Radiation Pressure: Proving Magnetic Field Reflection on Perfect Conductor

[megaflop]

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.

Thanks in advance for your answersPS: Why does it automatically go to a new line when i insert a Latex equation?

 

[jtbell]

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

 

[megaflop]

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

Thank you.
And what about the subject ?

 

[megaflop]

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 ?

 

[Meir Achuz]

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.

 

[megaflop]

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.

Yep I figured this out.
Thank you for you answer.