What is the force on a particle in an electromagnetic wave?

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Homework Statement


Wangsness Exercise 24-6
A particle of charge q and mass m is traveling with a velcoity u i nthe field of a plane EM wave in free space that itself is traveling in the z direction.

Find the force on the particle.

What does this become for the special case in whic hthe particle is traveling in the same direction as teh wave?

What is the direction of the force?

Under what circumstances (if any) will the force vanish in this case??

2. The attempt at a solution

First of all I am stumped as how to find the force. Would i just simply use the lorentz force law??

If that is the case then
Suppose we assume the E to point in teh X direction and the B to point in teh Y direction then

\vec{F} = q (\vec{E} + \vec{u} \times \vec{B})

suppose the particle was traveling in teh Z direction then
\vec{F} = q (\vec{E} + u\hat{z} \times B\hat{y} )

\vec{F} = q (\vec{E} + uB (-\hat{x}))

the force is pointing the x direction

The magnetic force would vanish if V is parallel to B
but the electric force will never vanish

is this even clseo to being right? OR am i sorely mistaken?

Thanks for your input!
 
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I think you are very close...

I don't know if this is correct (if you can neglect the field and radiation caused by the particle)
consider the case when
\vec{E}=-\vec{v}\times\vec{B}

however, for plane wave in free space,
E=cB

which implies v must be a certain (and impossible) value.
 
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the only way E would be equal to B is if the velocity of the particle was near the speed of light so its not possible for the force to ever be zero
 
I think you'll need oscillating terms like cos(k(ct-z)) won't you?
 
Tomsk said:
I think you'll need oscillating terms like cos(k(ct-z)) won't you?

yea i had forgotten to mention that

\vec{E} = \vec{E_{0}} \exp(i(kz-\omega t))
\vec{B} = \vec{B_{0}} \exp(i(kz-\omega t))
 
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