A cyclotron (Fig. 29.16) designed to accelerate protons has...

A cyclotron (Fig. 29.16) designed to accelerate protons has an outer radius of 0.350 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 T. (a) Find the cyclotron frequency for the protons in this cyclotron. Find (b) the speed at which protons exit the cyclotron and (c) their maximum kinetic energy. (d) How many revolutions does a proton make in the cyclotron? (e) For what time interval does the proton accelerate?

Key Concepts

Cyclotron Principle

A cyclotron is a type of particle accelerator that uses a constant magnetic field to bend the trajectory of charged particles into circular paths. An oscillating electric field accelerates the particles each time they cross the gap between the D-shaped electrodes (dees). The design ensures that particles gain energy incrementally as they spiral outward.

Cyclotron Frequency

This is the characteristic frequency at which charged particles orbit within the cyclotron under the influence of a magnetic field. It is determined by the charge-to-mass ratio of the particle and the strength of the magnetic field, and it remains constant as long as the particle’s speed is much less than the speed of light. The cyclotron frequency is essential for synchronizing the alternating accelerating voltage with the particle motion.

Uniform Circular Motion in a Magnetic Field

When a charged particle moves in a uniform magnetic field, the magnetic Lorentz force acts perpendicular to its velocity, producing a centripetal force that causes circular motion. The radius of the orbit increases as the particle gains speed. This concept is fundamental to understanding how the path and dynamics of particles change during acceleration in a cyclotron.

Energy Gain via Accelerating Voltage

In a cyclotron, particles gain kinetic energy each time they cross the gap between the dees where an accelerating voltage is applied. The energy gained per crossing is directly related to the voltage. Over many revolutions, this incremental energy addition results in the particle reaching very high speeds, with its maximum kinetic energy determined by the total energy imparted across all the gap crossings.

Kinetic Energy and Speed Relationship

The kinetic energy of a particle is linked to its speed by the standard classical relation (½mv²). As a charged particle gains energy in a cyclotron, its speed increases and its orbit expands. This relationship is used to determine the exit speed and the maximum kinetic energy of the particle upon leaving the cyclotron.

Revolution Count and Acceleration Time

The total number of revolutions a particle makes in a cyclotron, as well as the overall time it spends accelerating, depend on the energy increments received per gap crossing and the cyclotron frequency. The number of revolutions can be calculated by dividing the total energy gained by the energy per crossing, while the acceleration time is obtained by multiplying the number of revolutions by the period corresponding to the cyclotron frequency.

Transcript

00:01 We're told to refer to the cyclotron figure, and we're given that it has an outer radius of 0 .35 meters, and also that the voltage of the gap between the ds is 600 volts, and we're also told that the ds are between the poles of an electromagnet, which has the magnetic field of 0 .8 tesla's.

00:23 So we're going to start with part a, where we need to find the cyclotron frequency, or the protons that are traveling within the cyclotron.

00:33 So this frequency can also be said as the angular frequency.

00:44 So this angular frequency, we can calculate it by taking the charge of a proton and then multiplying it by the magnetic field of the gap and then dividing it by the mass of a proton.

01:08 So this angular frequency is going to be equal to the charge of a protein.

01:14 Which we know is 1 .6 times 10 to the negative 19 kulams times the magnetic field of the gap which is given at 0 .8 teslas and then dividing it by the mass of a proton which we know is 1 .67 times 10 to the negative 27 kilograms and then this will give us an angular frequency that is equal to 7 .66 times 10 to the 7 radiance per second.

02:01 Now for part b, we need to find what the speed is of a proton that is exiting the cyclotron.

02:16 So to find the speed, we're going to first write that the force of a proton that is in a magnetic field is equal to m of p times the velocity of that proton squared divided by the radius.

02:42 And so now we're going to write this force in terms of the magnetic field.

02:51 So i'm going to rewrite this, this entire equation, by replacing fb with v times q times v and equal to mfp times v squared divided by r.

03:08 And so we're going to take this equation now, and we can also solve for the velocity, b, and that's going to be equal to b, qr, divided by m of p.

03:28 And so we can now find what this velocity is by just plugging in the noun terms.

03:37 So the magnetic field we know is 0 .8 teslas, then the charge of a proton is 1 .6 times 10 to the negative 19 couloms.

03:50 The radius r is given as 0 .35 meters and then the mass of a proton we know is 1 .67 times 10 to the negative 27 kilograms.

04:10 And so this will give us a velocity that's equal to 2 .68 times 10 to the 7 meters per second.

04:27 Now for part c, we need to find the maximum value of the kinetic energy of a proton that is traveling in this cyclotron...

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