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I'm studying for my test on radiation for Tuesday. I came across this exercise. Thought it looked interesting but now I'm stuck and I can't move forward before I finish this one.

Exercise:

A particle with charge $q$ is initially at rest in the origin. An EM pulse travels in the direction of the y-axis. The electric field osscialtes around the z-axis and the magnetic field around the x-axis. Assuming that$v$ (speed of charge) $<< c$.

a) Write the charge's equation of motion in rectangular $xyz$-coordinates

b) What is the velocity as a function of time?

Questions:

In a) could I calculate the acceleration's z-component created by the electric field according to Newtons II, ma=qE. The acceleration's x-component created by the magnetic field like,
ma=qv x B. And then integrate them twice to get the position? I would however get some less beautiful constants. Is there a better way?

For b) I'm confused. The electric field can be written as $E(r,t)=E_0 \sin(ky-\omega t)$, an equation that depends on the time. But the only thing I can think of is to once again integrate the acceleration. However it would be strange for them to have a) and b) to be the same?

Super-grateful for your insight! :)

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  • $\begingroup$ In a) they ask for an equation, in b) for its solution. $\endgroup$
    – Ján Lalinský
    Commented May 6, 2016 at 20:42

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This is much more simple than you imagine. Yes of course you use Newton II to get the acceleration and the Lorentz force as the right hand side. However, the key here is the statement that the charge moves with $v \ll c$.

For an electromagnetic wave in vacuum then $B = E/c$. Thus is we look at the two terms in the Lorentz force, the first will have a magnitute $qE$, whilst the second will have a magnitude of $qEv/c$; so if $v \ll c$ the second (magnetic) term can be neglected and the acceleration only has a $z$ component.

For the second part, of course the velocity is simply the integral of the acceleration with respect to time.

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  • $\begingroup$ Thank you! I realized that intergrating the acceleration gives me the equation v(t)=((qE)/m)t + C. At t=0 the velocity should also be zero (the particle was initially at rest), and from that you can easily see that C should as well be zero. Thanks again! :) $\endgroup$
    – A. Fågel
    Commented May 8, 2016 at 10:20

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