Deriving the rate of change of the radius of a moving particle in a cyclotron.

[JHans]

The question:

A particle in a cyclotron gaines energy

[tex]

q \Delta V

[/tex]

from the alternating power supply each time it passes from one dee to the other. The time interval for each full orbit is

[tex]

T = \frac{2 \pi}{\omega} = \frac{2 \pi m}{q B}

[/tex]

so the particle's average rate of increase in energy is

[tex]

T = \frac{2 q \Delta V}{T} = \frac{q^2 B \Delta V}{\pi m}

[/tex]

Notice that this power input is constant in time. On the other hand, the rate of increase in the radius r of its path is not constant. Show that the rate of increase in the radius r of the particle's path is given by

[tex]

\frac{d r}{d t} = \frac{1}{r} \frac{\delta V}{\pi B}

[/tex]My thoughts:

I'm completely stuck on how to accomplish this. I have a general idea, but I don't know how to go about it. My thought is: if the radius of motion increases, then the velocity must increase. For the velocity to increase, the kinetic energy must increase. That increase in energy comes from the electric field between the dees. I should be able to model a function of the kinetic energy at some time t and then put it into an equation that shows how the radius depends on the velocity of the particle. From there, I can just differentiate the radius equation with respect to time and prove the law. How, though, can I go about showing that the energy of the particle depends on time?

 

[JHans]

Sorry if there's any confusion... the lowercase deltas should be capital deltas, i.e., delta V (electric potential). I tried to correct it, but Latex is being stubborn.

 

[gneill]

If the KE increases by 2q∆V each cycle, and each cycle has a fixed period, T, then the energy with respect to time should go as 2q∆Vt/T, then velocity becomes (at least approximately):

[tex]
v(t) = \sqrt{\frac{4 q \Delta V t}{m T}}
[/tex]

 

[JHans]

Wouldn't the expression you gave for energy only represent the INCREASE in energy? My thought is that:

[tex]

K (t) = K_i + \frac{q^2 B \Delta V}{\pi m} t

[/tex]

[tex]

v^2 (t) = v_i^2 + \frac{2 q^2 B \Delta V}{\pi m^2} t

[/tex]

Is that an incorrect assumption?

 

[gneill]

I assumed that the electron is starting essentially from rest, and get's started by its first dee crossing. After that it picks up 2∆E energy each cycle. Since the cycles are of fixed length T, the number of cycles executed in time t is t/T. So 2∆E(t/T) should approximate the total energy at time t.

I'm not sure about your expressions. I haven't "gotten my head around them" yet. But I don't have a good feeling.

 

[JHans]

Oh, okay! I hadn't thought of the fact that the electron is starting at rest, which would mean that the initial velocity is 0 m/s. That would also mean that the initial radius of the path in the cyclotron would be 0 m! Here's my general thought process:

Each time the particle exits a dee, it receives energy from the electric field:

[tex]

\Delta U = q \Delta V

[/tex]

In one full orbit, the particle crosses the dees TWICE. Therefore, in one full orbit:

[tex]

\Delta U = 2 q \Delta V

[/tex]

The energy delivered is constant in time, so we can model:

[tex]

\frac{\Delta U}{T} = \frac{2 q \Delta V}{T} = \frac{q^2 B \Delta V}{\pi m}

[/tex]

The above holds true because:

[tex]

T = \frac{2 \pi r}{v} = \frac{2 \pi}{\omega} = \frac{2 \pi m}{q B}

[/tex]

From here, the kinetic energy at some time t will be the sum of the initial kinetic energy PLUS the amount of energy delivered by the electric field as the particle crosses the dees. We've asserted that the initial kinetic energy is 0 J (because the initial velocity is 0 m), so:

[tex]

K (t) = \frac{q^2 B \Delta V}{\pi m} t

[/tex]

(That's the energy delivered per unit time, multiplied by the time elapsed.)

If we multiply by 2 and divide by the mass, we come across the square of the velocity at some time t.

[tex]

v^2 (t) = \frac{2 q^2 B \Delta V}{\pi m^2} t

[/tex]

Taking the square root of that function would yield the velocity at a time t. Plugging that answer into the formula for the radius of the path in the cyclotron:

[tex]

r = \frac{m v}{q B}

[/tex]

would yield a formula for the radius of the path at some point in time t, wouldn't it? And then you'd just differentiate with respect to time?

 

[gneill]

Suppose we go with my expresion for velocity:

[tex]

v(t) = \sqrt{\frac{4 q \Delta V t}{m T}}

[/tex]

and we also know that

[tex]
v = \frac{2 \pi r}{T}
[/tex]

This leads to

[tex]

r(t) = \frac{1}{\pi} \sqrt{\frac{q}{m} \Delta V T t}

[/tex]

Differentiating w.r.t. t I find

[tex]

\frac{dr}{dt} = \frac{1}{2 \pi^2 r} \cdot \frac{q}{m} T \Delta V

[/tex]

and since

[tex]

T = \frac{2 \pi m}{q B}
[/tex]

the desired result obtains. (whew!)

 

[JHans]

I follow your explanation up until you take the derivative of the function for the radius with respect to time. Can you show me how you got the derivative you've described?

 

[gneill]

When the derivative is taken, the sqrt() turns up in the denominator. It is just the original function for r with the 1/pi missing.

Sorry, but I've got to run to a meeting with a lovely beverage. I'll check back later to see how you've got on...

 

[JHans]

Oh, I see now! Everything makes sense.

I checked it, and my equation for modeling the velocity as a function of time works as well. It's a little bit messier, but it can be cleaned up at any point in time by substituting for B:

[tex]

B = \frac{2 \pi m}{T q}

[/tex]

The end result is the same either way, and you end up establishing the law asked for. Thanks a lot for your help!

 

[gneill]

Excellent. Glad I could help.