Solve Betatron Calculations Homework: (c) & (d)

In summary, we discussed the parameters and equations for a betatron and solved for the induced electric field and work done on an electron per revolution. We approximated the electron speed to be c in part (c) and discussed the calculation of the final kinetic energy, taking into account relativistic effects. In part (e), we solved for the change in magnetic field over time and used this value to calculate the induced electric field.
  • #1
silmaril89
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Homework Statement



I'm confused about part (c) and on in the following questions:

Consider a betatron with these parameters: the radius of the electron orbit is 0.5 m; the kinetic energy of an electron injected into the accelerator is 2.0 MeV; and the rate of increase of the magnetic flux through the area of the toroidal vacuum chamber is 25 Wb/s. The electrons are ejected after 4 ms of acceleration.

(a) Compute the magnitude of the induced electric field

(b) Compute the work done on an electron per revolution around the orbit.

(c) Compute the number or revolutions completed by an electron before ejection. (Approximate the electron speed by c.)

(d) Compute the final kinetic energy of the electron.

(d) In order to keep the radius r of the electron path constant, B at r must be one half of the average of B over the area enclosed by the circular orbit. What is dB/dt at r during the acceleration?

Homework Equations



curl E = -dB/dt

B(r = electron orbit radius, t) = 1/2 B(t) (average field inside electron orbit)

dW = F . dl

The Attempt at a Solution



I solved (a) using ampere's law, and found the electric field to be:

E = (-r/2) (dB/dt) in the phi direction, and B is the average field

I solved (b) by using the equation for W above:

W = pi * r^2 dB/dt, where B is again the average field inside the orbital radius of the electron


Now, I'm stuck on (c), I don't get why we would be approximating the electron speed to be c. If it was c, and we know it is in the betatron for 4ms, the calculation is trivial. But, the electrons would be accelerating the whole time, they couldn't just have one speed?

In part (d), I'm really not sure how to calculate the final kinetic energy. Do I have to take into account relativistic effects?

Thanks to anyone who replies.
 
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  • #2


Hello,

In part (c), we are approximating the electron speed to be c because the acceleration time is very short (4ms) compared to the speed of light. This approximation allows us to simplify the calculation and assume that the electron is moving at a constant speed throughout the acceleration process. However, if you want to take into account the acceleration time, you can use the equation for the velocity of a charged particle in a magnetic field:

v = (qBt)/m

where q is the charge of the electron, B is the magnetic field, t is the time of acceleration, and m is the mass of the electron. You can then use this velocity to calculate the number of revolutions completed by the electron before ejection.

In part (d), you can use the equation for the final kinetic energy of a charged particle in a magnetic field:

Kf = (1/2)mv^2

where m is the mass of the electron and v is the final velocity calculated in part (c). Yes, you do need to take into account relativistic effects in this calculation. You can use the relativistic equation for kinetic energy:

Kf = (mc^2)/(sqrt(1-(v/c)^2)) - mc^2

where c is the speed of light.

In part (e), you can use the equation for the change in magnetic field over time to solve for dB/dt at r:

dB/dt = (B(t) - B(0))/t

where B(t) is the magnetic field at time t and B(0) is the magnetic field at the beginning of the acceleration. You can then substitute this value into the equation for the induced electric field (E = (-r/2)(dB/dt)) to solve for the electric field at r.

I hope this helps clarify your confusion. Let me know if you have any further questions. Good luck with your calculations!
 

Related to Solve Betatron Calculations Homework: (c) & (d)

1. What is Betatron?

Betatron is a type of particle accelerator that uses alternating magnetic fields to accelerate charged particles, typically electrons, to high speeds.

2. What are "c" and "d" in Betatron calculations?

"c" and "d" refer to the constants used in the equations for calculating the magnetic field strength and particle trajectory in a Betatron. "c" is the speed of light and "d" is the radius of the particle's orbit.

3. How do you solve Betatron calculations?

To solve Betatron calculations, you will need to use the equations for calculating the magnetic field strength and particle trajectory, using the known values for "c" and "d" as well as the charge and mass of the particle. You can then use these values to determine the particle's speed and trajectory at different points in the Betatron.

4. What are some common applications of Betatron technology?

Betatron technology is commonly used in medical facilities for cancer treatment, as well as in scientific research for particle physics experiments.

5. Are there any limitations or challenges in Betatron calculations?

One of the main challenges in Betatron calculations is accurately accounting for all of the different factors that can affect the particle's trajectory, such as external magnetic fields or variations in the electric field. Additionally, Betatrons are limited in the types of particles they can accelerate and the maximum energy they can achieve, making them less versatile than other types of particle accelerators.

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