1. ### megaflop

10
Hi everyone ! This is my first post!

How can we prove that during the reflection of an electromagnetic wave on the surface of a perfect conductor, the magnetic field $\vec{b}$ acting on a surface element $ds$ is worth half the total magnetic field $\vec{B}$ using Ampere's Law. That is $\vec{b}=\frac{1}{2}\vec{B}$.
This in order to justify the one half factor in the expression of the force acting on $ds$ which is $d\vec{f}=\frac{1}{2}\vec{j_{s}}\times \vec{B} \cdot ds$, where $\vec{j_{s}}$ is the surface current density on the conductor.
A drawing would be welcome.

PS: Why does it automatically go to a new line when i insert a Latex equation?

Last edited: Feb 7, 2012

### Staff: Mentor

When you want an equation to appear "in line", use "itex" in the opening and closing tags, not "tex".

3. ### megaflop

10
Thank you.
And what about the subject ?

4. ### megaflop

10
There's maybe another way to justify the $\frac{1}{2}$ factor without using Ampère's law ? It's how it was justified in my textbook but I couldn't understand it.
Anyone ?

5. ### Meir Achuz

2,075
Consider a sheet of surface current K: KKKKKKKKKKKKKKKKKKKKKKKKKK
with the EM wave incident from above. The B field comes from two sources
1) the wave B_W, and 2)the current, B_K. Below the surface (in the metal) B is zero.
That mean the two components of B cancel so B_W=B_K in magnitude and opposite in sign.
Above the surface, B_K changes sign, but B_W doesn't, so the total B field outside is twice the field B_W which acts on the surface current.

6. ### megaflop

10
Yep I figured this out.