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Magnetic field inside a cylinder which is rotating in a nonconstant angular velocity 
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#1
Jun1711, 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 nonconducting 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 intime magnetic field, is not changing with time? Thanks in advance 


#2
Jun1711, 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) 


#3
Jun1711, 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 


#4
Jun1711, 07:36 AM

P: 15

Magnetic field inside a cylinder which is rotating in a nonconstant 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] . 


#5
Jun1811, 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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