Calculating Electric Field Inside a Wire with Resistivity and Current

[nikolafmf]

Homework Statement

Griffiths says in his "Introduction to Electrodynamics" that electric field inside a conductor is 0, but isnide a wire is different from 0. Since wire is also a conductor, how can that be possible? And how to calculate electric field inside a wire? Wire has resistivity \rho, radius a and current I trough it.

Homework Equations

E= 0 inside a conductor;
E = \frac{I\rho}{\pi a^2}z (How to calculate this?)

The Attempt at a Solution

Have no idea how to start...

 

[ideasrule]

The wire is not a perfect conductor. A perfect conductor has 0 resistivity, which implies no electric field via your second equation.

 

[nikolafmf]

Thank you ideasrule for your response. Any idea how to calculate field in a wire and get my second equation?

 

[Doc Al]

Griffiths says in his "Introduction to Electrodynamics" that electric field inside a conductor is 0, but isnide a wire is different from 0. Since wire is also a conductor, how can that be possible?

The electric field is zero within a conductor only in the electrostatic case. If you have a current in a wire, then you can certainly have a non-zero electric field.

 

[nikolafmf]

The electric field is zero within a conductor only in the electrostatic case. If you have a current in a wire, then you can certainly have a non-zero electric field.

Yes, it should be expected the field to set electrons in motion, or...? And how does this fields comes here in nonelectrostatic case?

 

[nikolafmf]

So, the question here arises is under what conditions is electric field inside a conductor zero and when is it nonzero? And why? Griffiths only explains that when we put conductor in an outer electric field, the field inside is still zero, as is zero without outer field. Then if there is current, the field is as in second equation. But he doesn't explain this.

 

[aim1732]

The basic question you leave unanswered is why does the field become zero inside an ideal conductor.It does not do that instantly.The external field sets charges in motion which,free to move,set up an electric field that exactly cancels the applied field.That takes time although that is measured on the nano scale.

In a wire the charges do not accumulate---necessary for setting up an electric field.The charges keep moving--drifting--to set up a current and not a field of their own. If you read good books such as Resnick-Halliday it will say that field inside is zero for an ideal ISOLATED conductor.

 

[yungman]

The electric field is zero within a conductor only in the electrostatic case. If you have a current in a wire, then you can certainly have a non-zero electric field.

i thought E is always 0 in a perfect conductor regardless of electrostatic or electrodynamic. Any field inside will immediately cause electrons to move in direction to cancel the field.

I understand about imperfect conductor have longitudinal E/B] and skin effect and all, but not with perfect conductor.

 

[tiny-tim]

Any field inside will immediately cause electrons to move in direction to cancel the field.

they'll only cancel the field if you don't replenish it!

as Doc Al says …

The electric field is zero within a conductor only in the electrostatic case. If you have a current in a wire, then you can certainly have a non-zero electric field.

so yes, a field inside will immediately cause electrons to move, but if you keep the field going (eg by using a battery), then the electrons will never cancel it!

 

[Doc Al]

i thought E is always 0 in a perfect conductor regardless of electrostatic or electrodynamic. Any field inside will immediately cause electrons to move in direction to cancel the field.

Yes, you are correct for a perfect conductor. But I thought we were talking about ordinary imperfect conductors, such as a copper wire. In such a case you can certainly have a non-zero field within the conductor to drive a current (as you well know).

aim1732 gave an excellent answer, which I agree with. (And I apologize to the OP for forgetting about this thread and not giving a more complete answer myself.)

To nikolafmf: To get your second equation, think of it as being equivalent to Ohm's law.

(Didn't see you sneak in there, tiny-tim. )