1
$\begingroup$

From what I understand, the point of cyclotrons is to accelerate charges without requiring a huge potential difference: e.g. to accelerate a proton to 10 MeV, instead of using a single 10-megavolt potential difference, we accelerate it incrementally.

We do so by alternating the voltage of the Dees. But since the voltage is alternating and not increasing, I don't see how the charges can keep accelerating.

Suppose the Dees have potential $V_A$ and $V_B$, with $V_A>V_B$. Initially the (positive) charges are in the Dee #1 with $V=V_A$ and they move to $V=V_B$ in the Dee #2, thus decreasing their potential energy by $\Delta U = q (V_B-V_A)$.

At some point (before the charges complete a half-cycle), we invert the polarity of the dees. The Dee #2 is now at $V=V_A$ and the other is at $V=V_B$, so that when the charges return to the initial Dee their potential energy decreases again by $\Delta U = q (V_B-V_A)$.

But shouldn't the charges gain potential energy (and slow down) as the voltage shifts inside the dees? The charges experience a change of potential $\Delta V = (V_A-V_B)$ during the polarity shift, which seems like it would counteract the acceleration between the dees.

To me it's like claiming that instead of climbing Mount Everest (= experience a huge change of potential) you can climb up and down a 1-meter step 8849 times and still end up at the same altitude. Obviously that's not how it works, so what's wrong with my reasoning?

$\endgroup$
2
  • $\begingroup$ When inside the dee, the charges are essentially in a field-free region (Faraday cage). The power supply can vary the voltage of the dee relative to ground and have no impact on the charges inside of it. $\endgroup$
    – Jon Custer
    Apr 12, 2022 at 15:04
  • $\begingroup$ Maybe it's because I don't understand how the alternating voltage works, but I would think that during the voltage shift there is an electric field (resulting from the emf of the power supply) that brings charges on the opposite dee (not the accelerated charges, the charges responsible for the potential difference) $\endgroup$
    – Jasmeru
    Apr 12, 2022 at 15:21

1 Answer 1

0
$\begingroup$

Keep in mind that the particles do not touch the electrodes, as they are confined with the magnetic field, so they never reach the potential of the electrodes. So your calculation about potential energy doesn't apply in this case.

From the point of view of the particles:

  1. In the gap: they experience an electric field of $E=(V_B-V_A)/d$ where $d$ is the distance of the gap, and so are accelerated via the Lorentz force ($F=qE$).
  2. Then there is no electrical field inside the Dee, (the whole Dee has a varying potential, but inside it, there is net difference), so there is no electrical acceleration on the particle (just the bending due to the magnetic field)
  3. By the time they cross the Dee, and reach the other gap, the field of the gap was inverted, but the particles are going in the other direction, so they still see the same field $E=(V_B-V_A)/d$ so more acceleration, which adds to the energy of the particles.
  4. In the other Dee, no electric field, so no acceleration.

This is why it's important to synchronise the frequency of the field with the movement of the particles. It's a similar principle in a synchrotron or a Linac: the particles only sees an electric field in the accelerating cavities, but never reaches the potential of any of the electrodes producing the fields.

To try to get back to your stairs analogy, by climbing up a single step 8849 times, you will still climb 8849m (ie: the particles are accelerated to 10MeV) but you will still have an altitude (~ gravitational potential) of 1m, just as the particles in the cyclotron don't have a potential of 10MeV at the end. (If they did, it would create a very strong field between the particles and the chamber)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.