A beam of protons traveling at 1.20 km/s enters a uniform magneti...

Question

A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E27.24) . The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

Accepted Answer

Hey everyone. So this problem is working with magnetic fields. Let's see what they are asking us and what they give us in the problem. So we have a beam of singly charged particles, it moves through a uniform magnetic field um at a given speed, they give us the mass of the particle. And they tell us that the magnetic field deflects the particles beam and the beam leaves that field region perpendicular to its initial direction for us. Then to determine the strength of the magnetic field, the distance traveled is this given distance here. So this is a tricky problem. The first thing that we need to recognize here is that if the beam leaves the magnetic field perpendicular from the initial direction, the distance traveled by the beam will be the circumference of a quarter circle of a given radius. So we can recall that the circumference of a circle is just given as two pi R. We're talking about a quarter circle. We're going to divide that by four. We can also recall that the equation for radius in a uniform magnetic field of a charged particle is given by R equals M the over Q B. And so this is good because this term here B is the strength of the magnitude of the magnetic field. So that's what we're going to be solving for. So we can rearrange this first equation, plug it into the second equation and then rearrange that for B and will actually have everything we need to solve the problem. So let's just step through what that looks like. So for this first equation R equals four D over two pi and then we can just simplify that so that it's two D over pi and then we'll rearrange this one for B our magnetic field magnitude and that becomes N V high over Q two D when we plug in this two D over pi for our. So from there, let's just step through each of these terms and make sure we have everything we need the masses given as 1. times 10 to the negative 26 kg. The speed is given as four kilometers per second. So watch those units, I'm going to rewrite that as four times 10 to the third meters per second. Pie is a constant Q. They tell us that it's a singly charged particle. So we can recall that that means that the charge of the particle is 1.6 times 10 to the negative 19 Fulham's, it's just a constant there. And then D the distance is also given to us in the problem statement as 3.24 centimeters. So I'm going to rewrite that as 0.303 to four m. And from there, we can just plug and chug we'll plug all of these in. So our mass times our velocity times pi all over our charge Times two times our distance. And we are left with a Magnetic field magnitude of 1.40 times 10 to the negative to Tesla. And so that is the answer for this one, we look at our choices and that aligns with choice D. And so that's the right selection. That's all we have for this problem. We'll see you in the next video.

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