# Prove that emf from E equals emf from dB/dt

1. May 14, 2017

Exercise:
I'm suppose to prove that e1 = e2
e1 = ∫E*dl
e2 = ∫dB/dt*dS

where S is the surface encircled by the conture c.
c is a box with with a length (in x axis) and b height (in y axis).

for an electromagnetic wave:
E = E0*sin(kx - wt) (in y axis)

I'm ASSUMING this means that
B = B0*sin(kx - wt) (in z axis)

Pathetic Attempt:
e1 calculated on the outside of c becomes
b*E(x + a,t) - b*E(x,t) =
E0*sin(kx + ka - wt) + E0*sin(kx - wt)
This expression is confusing and doesn't lead to any simplification at all.

e2 = ∫dB/dt*dS
dB/dt = B0*-w*cos(kx - wt)

Now heres the tricky part, how do I integrate this infinitly thing vector over a surface?
The wave has no thickness, shouldn't the integral be zero?!

2. May 14, 2017

I'm also suppose to prove this explicitly, I DONT EVEN KNOW WHAT THAT MEANS!!!

3. May 14, 2017

### TSny

What happened to the factor of b in going to the last line. Also, what happened to the negative sign between the two terms?

You've made a good start once you make the corrections above.

OK

Not sure what you mean by "this infinitly thing vector".
$\frac{\partial B}{\partial t}$ is a function of $x$ and $t$. $t$ is some fixed but arbitrary time. So, you can think of $\frac{\partial B}{\partial t}$ as a function of $x$ that you need to integrate over the rectangular area. What would be a good way to break up the area into infinitesimal elements $dS$? Recall how you did similar integrals in your calculus course.