When we calculate the emf induced by placing a loop of wire inside a changing magnetic field, why do we use the area bounded by the loop to calculate total flux and then the rate at which it changes? Why does flux outside of the loop not affect the emf?
1 Answer
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Because of Stokes’ Theorem in vector calculus, which relates a line integral around a closed curve to a surface integral over the interior of that curve (or, more generally, over any surface bounded by that curve).
Start with the differential form of Faraday's law of induction,
$$\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}.$$
Integrate it over a surface $S$ whose boundary is the closed loop $C$ of the wire, and then apply Stokes' Theorem to the left side. You get
$$\oint_C\mathbf{E}\cdot d\ell=-\frac{d}{dt}\iint_S\mathbf{B}\cdot d\mathbf{S}.$$
The line integral is the EMF around the wire loop, and the surface integral is the magnetic flux through the wire loop.
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$\begingroup$ Good math here, but I think you need to tie it back to the question. Your answer talks about choosing a surface with a closed loop boundary. But why do we end up only considering the wire loop? I think addressing this would get at the heart of what the OP wants to know, and you wouldn't need to add much more to the answer to address it. $\endgroup$ Commented Nov 15, 2018 at 3:46
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$\begingroup$ I made a few edits to try to clarify the connection to the wire loop, the EMF, and the flux. $\endgroup$– G. SmithCommented Nov 15, 2018 at 3:55
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$\begingroup$ Right. We have the freedom to choose any surface, but if we want the emf we have to choose the surface bound by the wire in order for Stoke's Theorem to give us what we want $\endgroup$ Commented Nov 15, 2018 at 3:59