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A rod of length $L$ has charge distributed on it such that the linear charge density $λ = λ_{0} x^{2},$ $λ_{0}$ is constant and $x$ is the distance of a point $A$ . The rod is being rotated about $A$ with constant angular velocity $ω$ , the equivalent current is

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$\frac{λ_{0} ω L^{3}}{2}$

$\frac{λ_{0} ω L^{2}}{6 π}$

$\frac{λ_{0} ω L^{3}}{6 π}$

$\frac{λ_{0} ω L^{3}}{3 π}$

answer is C.

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Detailed Solution

$d q = λ d x = λ_{0} x^{2} d x$

$q = λ_{0} \int_{0}^{L} x^{2} d x = \frac{λ_{0} L^{3}}{3}$

$T = \frac{2 π}{ω}$

$I = \frac{q}{T} = \frac{λ_{0} L^{3} / 3}{2 π / ω} = \frac{λ_{0} ω L^{3}}{6 π}$

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