An early cyclotron at Cornell University was used from the...

Name: Problem 114, Chapter 19: Physics
Uploaded: 2021-07-21T17:48:17-08:00
Duration: 9 min 39 s
Channel: Vishal Gupta
Description: An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of 1.3 T over a region in the shape of a flat cylinder. Two hollow copper dees

An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of 1.3 T over a region in the shape of a flat cylinder. Two hollow copper dees

Key Concepts

Resonance and Radio Frequency (RF) Acceleration

Resonance in the context of a cyclotron refers to the synchronization between the frequency of the oscillating electric field and the natural frequency of the charged particles’ circular motion. The radio frequency (RF) accelerating field is applied in phase with the particles' transit through the gap between the electrodes, allowing them to gain energy consistently on each pass.

Lorentz Force

The Lorentz force is the force exerted on a charged particle moving within electric and magnetic fields. In a cyclotron, the magnetic component of the Lorentz force causes the particles to travel in curved paths, while the oscillating electric field (applied at the correct phase) increases their speed. This force is fundamental to both the trajectory and the acceleration of the particles.

Cyclotron

A cyclotron is a type of particle accelerator that uses a perpendicular magnetic field and an oscillating electric field to accelerate charged particles along a spiral path. The design relies on the principles of circular motion and timed energy boosts, allowing particles to gain energy with each pass across the gap between the accelerating electrodes.

Uniform Magnetic Field

A uniform magnetic field is critical in the operation of a cyclotron because it ensures that the force experienced by the charged particles remains constant as they move through the device. This uniformity provides a constant bending force (via the Lorentz force) which keeps the particles in a predictable, circular trajectory, essential for the efficient acceleration process.

Transcript

00:01 All the given problem, magnetic field created in a cyclotron is given as 1 .3, tesla.

00:23 The radius of the cyclotron is r is equal to 16 centimeter or 0 .16 meter.

00:36 In the first part of the problem, you have to find frequency of oscillations of applied alternating voltage.

00:54 Frequency of oscillations of applied alternating voltage will be given by the expression f is equal to b, q, by 2, pi, m.

01:24 Q is the charge and m is the mass of the particle which is being accelerated within this cyclotron and that is given to be a proton.

01:35 So for proton this frequency will be given by magnetic field 1 .3 tesla charge over the proton 1 .6 into 10 to the power minus 19 column divided by 2 pi means 2 times.

01:57 Times of 3 .14 multiplied by mass of proton which is 1 .67 into 10 dash minus 27 kilogram.

02:09 So finally this frequency here comes out to be 1 .98 into 10 raised the part 7 hertz or approximately we can say this is 2 .0 into 10 into 10 dish part 7 hertz which is the answer for the first part of this problem and moreover we can write we can keep this frequency as 20 into 10 dash 6 which may be written as mega so this is 20 megahertz also now in the second part of the problem we have to find total kinetic energy gained by this proton when it comes out of the cyclotron.

03:08 So that total kinetic energy gained by the proton within cyclotron will be given by e k is equal to q square into b squared into r square divided by of mass of the proton.

03:45 So plugging in on one values per charge, this is 1 .6 into 10 kishpar minus 19 hula.

03:54 For b this is 1 .3 tesla.

03:58 The radius this is 0 .16 meter and divided by 2 times of the mass of the proton.

04:06 So here this is 2 times of 1 .67 into 10 against the power minus 27 kilogram.

04:14 So finally this kinetic energy here gained by the proton it comes out to be 2 .1 into 10 x to the power minus 11 june which is the answer for the second part of this problem...

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