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Radiation Pressure 
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#1
Feb712, 03:08 PM

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 [itex] \vec{b} [/itex] acting on a surface element [itex] ds [/itex] is worth half the total magnetic field [itex] \vec{B} [/itex] using Ampere's Law. That is [itex] \vec{b}=\frac{1}{2}\vec{B} [/itex]. This in order to justify the one half factor in the expression of the force acting on [itex] ds [/itex] which is [itex] d\vec{f}=\frac{1}{2}\vec{j_{s}}\times \vec{B} \cdot ds[/itex], where [itex]\vec{j_{s}}[/itex] 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? 


#2
Feb712, 03:34 PM

Mentor
P: 11,743

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



#3
Feb712, 03:51 PM

P: 10

And what about the subject ? 


#4
Feb812, 11:24 AM

P: 10

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


#5
Feb912, 10:42 AM

Sci Advisor
HW Helper
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P: 2,007

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
Feb1112, 11:56 AM

P: 10

Thank you for you answer. 


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