# Physical Explanation of Faraday's Law

by Lyuokdea
 P: 198 So we were going through Faraday's Law today in class: $$\int\vec{E}\cdot d\vec{l} = -\frac{d}{dt}\int\vec{B}\cdot d\vec{A}$$ Mathematically I'm fine with it, however, is there any good physical way to explain it, it seems very odd that if you had a field such as: $$\vec{B} = B_0cos(\omega t)\hat{z}$$ and that field was defined only in some circle of radius R that if you had a ring around R that the induced electrical current would be the greatest when the magnetic field doesn't exist, how does one object feel another objects change through time? I understand this in terms of a force for things like velocity, but that is only because the objects are in physical contact, however, the magnetic field isn't touching the right at all. I would assume that there are photons moving between the field and the wire, or virtual photons or something would be necessary for the wire to "see" the magnetic field at all, however when those photons don't exist, how does the magnet react to them? ~Lyuokdea
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P: 2,002
 Quote by Lyuokdea So we were going through Faraday's Law today in class: $$\int\vec{E}\cdot d\vec{l} = -\frac{d}{dt}\int\vec{B}\cdot d\vec{A}$$ Mathematically I'm fine with it, however, is there any good physical way to explain it, it seems very odd that if you had a field such as: $$\vec{B} = B_0cos(\omega t)\hat{z}$$
Your example is not possible, since it would violate $\nabla \cdot \vec B=0$.

Anyway, I don't really understand where the problem is in your understanding. The idea is that a change in $\vec B$ at a certain point induces an entire electric field which curls around -d/dt B. Maybe the differential form of Faraday's law is more intuitive (apply Stokes):
$$\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$$

As to why this is true I can't give a complete answer. Maybe you'll find it sufficient that, given an electric field and relativity, this must give rise to effects like Faraday's law.
 P: 1,295 There are two ways to look at faradays law. If you have a conducting ring and a changing B field (your example) then an "emf" appears in the ring. This can be physically explained: imagine that instead of B changing, it is the the ring which is moving in a static B field (this situation appears in Faraday's law problems anyway). Now, those are charges moving through a B field with a velocity, so they feel a force q(vxB). Now, here we go: q(vxB) is the emf! If you do the right hand rule, q(vxB) points in the direction of the emf. If you set up a calculation, you will get the same result this way as using Faraday's law. But Faraday's law is deeper than that, because there doesn't have to be a ring there or any charges! In otherwords, the electric field is a real entity because it shows up even with out any charges (we know it is there because thats why we have em waves). This is also easy to explain physically, all you need to know are the lorentz transformations.
P: 416
 Quote by Galileo Your example is not possible, since it would violate $\nabla \cdot \vec B=0$.
Not really.. $$\frac{\partial}{\partial z} [B_0 \cos(\omega t)] = 0$$
So you don't want to get caught in thinking the current is greatest when the field doesn't exist, but that it's greatest when the field is changing most rapidly... ie if your B field was like $$B = B_0 cos(\omega t) + B_1$$, $$B_1$$ could be really huge, and then the current would be greatest at some middle value of the B field.. crosson's explanation is maybe the best way to think about it.. you could have a coil of wire and push a bar magnet in and out of the coil, or you could reverse the situtation and have a bar magnet fixed and push the coil back and forth over the magnet. Both situations are equivalent, as long as the velocities are the same