A cylinder of electric field varying in time

  • #1

Main Question or Discussion Point

Hi, I have this problem:
In empty space there is an infinite cylinder, with its axis parallel to z axis and radius a, filled with an eletric field of equation
[tex]\vec{E}(t) = E_0 e^{\beta t} \hat{z}[/tex]
Now I put a rectangular wire on the plane yz out of the cylinder of side l and b (l lies on the y axis) and the question is: which is the current on the wire?

I try to attack the problem in the straightforward way. So I try to solve Maxwell's fourth equation
[tex]\vec{\nabla} \times \vec{B} = \frac{1}{c} \partial_t \vec{E}[/tex]
and initially I thought that I had to compute the flux through the wire of the varying magnetic field I get.....but I realize that this is impossible because this equation is only valid inside the cylinder, where I have a varying electric field, and not outside where there is only empty space!!!

Could anyone help...please?!?
 

Answers and Replies

  • #2
pam
455
1
Use Ampere's law (with the displacement current) to get B(t) outside the cylinder.
 
  • #3
What do you mean with Amperè's Law?? Maybe the fourth Maxwell's Equation?? This one?
[tex]\vec{\nabla} \times \vec{B} = \mu_0 \vec{J} + \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t}[/tex]
And if I do?? The displacement current is zero outside the cylinder...
 
  • #4
pam
455
1
Using Stoke's theorem, you can derive Ampere's law:
[tex]\oint{\bf dr}\cdot{\bf B}=\frac{1}{c^2}\int{\bf dS}\cdot\partial_t{\bf E}[/tex].
 
  • #5
Ok...I get what you mean. So I choose a cylinder of radius r>a and height z, evaluate the flux of [tex]\frac{\partial \vec{E}}{\partial t}[/tex] and thanks to Ampere's Law this is equal to the path integral of B over a circle of radius r in the at height z parallel to xy plane. May I say that B is always tangential so the path integral is easy to evaluate?? If yes why?? I thought in analogy with the magnetic field generated by a wire...

Thank you

Ciao!!
 
  • #6
pam
455
1
It is just like a wire, but it doesn't matter.
For the loop you have only B tangential enters the flux.
 

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