Respuesta :
Answer:
The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
Explanation:
Given that,
Radius = 2.2 cm
Charge [tex]q = 6.08\times10^{-6}\ C[/tex]
Angular speed = 2.01 rad/s
We need to calculate the time period
[tex]T = \dfrac{2\pi}{\omega}[/tex]
Now, The spinning produced the current
Using formula for current
[tex] I = \dfrac{q}{T}[/tex]
We need to calculate the magnetic moment
Using formula of magnetic moment
[tex]M = I A[/tex]
Put the value of I and A into the formula
[tex]M=\dfrac{q}{T}\times A[/tex]
[tex]M=\dfrac{q\times\omega}{2\pi}\times \pi\times r^2[/tex]
Put the value into the formula
[tex]M=\dfrac{6.08\times10^{-6}\times2.01\times(2.2\times10^{-2})^2}{2}[/tex]
[tex]M=2.96\times10^{-9}\ Am^2[/tex]
Hence, The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
