For this question. Rats to find the radius of circular orbit for do Theron in part A in terms of the radius of orbit of proton in a magnetic field where the velocity is perpendicular to the magnetic field. Okay, so to do this first, let's find out what the radius of the proton or the radius of a charged particle in general and a magnetic field is going to be. So the law of conservation of energy states that the absence of non conservative forces, the total energy of the system, remains constant. So therefore, the change in kinetic energy Delta K plus the change of potential bell to you is equal to zero. Or, in other words, the potential energy Delta U was equal to negative Delta K work Delta K is, um okay, final minus K initial finals. The final kinetic energy and K initial is the initial kinetic energy. Okay, but since the particle comes at rest, its final kinetic energy is zero, so we can let that go to zero. So therefore, the change of potential energy dull to you is equal to minus negative K I or just K II, which is the initial kinetic energy. So the energy possessed by a charged particle moving by virtue of motion. The kinetic energy here is equal to 1/2 in the squared. So dull to you, then is equal that, since it's equal to case of eye, is equal to also equal to 1/2 and the square okay. But the potential energy acquired by a charged particle that was accelerated through a potential difference is equal to Q times. The change in, uh, voltage potential Delta V. So we have dealt to you. There's also equal to the charge of the particle Q times The change in potential voltage Delta V We can set these equal to each other and sulfur V. Let me find that V here is equal to the square root two Claims Cube. I'm still TV, and this is velocity. So let's maybe indicated difference between velocity and voltage here, so we'll use little V for velocity and big B for voltage. That way you can see the distinction between them, so velocity is equal to two q Q times to change in the potential voltage Delta V, divided by the Mass m. Okay, but we're asked to find the radius of the orbit. So the magnetic field in this temple centripetal force balance in this situation. So therefore we have F sub C is equal to half seven, which f sub C is equal to in times V squared over our and F's and M is equal to the charge times. Um, the velocity V yeah, this is also velocity. So let's keep using little B times magnetic field times the side of the England between. But we're told the angle is data or is 90 degrees and the sign of 90 is one, so we can ignore that. We can rearrange this to Fox Sulphur, are we find that are here is equal to mm. Divided by Q B. Um, we can cancel out to use some of these voltages velocities here. Okay, And then for the velocity will plug in the term we found above. So this is multiplied by the square root of two times Q times. Delta V over in. Okay, and we're gonna use some simple algebraic properties here to simplify this a little bit more. So, for instance, route Q over Q is equal to one over route Q and em over route M is equal to room em. So using those facts we're gonna simplify are a little bit more We find that are here is equal to two times em times Delta V divided by q b squared all under the square root. So what we found at the bottom of this page is a public to that just with a little bit of algebraic rules there. So now we can substitute in our value for let's say, this is starting part ay, here we can substitute in our value for our sippy since you just replace everything in that with proton related. So this is gonna be two times the mass of the proton m stumpy times Delta v divided by, uh, the charge of the proton, which is the same as the charge of the electrons. So we could just write e times b squared. Okay, well, now, let's compare that to what would be the radius of orbit for the deuterium. Our sub d is going to be equal to the square root of two times the mass of its teary, um m sub d. The potential difference is the same Delta V divided by the charge of deuterium called excuse of D A times B squared. Okay, well, you can look these values up. Um, but I'll just tell them to. You hear the mass of deuterium is two times the mass of the proton. So this is two times s a p. I'm still TV divided by the charge of deuterium is the same as the charms of the proton. It's e times b squared. So if you compare that to arson p above you find that this is route two times more than next. You can pull that route to that gets multiplied by the M C. P out front so would take this route to we can pull it out front. And this is just gonna be rude to times arson p makin box. It is their solution for a for part B rasa do the exact same thing except for this time with an Alfa particle. So we have our save. Alfa is equal to the square root. It's right there of two times in some Alfa. Sometimes the change of potential divided by the charge of Alfa Times. Be square. Okay, so now let's put these in terms of the mass of the proton in the charge of the proton. So this is too times the mass of the alpha particle is four times the mass of the proton. Okay, time still to be divided by the charge of the Alfa particle is two times the charge of the proton. So this would be too time z times b squared. Okay, so we can pull that the extra four and the extra two that we get out. So we have root for over root to the root for over route to is just route to So this is also equal thio route too times the radius of the proton making box that in is their solution to part B.