Now that we know what the electromagnetic voltage is, we also can
find the current. We know that the current, which we would be
measuring here with our current meter, the current is equal to
VEMF times the resistance, like this. V is equal to IR. So
whatever resistance we've got across this, we can find both the
current and the VEMF. There's one other useful thing here. And
that's to recognize that the VEMF is given by the contour integral
of the electric field over the closed contour of the loop. This Z
is the loop contour. Now, putting this all together gives us
Faraday's Law. And we can see that the closed contour of the
electric field DL, which is equal to VEMF, is equal to minus the
integral of DB by DT dotted with DS. Now, this is the integral
form of Faraday's Law. If we would like the differential form of
Faraday's Law, we're just going to apply Stokes theorem which says
that we can integrate. Stokes theorem says that the integral over
a closed contour is equal to the integral over a surface of the
curl of that same vector. And then since we can see that we've
got an integral over a surface on both the left and right-hand
side, we can come up with the integral form which says the curl of
the electric field is equal to the minus of the time derivative of
the magnetic flux density. So let's talk about the two different
ways of applying Faraday's Law. This one right here is very
commonly used. Suppose that you have a loop, maybe a circular
loop. It has a magnetic field that's going through it, and you're
given some equation like B is equal to some magnitude times the
co-sine of omega T. And let's suppose that it's in the Z
direction. That means that phi is going around this way and Z is
coming out of the board. We can see that we have a time variation
because we have a co-sine in our B term. And the radius of the
loop, let's suppose that is staying the same. We need to know how
many turns of the loop we have, I don't know, perhaps 15 turns of
the loop. And in this case we could find the VEMF at two
terminals right here. It would be simply minus N, which is 15,
times the integral over the surface of the derivative with respect
to time of the B field dotted with DS. Now our DS here is going
to be in the Z direction. That's going to be RDRD phi. Phi is
going to go from 0 to 2 pi and R is going to go from 0 to A. The
time derivative is going to give us minus 15 times 10 times omega
times the sine of omega T. These are going to dot to 1. The R
term is simply going to integrate to R squared over 2. And the
phi term is going to integrate to phi from 0 to 2 pi. And this is
going to tell us the EMF voltage of the terminals of this loop.
If we wanted to know the current and the loop, we would take VEMF
and we would divide it by R. So I is going to be VEMF divided by
the resistance of the loop. If we wanted to find the electric
field, what we would do is integrate around this contour like so.
Whoops. I'm going in the wrong direction here. Integrate around
like that. So my contour integral, integral of E dot DL, is going
to be E phi in the phi direction, dotted with my phi directed DL
which is RD phi and we need to integrate that as phi goes from 0
to 2 pi. R in this case is the radius of the loop A. So we're
going to get A times E phi times phi from 0 to 2 pi and these dot
to one. So we're going to have 2 pi A times E phi. So if we
wanted to find E phi, we would be able to do it by just equating
these terms.
