Muons ( mass = 207 times the electron mass, same charge -e ) a...

Question

Muons ( mass = 207 times the electron mass, same charge -e ) are accelerated horizontally by 65 kV. They then pass through a uniform magnetic field B for a distance of 3.8 cm, which deflects them upward so they reach a detector 22 cm away, 11 cm above their original direction. Estimate the value of B.

Accepted Answer

Welcome back. Everyone in this problem. In an experimental setup, protons are accelerated using a 75 kilovolt potential difference. After acceleration, they pass through a uniform magnetic field. B this magnetic field deflects the proton sideways causing them to hit a detector positioned 25 centimeters horizontally and 13 centimeters vertically from their original path, determine the approximate value of the magnetic field. B to the nearest 10th where the mass of a proton is 1836 times the mass of the electron. The mass of the electron is 9.109 multiplied by 10 to the negative 31 kg. And the charge of a proton is 1.602 multiplied by 10 to the negative 19 coulombs. For our answer choices. A says it's 0.30 teslas, B, 0.40 teslas, C 0.50 teslas and D 0.60 teslas. If we're going to figure out the approximate value of our magnetic field be OK. Let's first make note of all the information that we have so far. We know that the protons were accelerated with a potential defense of 75 kilovolts or 75,000 volts. And we also know that our detector that they are deflected to sideways is at a horizontal position of 25 centimeters or 0.25 m away. And a vertical position of 13 centimeters or 0.13 m from their original path. In our, the rest of the information that's given up in the problem. We know that the mass of an electron is 9.109 multiplied by 10 to the negative 31st kilograms. We also know that the charge of a proton is 1.602 multiplied by 10 to the negative 19 coulombs. And we know that the mass of the proton M is 1836 times the mass of an electron. So when we multiply the mass of an electron by 1836 then we get that mass to be equal to 1.672 multiplied by 10 to the negative 27 kg. And now we want to use all of this information to help us figure out the approximate value of the magnetic field. B. Now how can we relate what we know about the magnetic field to the information that we have? Well, not here, we're talking about our protons in an accelerator. OK. And they are going to be moving in a circular path. We're told that the magnetic field deflects the proton sideways So, in other words, the force that's due to the magnetic field would have to be strong enough to deflect them from their circular path. So that magnetic force provides the necessary centripetal force for our charged particle moving in that circular path. So if we can figure out the forces due to the magnetic field and the centripetal force, then we'll be able to solve for the strength of the magnetic field. Now, what about that magnetic force? What do we know? Well recall, OK, that by Lawrence's force, the Law Lawrence force on a charge particle moving in a magnetic field, let's call it FB. OK is given by Q, the charge of the particle multiplied by V, the velocity of the particle multiplied by B the magnetic field. OK. Here we're talking about vector. So it's actually QV, actually the cross product of QV and B. But now if we're thinking about magnitude that tells us the magnetic force equals QVB. Likewise for our centripetal force, we know that the centripetal force required to keep the particle moving on a circular path is given by MV squared divided by R where M is the mass of the particle, V is the speed of the particle and R is the radius of the circular path. Now, since we said that we can equate both of those forces, that means then that FB or the magnitude of FP equals the centripetal force, which tells us that QV B QVB is going to be equal to MV squared divided by R. Now, if we go ahead and solve for Q, we can cancel V from both sides leaving us with V alone on the right hand side. And if we divide both sides by Q, that means B equals MV divided by QR, so we have a formula to solve for our magnetic field B, the strength of the magnetic field. But the only thing we need to do now is that we need to figure out the values of V, the speed of the particle and R the radius of the circular path. Let's first start by solving for the radios. OK. Now the radius of our circular path is going to be a half of the distance D from the accelerator to the detector. We don't know what that distance is, but we know it's horizontal and the vertical components. So we can solve for D by using the Pythagorean theorem because we know that our length D is going to be the square root of the X component squared plus the Y component squared. So that's the square root of 0.25 m squared plus 0.13 m squared. And now when we go ahead and solve, we should get D to be equal to 0.282 m. Therefore, then that tells us that our, our radius is gonna be a half of that which is 0.141 m. So that's the radius of our circular path. Next, let's solve for the speed of the particle V. How are we going to find how fast our particle is moving? Well, we know that the kinetic energy gained by the protons when accelerated through a potential V is given by the charge Q multiplied by the potential difference. V because as our problem told us that our protons are accelerated using the potential difference. But what else do we know about kinetic energy? Well, we also know that for our particle, its kinetic energy is going to be equal to a half of its mass multiplied by its speed squared. So since we're talking about the same kinetic energy here, we can equate both equations which tells us that QV equals a half of MV squared. So V sorry V squared rather is going to be equal to two QV divided by M. And thus our speed V is going to be equal to the square root of two QV divided by M. Now we can go ahead and solve. OK. Because no, that means it's gonna be the square root of two multiplied by 1.602 multiplied by 10 to the negative 19 coons, the charge multiplied by the potential different 75,000 volts. And all of that is divided by the mass of the proton 1.672 multiplied by 10 to the negative 27 kg. Now, when we go ahead and solve for our speed here. We should get it to be equal to 3.79 multiplied by 10 to the sixth meters per second. Now, we have values for our speed, OK? And the radius of our circular path. So we can go ahead and solve it, put it into our equation for A B or a magnetic field. I know when we do that, that tells us then that B is going to be equal to the mass of the proton 1.672 multiplied by 10 to the negative 27 kg multiplied by the speed of the proton 3.79 multiplied by 10 to the 6 m per second. The speed of the particle divided by the charge, the charge of a proton 1.602 multiplied by 10 to the negative 19 coulombs multiplied by the radius of its circular path. 0.141 m. Now we can go ahead and solve for the strength of our magnetic field. Now, when we do that, OK, when we put this expression into our calculator, we should get it to be equal to 0.2809 teslas or to the nearest 10th, approximately 0.30 teslas. Thus, this is the approximate value of the magnetic field B to the nearest 10th. If we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Lorentz Force

Kinetic Energy and Acceleration

Magnetic Field and Particle Motion

Watch next