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Question

A rod of length l having uniformly distributed charge Q is rotated about one end with constant frequency f. Its magnetic moment is

A
2πfQl2
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B
2πfQl23
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C
πfQl2
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D
πfQl23
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Solution

The correct option is D πfQl23

Let the linear charge density of the rod is λ.

So, charge on dx will be

dq=λdx

The magnetic moment of length dx is given by

dM=(dI)A

dM=(dqT)πx2=(fλdx)πx2 (f=1/T)

dM=πfλ(x2dx)

The dipole moment of the rod will be

M=dM=l0πfλ(x2dx)

M=πfλ[x33]l0=πfλ×l33

M=πQfl23 (λ=Q/l)

Hence, option (B) is correct.

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