Question

A uniform non-conducting ring of radius 2.2 cm and total charge 6.08 µC rotates with a constant angular speed of 2.01 rad/s around an axis perpendicular to the plane of the ring that passes through its center. What is the magnitude of the magnetic moment of the rotating ring?

Answer 1

Answer:

The magnitude of the magnetic moment of the rotating ring is $2.96\times10^{-9}\ Am^2$

Explanation:

Given that,

Radius = 2.2 cm

Charge $q = 6.08\times10^{-6}\ C$

Angular speed = 2.01 rad/s

We need to calculate the time period

$T = \dfrac{2\pi}{\omega}$

Now, The spinning produced the current

Using formula for current

$I = \dfrac{q}{T}$

We need to calculate the magnetic moment

Using formula of magnetic moment

$M = I A$

Put the value of I and A into the formula

$M=\dfrac{q}{T}\times A$

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

Put the value into the formula

$M=\dfrac{6.08\times10^{-6}\times2.01\times(2.2\times10^{-2})^2}{2}$

$M=2.96\times10^{-9}\ Am^2$

Hence, The magnitude of the magnetic moment of the rotating ring is $2.96\times10^{-9}\ Am^2$