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Electrons in a beam are accelerated from rest through a potential difference $\Delta V .$ The beam enters an experimental chamber through a small hole. As shown in Figure P29.59, the electron velocity vectors lie within a narrow cone of half angle $\phi$ oriented along the beam axis. We wish to use a uniform magnetic field directed parallel to the axis to focus the beam, so that all of the electrons can pass through a small exit port on the opposite side of the chamber after they travel the length $d$ of the chamber. What is the required magnitude of the magnetic field? Hint: Because every electron passes through the same potential difference and the angle $\phi$ is small, they all require the same time interval to travel the axial distance $d$ (FIGURE CAN'T COPY)

Uh huh. Well given that a cartoon ray beam and it's Catherine middleton with electron and when I try to be murder and mass is Of the electron is 9.1 time. Standard Power -31 kg. And the charges minus E. Charge which is equal to minus all went six times stands above minus 19 columns. Mm And that's all that is in the second part. So the force magnetic force we equal to change battle force regard to keep it in the circle. So can you quit those two Q V V. Science data? The quarter M. V. Squared or he constantly out on both sides and mhm V be a culture Q. B. Same data over. Mhm. And times are Mhm. Yeah. So we're given that B is equal to 4.5 times transfer power one ST Teslas or is a quarter 22 cm which is 0.02 substitutes. And we have Mhm 1 -1.6 10 cents above -19 thomas. 4.5 time stands about ministry sign. And the angle of the 19 the radical audacity magnetic field do lusty which is going to find and thomas do you? Radius 2 0.02. The mass is 9.1. To understand the ball -31. Is it called Sir? One went 58 understands above seven with every second approximate

Here we have to find the radius of the circle traced by an electron which is moving with a Cuban. Speed in acura and magnetic fields. And the magnetic field is normal too. The velocity of the microphone. So yeah, he had when the electron moves on a circle, the magnetic force F provides the centripetal force, centripetal force is envious. Where upon our mm isthmus fee is speed. And our radius now magnetic force F is given by Q B. B. Science theater. Q. E. The magnitude of charge an electron which is E be signed. Theta is the angle between I think electron's velocity and magnetic field Which is 90°.. So this gives us a the equal to This gives us a radius article two M. V upon E. This can be written as we upon E buy em who took it Now speed is 5.2 In 210 20 power 62 m/s and Charles to mass ratio upon M. S. Mhm. Mhm. Yeah. And magnetic field is 1.3 and two 10 to the power minus status left. Mhm. So this gives us 0.27 meter to convert it into centimeters Be multiplied by 100. This gives us 22.7 centimetre as the radius after electron. Now for part B, the energy of electron is 20 mega electron volt. So well. First try to find the speed of flip phone. Can I take and educate equal to have and we square. So he is going to stay on. Em Kinetic energy is 20 million electron volts. So 20 to 10 to the Power six electron world. And electron world is converted into Azul By 1.6 into 10 to the power -19 and mass is 9.11 in 2, 10 to the power minus 31 kg. So this gives us speed equal to 2.65- 10 to the power nine m/s. But the speed is greater than speed of light, which is C Equal to three and 10 to the power eight m/s, which is the speed of light in vacuum. According to a special theory of relativity, any speed cannot be better than the speed of light in vacuum. So this speed will not work here. And we have to use the relativistic formula where mass is given us. I am equal to I am not well, I'm not either rest mass, Which is a mass when the electron is at zero speed or electron is at rest minus V square upon C square To the power one x 2. Hard. It can return as I am not 1- The Square Fancy Square Square. So Using this master radios can be given by Article two. I am not the, sorry, radius we found about envy upon E. B. And we put the value of them here. I am not upon E B will remain the same and look under one minus the square ponsi square. This is a required radius. So we'll have to use a relativistic formula to find the radius when the energy is 20 mega electron world, which is much, much greater than the rest energy of the electron, which is 0.512 mega electron volts. So this completes the solution. Thank you.

This problem has a beam of electrons being shot between to charge plates of a parallel plate capacitor, and it wants to know how much kinetic energy each electron will have when it leaves the plates. So let's think about kinetic energy, right? Since we're given a kinetic energy initial for each electron, we confined kinetic energy by finding the work that the plates do on each electron and then adding the initial kinetic energy that's given so the work that each plate does on the electron. What's the simplest way to find work, the simplest way to find work? The most basic definition of work is that work is force times a distance. So bear with me here, this is going to get a little ugly, and we're gonna have to pick up pieces at the end. But we're gonna get smaller and smaller. We have the distance that the electrons get deflected. It's three millimeters. We don't have the force. How do we find force on a charge from electric field? It's the magnitude of the charge times, the magnitude of the electric field. Now it's happened again. We don't have electric field. How do we find electric field in a in a parallel plate capacitor. Specifically, electric field is the same as Volt over the distance between the plates. Now keep in mind here there's two different D's at play. The reason I made one capital on the other not is because they both refer to different distances. This is the distance that the electrons were deflected. This is the distance between the plates. So let's plug this W back in up here and then this f with this and then this e with this and I see what we get, our final kinetic energy is going to be equal to the charge times the voltage over the distance between the plates times the distance that the electrons are deflected, plus initial kinetic energy. Finally, we have something that we can solve for, So our que ee final is e the magnitude of the charge of electron times. It's 100 kilovolts, so 100,000 volts at the lot, divided by 12 millimeters times three millimeters, then finally plus two times 10 to the minus 15 jewels. Plug that big thing into your calculator and you get out 6.0 times 10 to the minus 15 jewels. And that's the final answer. It's a little ugly, but it works. Good luck

So for this problem were given the radius, which we can use to find the area by taking pi r squared and were given the current and the velocity. And so for part A, we want to find the current dense D J, which is equal to the current over the area, and we can substitute in those values to give us 2.55 amps per meter squared. That's part a for part B. We want to start by finding the amount of time it takes for one particle to pass. So you know that the current is equal to Delta Q over Delta T. And so Delta T is equal to still to Q over I that's a que is just go to be the charge of it electron. So 1.6 times sent to the negative 19 cool ums. So Delta T is going to be two times since the negative 14 seconds. So this is the amount of time it takes for one electron to pass. And from that we want to find the separation between the electrons. So we know that velocity is equal to the separation over time it takes one electron to pass. So our is equal to fi times Delta T, which is equal to six times 10 to the negative six meters. So this is a separation between each electron and the beam. Then, to find the density, you just need to take one over the volume, which is one over the separation times the area of the electron beam, and that's equal to 5.3 times 10 to the 10 electrons over readers cubed. So then, finally, for part C, we want to find how long it takes for a mole of electrons to emerge from the accelerator. So the distance that one electron travels through is our. If we multiply that by of a God dress number, that's the amount of time it takes for a mole of electrons to pass through a point. And so that's equal to 3.6 times 10 to 18 meters. So we want to find the amount of time it takes for one electron to travel 3.6 times into the 18 meters. So to do that will take Delta T is equal to Delta X over velocity, and we find that the time is equal to 1.2 times 10 to the 10 seconds, which is about 380 years


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