OpenStax College Physics for AP® Courses, Chapter 22, Problem 12 (Test Prep for AP® Courses)

This is College Physics Answers with Shaun Dychko. An electron is accelerated starting from rest through a potential difference of 240 volts, the charge on an electron is the elementary charge— 1.60 times 10 to the minus 19 coulombs— and after it's finished being accelerated, it enters a magnetic field with a strength of 3.5 millitesla, which is 3.5 times 10 to the minus 3 tesla and this field is directed straight down and the electron is moving to the right here in the positive x-direction. So I'll answer question (b) first because that's the quicker thing to answer what's the initial direction of the magnetic force on this electron when it enters this field? We'll use the right hand rule with our thumb pointing in the direction of the velocity, which is to the right our fingers will point in the direction of the magnetic field, which is into the page and our palm is in the direction of force that would be applied on a positive charge and the palm is facing upwards. Now since this is a negative charge, we take the opposite answer to what the force direction is on a positive charge being a negative charge, it will be directed down in other words. So thumb to the right, fingers into the page, palm faces up because this is negative take the opposite answer to the palm that is down. There! So part (a) then is what is the radius of curvature of this circle that the electrons will move in this uniform field? Because this force due to the magnetic field will always be perpendicular to the velocity and it is therefore a centripetal force and so this circle has some radius r that we are going to try and find here. Okay! (Let's put that in green... there!) The force due to the field is going to be the amount of charge q times its speed v times the magnetic field strength times sin of an angle between them but these things are perpendicular the velocity is in the plane of the page whereas the field is into the page and so those are perpendicular. Because this is a centripetal force, it can also be written as mass times speed squared divided by the radius so we can equate these two things and we have mv squared over r then is qvB and our job is to solve for r so we'll do that in two steps, let's multiply first by r over v and switch the sides around as well and we have rqB on one side and we have mv on the other side then divide both sides by qB and we get that the radius is then mass times speed divided by charge times magnetic field strength. Now we don't know what v is though, we know what the mass of an electron in the charge is and we are given the magnetic field strength but we need to use this potential difference through which the electron has accelerated in order to determine the speed with which it's entering this magnetic field. So you know, energy is conserved here: we have some initial kinetic energy plus initial potential energy is going to equal the final kinetic energy plus the final potential energy. So we'll subtract the initial kinetic energy from both sides and also subtract the final potential energy from both sides and then we get this line here, which is kinetic final minus kinetic initial equals potential initial minus potential final and because this i and the f are switched around here compared to where they are on the kinetic energy side, we can say that this is the negative of potential final minus potential initial and this bracket here can be substituted with ΔPE so change in something is its final state minus its initial state so we have the change in kinetic energy is the negative of the change in potential energy but since there is no initial kinetic energy at all because the electron begins at rest, we can just write kinetic energy final on this side. So we have one-half mass times final speed squared equals negative of the change in potential energy, which is charge times potential difference and we can solve this for v by multiplying both sides by 2 over m and then take the square root of both sides as well and we get v is the square root of negative 2qV over m. Now you could plug in some numbers here and get an answer as a number and then plug that number in for v in our formula for radius but I don't know... I always have a habit of getting an algebraic answer and substituting numbers at the very last step. So we are going to replace v then with this square root negative 2qV over m and so we have m to the power of 1 divided by m to the power of a half, which is m to the half or m in the square root in other words and we have q to the power of a half, which is to say it's in the square root sign divided by q to the power of 1 which is... we are going to end up with q in the square root sign or q... not sure how to explain the division there but you know, you have q to the power of one-half divided by q to the power of 1 which is q to the power of one-half minus one which is q to the power of negative a half which is 1 over square root q and this explains why it is in the denominator position in the square root. Okay! And the magnetic field is not part of this radical and so we just have 1 over B outside the radical. So 1 over B times square root negative 2 mass times potential difference divided by charge. So that's 1 over 3.5 millitesla, written as 3.5 times 10 to the minus 3 tesla times square root of negative 2 times 9.11 times 10 to the minus 31 kilograms— mass of an electron— times 240 volts divided by the charge on an electron, which is negative 1.602 times 10 to the minus 19 coulombs and this works out to 15 millimeters when you round it to two significant figures.