Faraday's law. Induced electric field.

by disknoir
Tags: electric, faraday, field, induced
disknoir is offline
Apr20-13, 10:46 AM
P: 19
When you use Faraday's law to calculate the induced electric field due to a changing magnetic flux, you integrate over a loop defined by the circuit you're interested in.

Why is the electric field confined to the circuit? Couldn't I just pick a random loop in space and integrate over that?

Why can't free space support the magnetic field?

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disknoir is offline
Apr20-13, 11:06 AM
P: 19
What I meant to say was, why can't free space support the electric field?
vanhees71 is offline
Apr20-13, 04:18 PM
Sci Advisor
P: 2,132
The answer is that you can use any loop in Faraday's Law. There's no need for a wire or something. Often, however it's convenient to choose the path through a current conducting wire since, if it's thin enough, the electric field is homogeneous inside it in very good approximation.

One should keep in mind that there is a lot of confusion concerning Faraday's Law in the literature, because often they present the integral form of this law first, and usually they present it incompletely or for special cases. The basic laws of electromagnetism, valid in any inertial frame of reference are Maxwell's equations in differential form. Faraday's law reads
[tex]\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}=0.[/tex]
This holds true in any case. It's a fundamental law of physics!

Now you can integrate the whole equation over an arbitrary area and use Stokes's integral theorem to get:
[tex]\int_{\partial A} \mathrm{d} \vec{x} \cdot \vec{E}=-\frac{1}{c} \int_{A} \mathrm{d}^2 \vec{F} \cdot \vec{B}.[/tex]
Here, you can choose any area with any boundary you like. It doesn't need to be a wire or anything else physical. It can be just abstract mathematical areas and boundary lines.

The confusion now comes from the fact that usually one tries to write the time derivative outside of the surface integral and then makes more or less explicitly assumptions, which are not always valid. It's way easier to first look at the most general case of a moving surface with moving boundary. You can prove that in this most general case the Faraday law takes the integral form
[tex]\int_{\partial A} \mathrm{d} \vec{x} \cdot \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{b} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d}^2 \vec{F} \cdot \vec{B}.[/tex]
Here [itex]\vec{v}=\vec{v}(t,\vec{x})[/itex] is the velocity of each point at the surface-boundary line. The left-hand side thus gives you the complete electromotive force, including the one due to the magnetic field.

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