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Here is another old exam question I'm wondering about:

A proton moving in the $-x$ direction encounters a region of space with a magnetic field that randomizes the direction of the particle. The magnetic field is at rest. After scattering off the magnetic fields the proton is ejected from the region of space and is moving in the $+x$ direction.

A) If the original energy of the proton is $10^{12} \, \mathrm{eV}$ and the mean strength of the magnetic field is $1\, \mathrm{mT}$, what is the final energy of the proton after scattering?

Well, here I thought that magnetic fields don't do work, so the energy would be unchanged? But on the other hand, an accelerated charge emits radiation, and of course I must accelerate the proton to turn its direction by 180 degrees. But if it's not a trick question where the answer is just $10^{12}\, \mathrm{eV}$, then how can I say anything meaningful at all when the field is random?

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What level of student is this exam intended for? –  David Z Aug 22 '11 at 23:53
Comprehensive exam for beginning graduate (PhD) students –  Lagerbaer Aug 23 '11 at 0:05
If they wanted to consider radiation, they would at least have to give you the scale at which the random field varies. –  Ron Maimon Aug 23 '11 at 8:18
@Ron Yes, that's true. But since the proton is ultra-relativistic at $0.9995 c$, how can I neglect radiation? –  Lagerbaer Aug 23 '11 at 15:33

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Hmmm... this is a confusing question. As far as I can tell, it's not clear exactly what they mean by "magnetic field that randomizes the direction of the particle." And unless "the magnetic field is at rest" is supposed to mean that it's static, I'm not sure what they're getting at with that either.

Anyway, the physical phenomenon in question is synchrotron radiation. This is described by the relativistic Larmor formula,

$$P = \frac{q^2}{6\pi\epsilon_0 c}\gamma^6\Bigl[\dot\beta^2 - \Bigl(\vec{\beta}\times\dot{\vec{\beta}}\Bigr)^2\Bigr]$$

For this case, since we have only static magnetic fields and the particle is highly relativistic, it's approximately true that $\vec\beta\perp\dot{\vec\beta}$ (in other words, the component of acceleration parallel to the velocity is relatively negligible, c.f. Jackson p.667). Using this and the formula for the cyclotron radius $R = \frac{\gamma mc\beta}{qB}$, I get it down to

$$P = \frac{q^2}{6\pi\epsilon_0 c^3}\frac{\gamma^2\beta^2B^2}{m^2}$$

From here, I guess you would be expected to figure out the total energy loss while the particle is in the magnetic field. However, I don't see how to do that, for a couple of reasons: first of all, although you know the mean value of the magnetic field $\langle B\rangle$, you don't know enough to figure out $\langle B^2\rangle$, which is what you need to figure out the mean radiative power loss. And even if you did have that, you don't know how much time the particle spends in the magnetic field, which you would need to go from power to energy.

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