A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34x10^-27 kg and a charge of +e. The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

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Hey everyone. So this problem is working with magnetic fields, let's see what they're giving us and what they're asking of us. So in a nuclear experiment, we have a proton of a given mass in charge that is moving in a circular trajectory in a uniform magnetic field. The radius of that trajectory is given to us as well. And the magnitude of the magnetic field in which it is moving is also given to us in the problem. They're asking us to calculate the speed the period of revolution of the proton and the difference in potential that the proton would need to overcome to reach the speed. So it is a three part question and it's a little overwhelming at first, but we're going to break it down part by part and show you how the information that you need to solve each of the parts is contained in this problem. So for part one, the first thing we're going to do is recall that the radius of travel for a charged particle in a uniform magnetic field is given to us in the equation R equals MV over the absolute value of Q times B. So we're gonna take each of these terms one by one. So R is the radius that's given to us in the problem as 8.7 millimeters. I'm gonna rewrite that as 8.7 times 10 to the minus 3 m to keep everything in standard units. Our mass next is also given to us in the problem as 1.67 times 10 to the minus 27 kg V is our velocity or our speed. That's what we're solving for Q is the charge. And that is given to us. And the problem as well is positive 1.607 times 10 to the minus 19 columns. So it's positive. So the absolute value of a positive is the positive. So that's good. And then B is the magnitude of our magnetic field. So it's 1.25 teslas. So from here you can see that we have everything we need to solve for velocity. We'll rearrange this equation to be RQB over M and then plug and chug from there. So R is 8.7 times 10 to the minus 3 m Q. Our charge is 1.607 times 10 to the minus 19 columns. And then the magnitude of our magnetic field is 1.25 teslas, all of that over our mass of 1.67 times 10 to the minus 27 kg. We plug that into our calculator and get a speed of 1.05 times 10 to the sixth meters per second. And so that's the answer for part one. So when we look at our choices, we can actually tell that the only correct answer is D but let's work through parts two and three to make sure that we're on the right track. So part two is asking for the period of revolution of the proton. So we can recall that the period of revolution is the time it takes to go around in a circle once. And that equation for time is given as T equals two pi R over lean. We have, we are solving for the time we have the radius and we've just solved for the velocity, we have that as well. So we are in another plug and CHG situation. So T equals two pi times 8.7 times 10 to the minus 3 m all over 1.05 times 10 to the 6 m per second. We plug that in and we get 5.21 times 10 to the minus eight seconds. And that is the answer for part two. So we go into choice D and yes, that is correct. So we looks like we are on the right track. So for part three, they are asking us for the potential, the difference in potential that the proton would need to overcome to reach that speed, the speed that we found in part one. So the potential is a hint to use the conservation of energy for potential energy. So we can recall that the conservation of energy is given as delta K plus delta U equals zero. And delta K is going to be one half MV squared and delta U for a charged particle, it is going to be Q times V or voltage. So notice here we're working in lower case V and upper case V. Um So let's make sure we don't get those confused. So when we plug that in, we'll have one half M or lower case V squared plus QV equals zero, we will solve that for V MV squared. We'll so that for upper case BM lower case V squared over two Q and we have everything we need to solve for that. So our mass is 1.67 times 10 to the minus 27 kg. Our velocity is 1.05 times 10 to the 6 m per second. And that term is squared and two times our charge, which is 1.607 times 10 to the minus 19 poum plug that into our calculator and we get 5.73 times 10 to the third volts. And that is the answer for part three. So we go back up to D and we see that yes, that is correct. So we were tracking along here. The correct answer for this problem is D that's all we have for this one, we'll see you in the next video.

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Key Concepts

Lorentz Force

Centripetal Force

Kinetic Energy and Potential Difference