An insulating thin rod of length $l$ has a linear charge density $ρ (x) {=ρ}_{0} \frac{x}{l}$ on it, where $x$ is distance measured along the rod from one end of the rod and $ρ_{0}$ is a positive constant. The rod is rotated about an axis passing through the end at (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

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${nρ}_{0} l^{3}$

$π {nρ}_{0} l^{3}$

$\frac{π}{2} {nρ}_{0} l^{3}$

$\frac{π}{4} {nρ}_{0} l^{3}$

answer is D.

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Detailed Solution

Consider a small element of charge of thickness $d x$ at a distance $x$ from the end at $x = 0$ .
Charge on the element, $d q = ρ d x = \frac{ρ_{0}}{l} d x$
Current due to the element, $d i = \frac{d q}{T} = \frac{n ρ_{0} x}{l} d x$
Magnetic moment due to the elemental current, $d M = di (area) = \frac{n ρ_{0} x}{l} d x (π x^{2})$
Total magnetic moment, $M = \int d M = \frac{n ρ_{0} π}{l} \int_{0}^{l} x^{3} d x = \frac{n ρ_{0} π}{l} \frac{l^{4}}{4} = \frac{π}{4} n ρ_{0} l^{3}$