# Particle beam in infinite magnetic field

In a physics exam at university I had the following problem:

A proton-deuteron beam is accelerated by a $\Delta V = 10^7\,\mathrm{V}$ difference of potential. At some point, the particles enter a uniform, infinite, magnetic field with $B = 2 T$, perpendicular to the beam's direction. Calculate, at the exit of the magnetic field, the distance between protons and deuterons.

My approach was that, in the exam's short available time, since the magnetic field is so strong, the forces exerted by the particles were negligible; also that the beam was accelerated before entering the field, so moving with constant speed. So the only force I thought would be significant was the Lorentz force; since this force does no work on the particles, these may only change the direction of their speed, not the module, and so they start rotating in a cycloidal motion along the direction of the beam. My result so was:

$x_A(t) = \frac{v_0}{\omega} [1-cos(\frac{\omega t}{A})]$

$y_A(t) = \frac{v_0}{\omega} [sin(\frac{\omega t}{A})]$

$z_A(t) = 0$

$\omega = \frac{qB}{m}$

$v_0 = \sqrt{\frac{2q\Delta V}{m}}$

With y being the direction of the beam, z the position along the direction of B, the x axis perpendicular to both, and A the atomic mass. My professor didn't tell me what was wrong, but implied that it was all wrong, and I have no idea where. I don't want to make wrong assumptions that may bring me even further from the right solution.

What's wrong in my approach, and what is the right solution/way to solve this problem?

• We don't normally answer check-my-work questions here, but this looks like a solid answer. Commented Jun 17, 2015 at 17:00
• I know, sorry for that. I should have asked the same question in general terms, without reference to my results. If the question is out of place, feel free to close it, but I want at least to know what I'm missing so I can have a better understanding. Commented Jun 17, 2015 at 17:06

Maybe this picture will help you get some intuitive handle on what was asked in the problem

There exists a strong magnetic field perpendicular to the plane and to the direction of the incoming beam, so it is the set up of your problem.

Charged particles traversing the liquid of the chamber ionize the molecules of the liquid and thus the tracks become visible. Note the helices . These are electrons knocked out of the atoms of the liquid at higher momentum than the little dots the main bulk of ionization electrons make.

Helices because the electrons not only lose energy by the ionisation and thus their circle ( Bqv=mv**2/r) loses momentum , but also they may have a momentum perpendicular to the plane which will generate a helix anyway.

The beam has very high energy and thus the momentum (mv) is large and the radius very large. If teh beam is composed of two particles with different masses you can see that the two components will be separated if the field is infinite in extent as the problem states.

So the lorenz force is balanced with the cetrifugal force to get the radius of the track.

I think your mistake is in the direction you are applying the Lorenz force.

• The question's phrasing (i.e. just the distance between the two beams) is fairly insensitive to the direction of the Lorentz force, though. Commented Jun 17, 2015 at 17:21
• Thank for you reply. I calculated the Lorentz force as $F = qv$ x $B = m\frac{dv}{dt}$, from which $dv_x = \frac{qBv_y}{m}$ and $dv_y = -\frac{qB v_x}{m}$. These two combined and derived with respect of time yield a diff. equation with solutions x(t) and y(t) above. So I'm not applying the force at any direction, which is a priori not known at any point as it changes with time. Commented Jun 17, 2015 at 17:33
• @EmilioPisanty For the beam it is small but the B is posited as infinite in extent ( I assume since it is given a value) after the beam enters it, so at large distances compatible with the radius of the equation I gave due to the masses the beams will separate. Commented Jun 17, 2015 at 17:39
• Commented Jun 17, 2015 at 18:01
• I see. That appears in the result however, as the module of x(t) is the radius, so $R = \frac{v_0}{\omega} = \frac{m v_0}{qB} => v = \frac{BqR}{m}$. I guess there's nothing wrong in my result, or am I missing something else? Maybe my professor didn't pay enough attention... but I'm seeing the whole picture a lot better now. Commented Jun 17, 2015 at 18:11

It looks like you were on the right track but weren't careful to label the velocities and frequencies with the associated particle species. $v_0$ and $\omega$ have different values for each beam. Additionally, the atomic mass $A$ shouldn't be in your formula for $x_A(t)$ because it's already in $\omega$. These may be the problems your professor saw.

Presumably the infinite magnetic field only occupies the y>0 region. So the beams enter the field at the same point, follow two semicircles of different radii, and leave. You just need to find the distance between the these exit points, and you've already calculated their paths $x_A(t), y_A(t)$. The distance is $$D = x_p(\pi/\omega_p) - x_d(\pi/\omega_d) = 2\left(\frac{v_p}{\omega_p} - \frac{v_d}{\omega_d}\right) = \frac{2(m_p v_p - m_d v_d)}{qB}$$

It's also possible your professor was expecting you to "just know" the gyroradius formula $R= |\frac{mv}{qB}|$ leading to $D = 2(R_p - R_d)$, and wasn't looking for more detailed answers.

• Right, I guess the "region" was to be interpreted as a infinite plane, not a generic region. Though he should have been more clear, but he may have expected me to give some hypothesis about the problem. Commented Jun 17, 2015 at 19:29