# Magnetic Force from an EM Wave

1. Dec 2, 2005

### eok20

I'm not sure if this belongs here or in the relativity forum but associated with an electromagnetic wave is a magnetic field B and the force of this field on a charge q with velocity v is qv x B. Since any charge will be moving with speed c relative to the wave, will the magnetic force on a charge from an EM wave always be qc x B?
Thanks.

2. Dec 2, 2005

### Tide

NO! I'm not sure how you would even write the equations of motion in the frame of reference moving with the light wave. Stick with the "lab frame" and the Lorentz force in its usual form.

3. Dec 2, 2005

### mmmusthafa

force in em wave

No. Do you think that the field is moving along with the wave w.r.t charge? Only the field is sccessively being induced hence the charge or the field does not have relative motion. Hence will not ihave that type of nteraction with stationary charge

mm_musthafa @rediffmail.com

4. Dec 5, 2005

### kmcf

An electromagnetic wave has two components: electric (E) and magnetic (B). So the force on a charge is F = qE + qvxB. For a plane wave, E and B are perpendicular. Both oscillate at the frequency of the wave. The electric field will cause the charge to oscillate perpendicular to the wave direction, and the resulting motion will induce a magnetic force on the particle. The net (classical) effect is that the charge will pick up speed in the direction of the wave. (I'm assuming that the mass of the particle is such that the velocity gained from absorption of a single photon from the wave is very small compared with any velocity of interest. Otherwise, you need to look at Compton scattering.)

5. Feb 18, 2006

### kuntal kumkar ghosh

kuto1

hey eok20 i congratulate you for your good question.well,the force won't be qc x b always thats because the speed of light on different mediums is different.

6. Feb 18, 2006

### gulsen

Weird approach, but i think it's just not fine, because:

-if you take a photon as a frame of reference, time will not flow, and the whole universe will be a single point; given these, what can you calculate!
speed of electron?

$$\frac{dx'}{dt'} = \frac{0}{0}$$

ugh! disaster! of course, one can limit "actual" quantities.

7. Feb 18, 2006

### Staff: Mentor

Sorry, you can't do that. As has been stated over and over and over again in the relativity forum, there is no inertial reference frame in which light is stationary.

8. Feb 18, 2006