A cyclotron with a radius of 1.0 m is to accelerate...

Key Concepts

Cyclotron Operation

A cyclotron uses a perpendicular magnetic field to bend charged particles into nearly circular paths while an alternating electric field accelerates them each time they cross the gap between the dees. This combination of fields enables the particles to gain energy progressively until they reach the desired kinetic energy before being extracted.

Lorentz Force and Circular Motion

Charged particles moving in a magnetic field experience the Lorentz force, which acts perpendicular to their velocity and causes them to follow circular trajectories. The relationship between the magnetic force (qvB) and the required centripetal force (mv²/r) is central to determining the particle’s radius of curvature within the cyclotron.

Deuteron Properties

Deuterons, being nuclei of deuterium, have a charge equal to that of a proton but roughly double the mass. Understanding the mass and charge of deuterons is critical when applying formulas involving momentum, force, and energy, as these factors directly impact the cyclotron's design parameters.

Cyclotron Frequency

The cyclotron frequency is the constant rate at which a charged particle spirals in a uniform magnetic field under non?relativistic conditions. It is given by the formula f = qB/(2?m) and determines the optimal frequency for the alternating voltage applied between the dees, ensuring that the particle receives a consistent energy boost at the right phase of its circular motion.

Acceleration and Energy Gain per Revolution

In a cyclotron, each time a charged particle crosses the gap between the dees, it experiences a potential difference that increases its kinetic energy. The energy gained during each pass is used cumulatively to calculate the total number of revolutions required for the particle to reach the target energy.

Time and Path Length Calculations

The period of each revolution, derived from the cyclotron frequency, multiplied by the number of revolutions gives the total acceleration time for the particles. Additionally, knowing the circumference of the particle's path allows estimation of the total distance traveled during the acceleration process.

Transcript

00:01 In this problem, we're looking at duteron's accelerated in the cyclotron.

00:05 So we have a cyclotron as two halves.

00:09 They're called d's.

00:10 There's a gap between them.

00:12 Each time the duteron cycles and passes through the gap, it adds an acceleration to the voltage.

00:21 The voltage is 22 ,000 volts.

00:25 We're given the mass of the duteron.

00:28 It's 2 .014.

00:31 Atomic mass units, so it's 3 .3, 4 times 10 to the nickel 27th kilograms.

00:36 It gives us the maximum connect energy.

00:39 So when the radius is maximum, this capital r is 1 meter.

00:44 The connect energy is 12 million electron volts.

00:51 The deuteron is implicitly singly ionized.

00:56 So the charge, or magnitude of the charge is 1 .6 times the negative 19.

01:02 Coals.

01:05 Cyclotrons cannot accelerate neutral atoms, neutral particles.

01:11 So the first thing it wants to know is what is the magnetic field strength be.

01:17 So for this, first thing we're going to know that this is on each cycle, it's approximately centripetal acceleration, centripetal force, a circular rotation.

01:33 But then that's expanded with each cycle when it adds connect energy.

01:41 So we have centripetal force.

01:54 So that is qvb, so it's charge times the speed, times magnetic field strength equals mv squared divided by r.

02:06 We know that the maximum kinetic energy is 1 .5m v squared.

02:20 So we can say 1 .5 times times 1 .5m like this.

02:51 So r equals r at k max.

03:03 So that works out to be q squared, b squared, r squared divided by 2m.

03:21 So it follows from that, that 2m k max divided by q squared r squared square root equals b.

03:47 Okay, since it, the problem wants a numerical value, that's 2 times 3 .34 times 10 to the minus 27 kilograms.

04:06 Let's see, kmax.

04:09 So 1 .6 times 10 to negative 19 joules per electron volt into 12 million electron volts is 1 .92 times 10 to negative 12 joules.

04:35 It's just doing that conversion because this other stuff is in si units.

04:58 So square root of that whole thing equals 0 .71 tesla is the magnetic field.

05:15 Strength.

05:17 So then what is the synch cyclotron frequency f? so the textbook derives the frequency relationship.

05:29 It actually doesn't depend on the radial distance.

05:34 It's the same throughout the cyclotron.

05:36 It also doesn't depend on the voltage.

05:38 The voltage just reduces the number of periods for the particle reach maximum kinetic energy, but the frequency stays the same regardless.

05:52 So f equals qb by 2f pi m.

06:01 And this problem m refers to the mass of a deuteron.

06:06 So it follows from that.

06:10 1 .60 times 10 to the minus 19.

06:14 Kulums times 0 .71.

06:19 Tesla.

06:24 I'm writing tesla out because i'm going to use the capital t for to refer to the period a little bit later.

06:33 So 2 pi times 3 .34, 10 to the negative 27 kilograms.

06:47 And that is 5 .40 times 10 to the 6 hertz.

06:55 It's our inverse seconds.

06:58 So f equals 5 .40 megahertz.

07:09 Okay.

07:10 So then see, given that the gap voltage is 22 ,000 volts, how many revolutions by the time the duteron exits the cyclotron is when it's at maximum connect energy.

07:52 So each time it passes the gap, it gains plus qv energy.

08:01 It gains 22 ,000 electron volts or i should say per period.

08:32 So n number of revolutions is the k max.

08:40 Each cycle passes, sorry, passes through the gap twice.

08:48 It goes one way, then the other way...

Please give Ace some feedback