A cyclotron designed to accelerate protons has an outer...

A cyclotron designed to accelerate protons has an outer radius of 0.350 $\mathrm{m}$ . The protons are emitted nearly at rest from a source at the center and are accelerated through 600 $\mathrm{V}$ each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 $\mathrm{T}$ (a) Find the cyclotron frequency for the protons in this cyclotron. (b) Find 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 one proton accelerate?

Key Concepts

Acceleration Time Interval

The total time a particle spends being accelerated in the cyclotron can be deduced by combining the constant cyclotron frequency with the number of revolutions completed. Given that the period of one revolution is the inverse of the cyclotron frequency, multiplying this period by the total number of revolutions provides the overall acceleration time.

Revolution Count in a Cyclotron

This concept involves determining the number of times a particle circulates within the cyclotron before reaching a certain energy. By understanding how much energy is gained per gap crossing (or per revolution) versus the total energy required for the particle to exit the device, one can calculate the number of revolutions made during the acceleration process.

Cyclotron Frequency

This concept involves the constant frequency at which charged particles orbit in a cyclotron when moving in a uniform magnetic field. The cyclotron frequency is given by the formula f = qB/(2?m), where q is the particle’s charge, B is the magnetic field, and m is the mass of the particle. This frequency remains constant for non?relativistic speeds, enabling synchronized acceleration in the cyclotron.

Uniform Circular Motion in a Magnetic Field

Charged particles moving perpendicular to a uniform magnetic field experience a magnetic (Lorentz) force that always acts perpendicular to their velocity, causing them to move in a circular path. The balance between the magnetic force, qvB, and the centripetal force, mv²/r, allows one to relate the radius of the particle’s circular trajectory to its velocity, magnetic field, and intrinsic properties.

Energy Gain from Accelerating Voltage

In a cyclotron, particles gain kinetic energy each time they cross the gap between the dees due to the applied accelerating voltage. The energy acquired per crossing is given by the product of the particle's charge and the voltage (qV). The cumulative energy gain, which determines the maximum kinetic energy, is the sum of these individual increments over successive revolutions.

Transcript

00:01 Hi everyone in this problem.

00:04 Cyclotron is designed to accelerate the proton.

00:07 So cyclotron is accelerating proton.

00:22 The radius of cyclotone is given 0 .350 meter.

00:32 The proton are emitted nearly at rest.

00:34 So initial velocity of proton is zero.

00:39 From a source at the center and accelerating potential is 600.

00:46 600 volt this is the accelerating potential of proton these are between the poles of electromagnetic where the field is so in the cyclotron magnetic field of 0 .80s is used we have to calculate first there are number of parts in the solution question which is to be calculated so we are calculating one by one find the cyclotron frequency the frequency of proton we have to calculate and in second part we have to calculate speed at which proton accessed so the final velocity of proton we have to calculate third maximum kinetic energy of proton how much energy it will gain that we have to calculate d how many revolution does the proton make in the cyclot number of revelations in cyclotron and and last part, time for time for how long it was inside the cyclotone.

02:35 So these are the parameter we need to calculate.

02:39 We start solving them 1 by 1.

02:42 That is the start solving.

02:45 First part frequency is given by qb by m.

02:55 Substitute the value.

02:56 This is proton's charge is magnetic wheel is given .8 -00.

03:04 Tesla and mass is 1 .67 into 10 to the power of minus 27 kg.

03:11 So on solving the frequency you will get 7 .66 into 10 to the power of 7 radian per second.

03:21 So this is one of the answer.

03:23 This is the answer of first part...