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Question

Physics Question on Electromagnetic induction

An insulating thin rod of length \ell has a x linear charge density p(x)=ρ0xp(x) = \rho_0 \frac{x}{\ell} on it . The rod is rotated about an axis passing through the origin (x=0)(x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is :

A

π4nρ3\frac{\pi }{4} n \rho \ell^3

B

nρ3 n \rho \ell^3

C

πnρ3 \pi n \rho \ell^3

D

π3nρ3\frac{\pi }{3} n \rho \ell^3

Answer

π4nρ3\frac{\pi }{4} n \rho \ell^3

Explanation

Solution

M=NIA\because M = NIA
dq=λdx&A=πx2dq =\lambda dx \& A =\pi x ^{2}
dm=(x)ρ0xdx.πx2\int dm = \int\left(x\right) \frac{\rho_{0}x}{\ell}dx . \pi x^{2}
M=nρ0π.0x3.dx=nρ0π.[L44]M = \frac{n \rho_{0}\pi}{\ell} . \int^{\ell}_{0} x^{3} .dx = \frac{n \rho_{0} \pi}{\ell} . \left[\frac{L^{4}}{4}\right]
M=nρ0π34M = \frac{n\rho_{0} \pi \ell^{3}}{4} or π4nρ3\frac{\pi}{4} n\rho\ell^{3}