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A charge Q is uniformly distributed over the surface of a right circular cone of semi-vertical angle $θ$ &height h. The cone is uniformly rotated about its axis at angular velocity $ω$ calculate magnetic moment.

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\frac{Q ω h}{4} {Tan}^{2} θ

\frac{Q ω h^{2}}{3} {Tan}^{2} θ

\frac{Q ω h^{2}}{4} {Tan}^{2} θ

\frac{Q ω h^{2}}{4} Tan θ

answer is C.

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Detailed Solution

$μ = N I A = \frac{I ω}{2 π} . A$

Dipole moment of ring

$d μ = \frac{θ}{2 π R l} 2 π r× \frac{d x}{\cos θ} \times π r^{2}$

$and \frac{R}{r} = \frac{h}{h - x}$

$\int d μ = \frac{Q ω R^{2}}{h^{4}} \int_{θ}^{h} {(h - x)}^{3} d x$

$= \frac{Q ω R^{2}}{4} = \frac{Q ω h^{2}}{4} {Tan}^{2} θ$

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