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


by cadmus
Tags: angular, cylinder, field, inside, magnetic, nonconstant, rotating, velocity
cadmus
cadmus is offline
#1
Jun17-11, 12:12 AM
P: 4
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
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Planck const
Planck const is offline
#2
Jun17-11, 03:00 AM
P: 15
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
cadmus is offline
#3
Jun17-11, 05:24 AM
P: 4
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
Planck const is offline
#4
Jun17-11, 07:36 AM
P: 15

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


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
cadmus is offline
#5
Jun18-11, 02:48 AM
P: 4
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]


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