In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV. An alpha particle has charge q = +2e and mass m = 6.64x10^-27 kg. If the magnetic field isn't changed, what will be the orbital radius of the alpha particles?

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Hey everyone. So this problem is working with magnetic fields. Let's see what they are asking us and what they're giving us. So we have um aspect that separates ionized particles, a single ionized particle of a given mass and a given energy moves in a circle of a given radius and a uniform magnetic field. They're asking us what the orbital radius of a doubly ionized particle of a different given mass entering the same magnetic field with the same energy would be. So we first need to recognize that we have two separate scenarios here. So we're going to focus first on the first scenario and we can do that by recalling that the radius of a charged particle in a uniform magnetic field is given by R equals M V over Q B. And so let's see what we have for this first scenario. We are given R the radius is 18 centimeters. I'm gonna rewrite that as 1.18 m, we are not given velocity. The charge is a single ionized particle. So we need to recall our electric constant Is going to be 1.6 times 10 to the - columns. And then we are also given the mass of the first particle, Which is 3.2, 7 times 10 To the -25 kg. So we've got two unknowns. We don't know velocity and we don't know the magnetic field. They do give us energy though. And so that's a hint that we can use our kinetic energy equation where K equals one half M V squared. So we're going to solve this equation for velocity and then plug that into this first equation. So that looks like B one is equal to the square root of two K over M one because our kinetic energies are the same between the two scenarios. So we don't need to denote that. And so when we plug that into our first equation, we get R equals square root of two KM 1 over Q B. And now we again, we were given K is one mega electron volt. So I'm just gonna rewrite that as one times 10 to the sixth electron volts. But other than that, we have everything we need to solve for the magnetic field. So B equals Square root of two. Again, times are kinetic energy, one times 10-6 electron bulbs times our mass, 3.2, 7 times 10 to negative kg all over The charge, 1.6 times 10 to the -19 columns, times our radius .18 m. We plug that in and we are left with a magnetic field of 2. Times 10 to the -10 Tesla. Okay. So now we're looking at the second scenario, so We know that the second scenario has the same magnetic field. So that's, that's good B2 equals 2.808 times 10 to the -10. We know the mass in the second scenario is one point times 10 to the - kg. And the charge in the second scenario, it's a doubly ionized particle. So it's just gonna be two times our electric constant, 1.6 times 10 to the -19 woolens. So we are asked for the radius. So again, we can write that as R two equals M two V two over Q two B two, but we still don't have velocity. So we'll write our kinetic equation or scenario two K two equals one half and the two squared and sub that in for our velocity. So again, this looks like the two is two K over M two And then our two becomes A square root of two K and two over Q to be too. When we plug in all of those knowns there, we get to times one times 10 to E V times 1.59 times 10 to minus 25 kg all over two times 1.6 times 10 to the minus 19 columns times our magnetic field that we found from the first scenario 2.8 oh eight times 10 to the minus 10 Tesla. And we plug all of that in and come up with our radius of 0.06, 2, 8 m or 6.2, 8 cm. And so that is our answer and that aligns with answer A. So a is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

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Key Concepts

Cyclotron Motion

Lorentz Force

Radius of Circular Motion