by megaflop
 P: 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. Thanks in advance for your answers PS: Why does it automatically go to a new line when i insert a Latex equation?
 Mentor P: 11,622 When you want an equation to appear "in line", use "itex" in the opening and closing tags, not "tex".
P: 10
 Quote by jtbell 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 ?

 P: 10 Radiation Pressure 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 ?
 Sci Advisor HW Helper PF Gold P: 1,991 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.
P: 10
 Quote by 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.
Yep I figured this out.