An electron in a cathode-ray tube is accelerated through a pot...

Question

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the 2.0-cm-wide region of uniform magnetic field in FIGURE P29.60. What field strength will deflect the electron by 10°?

Accepted Answer

Welcome back. Everyone. In this problem, a proton in a particle accelerator is speeded up using a potential defense of eight kilovolts and then sent into a three centimeter wide space containing a uniform magnetic field. If the proton experiences a deflection of 15 degrees due to the magnetic field, calculate the magnitude of the magnetic field. For our answer choices. A says it's 0.077 teslas B says 0.11 teslas C says 0.21 teslas and D says 0.05 teslas. Now, in this problem, we are trying to find the magnitude of the magnetic field for a proton in a particle accelerator. And that proton takes on a cyclotron motion. Now recall that for a cyclotron motion in the magnetic field, then the radius is given by MVK divided by QB where M is the proton's mass V is the proton's velocity Q is the proton's charge and B is the magnetic field. Therefore, that tells us then that to find the magnitude of the magnetic field, we can rearrange our equation to become B equals MV divided by QR. Now we're almost there. But the problem is we don't know what the velocity of our proton is and we don't know what the radius of our uh the radius of our cyclotron motion is. So if we can figure both of those out, then we'll be able to find the magnitude of the magnetic field. So first, let's try to figure out the velocity of the proton protons velocity. So solving for V, OK, when we think about it, our proton is moving. So in this problem, we can use the conservation of energy principle to find the proton's velocity in the magnetic field. So solving four V using, let me write that out using the principle of the conservation of energy. OK. Then what do we know? Well, we know that that principle tells us that the final kinetic and potential energy is equal to the initial kinetic and potential energy. OK. So the final, let me write it here. KF plus UF where F represents final equals K I plus UI where I represents initial. Now energy is conserved as the proton moves across the space containing the uniform magnetic field, we also assume that the proton starts from rest. And we know that the proton will follow a circular path as it enters the magnetic field. So the final kinetic energy, we know that the formula for kinetic energy is half MV squared. So because it will be moving that final kinetic energy is half MV squared. OK? Because it's moving the final potential energy will be zero and the initial kinetic energy was zero because as we said, we assume that a proton started from rest and it had that potential energy which equals Q multiplied by V. OK. That's the potential of our, of our charged particle. So if that's the case, then we can now solve for our velocity. OK? Because now this tells us by doing some transposition here, our velocity OK is going to be equal to or rather let me write a code step by step just for us to see it clearly. So here we have half MV squared being equal to Q multiplied by our voltage. So that means that V squared is going to be equal to two multiplied by Q multiplied by a voltage divided by the mass. Therefore, our velocity VF is going to be equal to the square root because if we find the square root of both sides, so that will be the square root of two QV divided by M. Now we can work that out because we know, OK, we know that our charge Q, the charge of a particle is 1.6 that's a constant, 1.6 times 10 to the negative 19 coulombs. OK. We know that we uh have a potential difference of 8 kg votes. So that's eight times 10 to the third votes. And we're dividing that by the mass of the proton, which is also a constant because the mass of A proton is 1.67 times 10 to the negative 27 kg. OK. So now we can solve for our protons velocity. And when we put that into our calculator, we get the velocity of our proton to be equal to 1.24 times 10 to the 6 m per second. OK. So now we have the velocity of our proton. We're at a good start, know that we've solved for that velocity. The next thing we need to find is the radius of the circular path. Now, we know that it's sent into a three centimeter wide space. So that gives us an idea then that tells us, OK. That tells us that the radius of our circular path, OK. R the radius of our circular path is going to be equal to that width three centimeters divided by the sine of the deflection, the angle of deflection due to the magnetic field. So that's gonna be the sine of 15 degrees right? In that in si units, that would be 0.03 m divided by the sine of 15 degrees. And now we have an expression to solve for the strength of our magnetic field because no substituting it into our equation that we transposed, that means the strength of our magnetic field B is going to be equal to our mass, which is, which just to make sure we, we, we remember, let me just make sure that we have that note here recall that we said that the strength of the magnetic field equals MV divided by QR. And now we're just substituting all of that information that we know how. So we know that the mass of the proton is still 1.67 times 10 to the negative 27 kg. OK. We also now know that the velocity is 1.24 times 10 to the 6 m per second. OK. And now we're going to divide that by the charge of the proton, which is 1.6 times 10 to the negative 19 columns multiplied by the radius of our circular path, which above we found to be 0.03 m divided by the sine of 15 degrees. And know that we have all this information, we can put it into our calculator. When we do that, we find that the strength of the magnetic field will be equal to 0.11 teslas. So the magnitude of the magnetic field is 0.11 teslas. If we look back in our answer choices, that would have been answer choice. B thanks a lot for watching everyone. I hope this video helped.

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the 2.0-cm-wide region of uniform magnetic field in FIGURE P29.60. What field strength will deflect the electron by 10°?

Key Concepts

Electric Potential and Kinetic Energy

Magnetic Force on a Charged Particle

Deflection Angle in a Magnetic Field

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