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Verdich [7]
4 years ago
11

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 arou

nd 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?
Physics
1 answer:
Alex787 [66]4 years ago
5 0

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

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