Question

A charge of e coulomb is uniformly distributed over the surface of a spherical shell of radius r. If the sphere rotates with angular velocity omega, find the magnetic dipole moment developed.

Answer 1

Answer:

Magnetic moment of the sphere is given as

$M = \frac{er^2\omega}{5}$

Explanation:

As we know that the ratio of magnetic moment and angular momentum is always conserved

so we have

$\frac{M}{L} = \frac{q}{2m}$

now we have

$M = \frac{q}{2m} \times L$

here we know that angular momentum of the sphere is given as

$L = I\omega$

$L = \frac{2}{5}mr^2 \omega$

now we have

$M = \frac{e}{2m} (\frac{2}{5} mr^2\omega)$

$M = \frac{er^2\omega}{5}$

#Learn

Topic : magnetic Moment

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