A long dielectric cylinder of radius `R` us statically plartized so that at all the its points the polarization is equal to `P = alpha x` , where `alpha` is a positive constant, and `r` is the distance from the axis. The cylinder is set into ratation about its axis with an angular velocity `omega`. FInd the magnetic induction `B` at the centre of the cylinder.

Explanation

Because of polarization a space charge is present within the cylinder. It's density is
`rho_(p) = -"div" vec(P) = -2 alpha`
Since the cylinder as a whole is neutral a surface charge density `sigma_(p)` must be present on the surface of the cylinder also. This has the magnitude (algebratically)
`simga_(p) xx 2pi R = 2 alpha pi R^(2)` or, `sigma_(p) = alpha R`
When the cylinder rotes, currents are set up which give rise to magentic fields. The contributes of `rho_(p)` and `sigma_(p)` can be calculated separaely and then added.
For the surface charge the current is (for a particular element)
`alpha R xx 2pi R dx xx (omega)/(2pi) = alpha R^(2) omega dx`
Its contribution to the magentic field at the centre is
`(mu_(0) R^(2) (alpha R^(2) omega dx))/(2(x^(2) + R^(2))^(3//2))`
and the total magentic field is
`B_(s) = int_(oo)^(-oo) (mu_0) R^(2) (alpha R^(2) omega dx)/(2 (x^(2) + R^(2))^(3//2)) = (mu_(0) alpha R^(4) omega)/(2) int_(-oo)^(oo) (dx)/((x^(2) + R^(2))^(3//2)) = (mu_(0) alpha R^(4) alpha)/(2) xx (2)/(R^(2)) = mu_(0) alpha R^(2) sigma`
As fpr the volume charge density consider a circle of radius `r`, radial thickness `dr` and length `dx`.
The current is `-2a xx 2pi r dr dx xx (omega)/(2pi) = -2 alpha r dr omega dx`
The total magnetic field due on the the volume charge distribution is
`B_(v) = -int_(0)^(R) dr int_(-oo)^(oo) dx 2pi r omega (mu_(0) r^(2))/(2(x^(2) + r^(2))^(3//2)) = -int_(0)^(R) alpha mu_(0) omega r^(3) dr int_(-oo)^(oo) dx (x^(2) + r^(2))^(3//2)`
`= -int_(0)^(R) alpha omega r dr xx 2 = -mu_(0) alpha omega R^(2)` so, `B = B_(s) + B_(v) = 0`