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

The magnitude of the magnetic moment of the rotating ring is
`2.96*10^(-9)\ Am^2`

Step-by-step explanation:

Given that,

Radius = 2.2 cm

Charge
`q = 6.08*10^(-6)\ C`

Angular speed = 2.01 rad/s

We need to calculate the time period

`T = (2\pi)/(\omega)`

Now, The spinning produced the current

Using formula for current

`I = (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=(q)/(T)* A`

`M=(q*\omega)/(2\pi)* \pi* r^2`

Put the value into the formula

`M=(6.08*10^(-6)*2.01*(2.2*10^(-2))^2)/(2)`

`M=2.96*10^(-9)\ Am^2`

Hence, The magnitude of the magnetic moment of the rotating ring is
`2.96*10^(-9)\ Am^2`