A cylinder of electric field varying in time

1. May 15, 2008

alle.fabbri

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
$$\vec{E}(t) = E_0 e^{\beta t} \hat{z}$$
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
$$\vec{\nabla} \times \vec{B} = \frac{1}{c} \partial_t \vec{E}$$
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?!?

2. May 15, 2008

pam

Use Ampere's law (with the displacement current) to get B(t) outside the cylinder.

3. May 15, 2008

alle.fabbri

What do you mean with Amperè's Law?? Maybe the fourth Maxwell's Equation?? This one?
$$\vec{\nabla} \times \vec{B} = \mu_0 \vec{J} + \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t}$$
And if I do?? The displacement current is zero outside the cylinder...

4. May 15, 2008

pam

Using Stoke's theorem, you can derive Ampere's law:
$$\oint{\bf dr}\cdot{\bf B}=\frac{1}{c^2}\int{\bf dS}\cdot\partial_t{\bf E}$$.

5. May 15, 2008

alle.fabbri

Ok...I get what you mean. So I choose a cylinder of radius r>a and height z, evaluate the flux of $$\frac{\partial \vec{E}}{\partial t}$$ 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. May 16, 2008

pam

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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