = \dfrac{q}{t} \times \pi {r}^{2} }$

$\sf{ = > M = \dfrac{q}{ (\frac{2\pi}{ \omega}) } \times \pi {r}^{2} }$

$\sf{ = > M = \dfrac{ \omega q}{ 2\pi} \times \pi {r}^{2} }$

$\sf{ = > M = \dfrac{ \omega q {r}^{2} }{ 2 }}$

Let angular momentum be L

$\sf{L = moment \: of \: inertia \times \omega}$

$\sf{ = > L = \dfrac{2m {r}^{2} }{5} \times \omega}$

$\sf{ = > L = \dfrac{2m {r}^{2} \omega}{5} }$

So, required ratio:

$\sf{ = >M : L = \dfrac{ q\omega {r}^{2} }{2} : \dfrac{2m {r}^{2} \omega}{5} }$

$\boxed{ \sf{ = >M : L = 5q: 4m }}$