Q charge is uniformaly distributed over the same surface of a right circular cone of semi -vertical angle theta and height h The cone is uniformly rotated about its axis at angular velocity omega Calculated associated magnetic dipole moment
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Answer

`(Q_omega)/(4) h^(2) tan^(2)theta`

Explanation

`dq = (q)/(piRh) . 2 pi y dx`
`(h)/(a) = tan theta implies h =R tan 0…(2)`
`(y)/(x) = tan 0 implies y =xtan 0..(3)`
From (1)(2) and (3)
`dq = (q)/(pih^(2) tan theta) 2pi tan theta dx`
`di = ((dq)omega)/(2pi) = (q^(2) pi xx tan theta dx)/(pih^(2) tan theta) . (omega)/(2pi) = (qomega)/(pih^(2) xdx`
`dm = di A = (qomega)/(pih^(2)) xdx. (piy^(2)) = (qomega)/(pih^(2)) xdx pix^(2) tan^(2)0`
` = m (qomega)/(h^(2)) tan^(2)theta.underset(0)overset(h)intx^(3) dx`
`=(qomega)/(h^(2)) tan^(2)theta. (h^(4))/(4) = (qomegah^(2)tan^(2)theta)/(4)`
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