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An insulating thin rod of length $l$ has a linear charge density $ρ (x) = ρ_{0} \frac{x}{ℓ} on it. The rod is rotated about$ an axis passing through the origin (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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$\frac{π}{4} n ρ ℓ^{3}$

$π n ρ ℓ^{3}$

$\frac{π}{3} n ρ ℓ^{3}$

$n ρ ℓ^{3}$

answer is A.

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Detailed Solution

$ω = 2 π nrad / s$

$\begin{aligned} d Q = ρ \cdot d x = \frac{ρ_{0}}{ℓ} x d x \\ d I = \frac{d Q \cdot ω}{2 π} \end{aligned}$

$\begin{aligned} d M = d I \times A \\ \int d M = \int_{0}^{ℓ} \frac{ω}{2 π} \cdot \frac{ρ_{0}}{ℓ} x \cdot π x^{2} d x \end{aligned}$

$M = \frac{π n ρ_{0} ℓ^{3}}{4}$

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