- #1
Addez123
- 199
- 21
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 here's 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?!
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 here's 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?!