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
.

Options

A. `(Qomega tan^(2)thetah^(2))/(4)hat(n)`
B. `(Qomega tan^(2)thetah^(2))/(2)hat(n)`
C. `(Qomega tan^(2)thetah^(2))/(3)hat(n)`
D. `Q omega tan^(2) thetah^(2)hat(n)`

Answer

A

Explanation

Surface area of the cone `=pih^(2) tan theta sec theta` hence , charge per unit surface area
`=(Q)/(pi h^(2)tan theta sec theta)`
`=(Q)/(pi h^(2)tan theta sec theta)(2pi)(x tan theta )(dx)/((cos theta))((2Q)/(h^(2)))xdx`
The current `dl` due to the rotation this charge is given by

`dl=((omega)/(2pi))((2Q)/(h^(2)))xdx`
Hence, the magnetic moment `(d mu)` associated with this current would be.
`dvec(mu)=((Qomega)/9pi h^(2))(x)(pi)(x than theta)^(2)(dx)hat(n)`
`=((Qomega tan^(2)theta ))/(h^(2))x^(3)dxhat(n)`
`(hat(n)=` unit vector along the axis )
`vec(mu)=((Qomega tan^(2)theta)/(h^(2)))(int_(0)^(h)x^(3)dx)hat(n)`
`=(Qomegatan^(2)thetah^(2))/(4)hat(n)`