# EM problem

## Main Question or Discussion Point

Consider a wire, if I suddenly pass current on it. There would be an instant localized B field around the wire. Since there is sudden change of B field, there will also be a finite E field. According to Maxwell equation E=B/c.

As the B field spread out the E field will follow, even when the current is stable and B field stop changing the E field will still be there with the B field. But this has got to be wrong right? Otherwise whenever we turn on a solenoid there will be an observable E field around it!!??

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neu
The equation you mentioned is not valid.

I presume you mean:
$$\nabla\times\textbf{E}=-\frac{1}{c}\frac{\partial E}{\partial t}$$

from Wikipedia:
"Faraday's law of induction states that the induced electromotive force in a closed loop of wire is directly proportional to the time rate of change of magnetic flux through the loop."

So you would get a current in a conductor MOVING through the B-field of a solenoid, but not when static.

Andy Resnick
Neu,

I think you made an error in the LaTex formula (enroger's is not correct, either). The appropriate equation is:

$$\nabla\times\textbf{H} = \frac{1}{c}( \frac{\partial}{\partial t}\textbf{E}+4\pi \textbf{i})$$

Differentiating this, to get a time-varying current, and I think you will find that when a solenoid is turned on or off, a pulse of EM radiation is emitted, in accordance with everyday experience. The EEs in the audience should be able to give a better idea of what happens

neu
yeah sorry i meant:
$$\nabla\times\textbf{E}=-\frac{\partial \textbf{B}}{\partial t}$$

The fact that there is a pulse of E field accompany with B field when the solenoid got turn on has no problem. What I want to know is: When there is no change in current a long time after turn on, therefore no change in B field around the wire, would there still be a E field around the wire?

I think the answer is no, but I can't derive that out of maxwell equation. help?