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Q.

A rod has total charge ‘Q’ uniformly distributed along it length ‘L’. If the rod rotates with angular velocity ‘ $ω$ ’ about one end, its magnetic moment is

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a

$\frac{3 {qω}^{2} L}{2}$

b

$\frac{{qωL}^{2}}{4}$

c

$\frac{5 {qωL}^{2}}{4}$

d

$\frac{{qωL}^{2}}{6}$

answer is D.

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Detailed Solution

For small length ‘dx’ at ‘x’ distance, $dq = \frac{q}{L} dx$
Current, $i = \frac{dq}{T} = \frac{Q \cdot dx / L}{2 π / ω} = \frac{q ω dx}{2 πL}$
$dM = i. {πx}^{2} = \frac{qω}{2 L} x^{2} \cdot dx$
Total $M = \int_{0}^{L} dM = \frac{q ω}{2 L}, \frac{x^{3}}{3}$
$M = \frac{q ω L^{2}}{6}$

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