 #26
fisher garry
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I think I understand. Let me change the scenario. You have a circular current coil and apply an external B field. I will rewrite the relations of the kinetic energy produced:Why in the world are we dealing with radiation pressure in the first place? Why do you think it has any relevance in describing Faraday's law? Why haven't you answered that?
Zz.
When the coil has turned 90 degrees it reaches equilibrium and deaccelerates to 0 velocity. Lets assume hypotetically that all this energy is released as one photon. Now I will try to introduce radiation pressure again:
So we hypotetically want the electron that has been deaccelerated to 0 velocity (I will ignore the current direction) to obtain the same kinetic energy and velocity. We want it to absorb a photon to do so. If we look at one half cycle of an EM wave that would be half of the energy of one photon if all of the energy of the photon is contained in one cycle. The EM wave exert radiation pressure as described above. It is also said that E=pc where p is radiation pressure density and E is energy density.
Over a half cycle of an EM wave the avearge cycle can be obtained the following way:
In order to calculate the amplitude of the EM wave I try the following:
First I calculate the kinetic energy from the 90 degrees turn of the circular coil
Then I try to find the amplitude of the EM wave and uses the theory for the radiation pressure above along with E=pc:
Is this also a wrong usage of the radiation pressure?
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