Solving Cyclotron Movement Problem with 500V and 0.790B

[Canadian]

Homework Statement

A 61.0 cm diameter cyclotron uses a 500 V oscillating potential difference between the dees. The magnetic field strength is 0.790 ?

How many revolutions would a proton make before leaving the cyclotron?

Homework Equations

v/r=Bq/m Ke=1/2mv^2

The Attempt at a Solution

I am completely stumped on this problem, I have calculated the max speed, and therefor max kinetic energy. But cannot figure out how to derive number of revolutions from there.

Any hints would be greatly appreciated.

Thanks

 

[lalbatros]

Canadian,

Why would the electron leave the cyclotron,
Describe first how this device works.
How does the electron reach the escape speed?

Michel

 

[m2003]

for your question. I understand your struggle with this problem. Let's break it down and see if we can come up with a solution together.

First, let's review the basics of a cyclotron. A cyclotron is a type of particle accelerator that uses a combination of electric and magnetic fields to accelerate charged particles, such as protons, to high energies. The particles are injected into the center of the cyclotron and are then accelerated by the electric field between two metal plates called "dees" that have a potential difference between them. As the particles gain speed, they are deflected by a magnetic field, causing them to spiral outward until they reach the outer edge of the cyclotron.

Now, let's apply this knowledge to the given problem. We know the diameter of the cyclotron (61.0 cm) and the strength of the magnetic field (0.790 ?). We also know that the potential difference between the dees is 500 V. Using the equation Ke=1/2mv^2, we can calculate the maximum kinetic energy of the proton as it reaches the outer edge of the cyclotron. However, we also need to consider the magnetic force acting on the proton as it spirals outward. This force is given by F=qvB, where q is the charge of the proton, v is its velocity, and B is the magnetic field strength. This force must be equal to the centripetal force acting on the proton, which is given by F=mv^2/r, where m is the mass of the proton and r is the radius of its circular path.

Putting these equations together, we can set F=qvB=mv^2/r, and solve for v to get v=Br/q. Now we have an expression for the velocity of the proton in terms of the magnetic field strength and the charge of the proton. Substituting this into the equation for kinetic energy, we get Ke=1/2m(Br/q)^2.

We also know that the maximum kinetic energy of the proton is equal to the potential difference between the dees, so we can set Ke=500 V. Solving for r, we get r=qB/500. This tells us the radius of the proton's circular path at the outer edge of the cyclotron.

Now, we can use the circumference formula, C=2πr, to calculate the distance the proton travels in one revolution.