Magnetic field inside a cylinder which is rotating in a non-constant angular velocity

cadmus

1. The problem statement, all variables and given/known data

A hollow cylinder of length L and radius R, is madeout of a non-conducting material, is charged with a constant surface charge σ, and is rotating, along its axis of symmetry, with an angular velocity w(t) = αt.

Q:What is the magnetic field inside the cylinder?


2. Relevant equations

Maxwell correction for Ampere law.

3. The attempt at a solution


The answer in the manual is B = μαtRσ

Where μ is ofcurse μ zero. [ the magnetic constant ].

The manual's solution makes perfect sense if I knew that the circular electric field which is induced by the fact that the magnetic field is changing in time is constant.

because then i could say that that the displacement current density is zero.

Q: How can derive that the circular electric field, induced by the changing -in-time magnetic field, is not changing with time?


Thanks in advance

Planck const

What is I?
its the relation betwwen the charge (you know it from sigma) and the period (you know it from w)
The charge inside the cylinder its Sigma*A(r) (and not all A!!)
What is the integral of B*ds ?
its the product of B and the scale circuits that thir radius its r (r<R)

You need to replace those sizes into Amper equation.. and get B(r)

cadmus

Thanks for the reply.

But it did not address my question,

I would like to know why in this problem there is a certainty that the Electric field is not changing with time ?

id est, look at Ampere's Law after Maxwell correction:

[itex]\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}\ [/itex]

and of course, the integral form of this equation:

[itex]\oint_{\partial S} \mathbf{B} \cdot \mathrm{d}\mathbf{l} = \mu_0 I_S + \mu_0 \varepsilon_0 \frac {\partial \Phi_{E,S}}{\partial t} [/itex]

I could, with rather ease, derive the manual's solution if I knew that [itex] \frac{\partial \mathbf{E}} {\partial t} \ [/itex] is zero.

Any notions about why E is constant in time?

Thanks in advance

Planck const

The answer is simple. (I will call sigma -> rho)
if dE/dt (partial derivative) = 0

==> (Gauss law)

d[itex]\rho[/itex]/dt = 0

==> (math)

[itex]\rho[/itex] is constant in time. (stationary current)

And you can see in the problem data, that they didn't say anything about the function [itex]\rho[/itex] .

cadmus

Again, thanks for the reply.

The [itex] \nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}[/itex] formula applies only to E filed that are not circular. [ I mean in order to derive the total electric field inside the cylinder you will have to find the E in the theta direction as well]

According to Faraday's Law:
[itex] \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}[/itex]

Which means that a changing in time magnetic field induces a circular E field.

How can I infer that the circular E field is not changing by time? [ prior to calculating the magnetic field - because then i just use faraday law to see that]