(II) An electron enters a uniform magnetic field B=0.28 T at a...

Question

(II) An electron enters a uniform magnetic field B=0.28 T at a 45° angle to B (→ above B). Determine the radius r and pitch p (distance between loops) of the electron’s helical path assuming its speed is 2.2 x 10⁶ m/s . See Fig. 27–48.

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Accepted Answer

Welcome back. Everyone. In this problem. A proton is launched from a research module near a space station into a uniform magnetic field generated by the station's power grid. Magnetic field strength is 0.3 teslas. And the proton moves at an angle of 50 degrees to the magnetic field with a speed of 3.4 multiplied by 10 to the sixth meters per second. Calculate the radius R of the protons helical path and the PHP the distance between loops as it spas through the magnetic field. Here we have a diagram of our proton. And for our answer choices, A says the radius of the protons helical path is 9.1 multiplied by 10 to the negative 2 m. While the pitch is 1.5 multiplied by 10 to the negative 1 m. B says they are 9.1 multiplied by 10 to the negative two and 3.3 multiplied by 10 to the negative 1 m respectively. C 9.1 multiplied by 10 to the negative two and 4.8 multiplied by 10 to the negative 1 m and D 9.9 multiplied by 10 to the negative two and 4.8 multiplied by 10 to the negative 1 m respectively. Now, before we calculate the radius and the pitch, let's make note of all the information we have so far. Ok. So for starters, we know we're talking about a proton and the mass of a proton is equal to 1.67 multiplied by 10 to the negative 27 kg. While its charge Q is equal to 1.6 multiplied by 10 to the negative 19 coulombs. We also note that a magnetic field strength B is equal to 0.3 teslas. Now when it comes to our speed that the proton is moving at notice, our problem tells us that the proton moves at an angle of 50 degrees to the magnetic field. OK. And when we think about the proton's velocity, it has two components, the perpendicular component will cause circular motion and the parallel component which result will result in linear motion along the magnetic field creating a helical path. OK. So that tells us then that it might be a little more efficient to figure out the components of our velocity here. So in this case, OK, the perpendicular component is going to be equal to V sine theta. So that's gonna be 3.4 multiplied by 10 to the 6 m per second multiplied by the sine of 50 degrees. OK? Which is gonna be approximately equal to 2.6 multiplied by 10 to the sixth meters per second. Likewise, the parallel component is going to be equal to the cost theta. So that's 3.4 multiplied by 10 to the 6 m per second multiplied by the cost of 50 degrees, which would be approximately equal to 2.19 multiplied by 10 to the sixth meters per second. So it's very important for us to keep these values in mind as we go along. Now, let's start with the first part of our problem. How can it help us to calculate the radius R of the protons heal path? Well, we know that the magnetic force provides the centripetal force for circular motion. So in this case, our force is going to be equal to the magnetic force which is Q multiplied by the perpendicular component of V multiplied by our magnetic field strength B and it's gonna be equal to our centripetal force which is gonna be MV squared divided by R. Now, remember here, we're trying to solve A R. So in this case, we could cancel V from the left hand side and V into V squared leaves V on the right hand side and no solving for R. OK. That means R is going to be equal to MV divided by UB. Now we can go ahead and substitute our values. Our mass is 1.67 multiplied by 10 to the negative 27 kg. The perpendicular component of our velocity is 2.6 multiplied by 10 to 6 m per second. Now we're dividing all of that by the charge. OK, which is 1.6 multiplied by 10 to the negative 19th columns, multiplied by a magnetic field strength of 0.3 teslas. When we go ahead and solve here, then we should get R to be approximately equal to 9.1 multiplied by 10 to the negative 2 m. Let's move on to the second part of our problem. We would like to find a pitch p, the distance between loops as it spirals through the magnetic field. Now, what do we know about our pitch? Well, the pitch is the distance between the loops and it's given by the velocity. In this case, the parallel velocity or sorry, the parallel component of our velocity that results in a linear motion along the magnetic field multiplied by the period of circular motion. T Now we know what our parallel velocity is. Our part, the parallel component of our velocity is. But do we know what the period is? Well, recall that a period. OK, is equal to two pi R divided by the perpendicular component of our velocity. But we already have an expression for R here. OK. Because if we substitute that expression, it's going to be equal to two pi OK. Divided by V or perpendicular component of V. OK. Multiplied by MV. OK. Per perpendicular component. I don't know why that word gives me such a hard time divided by QB. Now we can cancel our velocities. And thus the formula for the pitch will be equal to the parallel velocity multiplied by two pi M divided by QB. Let's go ahead and substitute our values. We know then that's going to be equal to 2.19 multiplied by 10 to the 6 m per second, multiplied by two pi multiplied by a mass of 1.67 multiplied by 10 to the negative 27 kg. I know all of that is being divided by QB 1.6 multiplied by 10 to the negative 19 Coulombs multiplied by 0.3 Teslas. Now when we go ahead and solve here, then we should get our pitch to be approximately equal. OK, approximately equal to 4.78 multiplied by 10 to the negative 1 m which to two significant figures would be 4.8 multiplied by 10 to the negative 1 m. Therefore, that tells us the radius R of the protons helical path is approximately 9.1 multiplied by 10 to the negative 2 m. And the PP is approximately 4.8 multiplied by 10 to the negative 1 m. If we look back at our answer choices that tells us, then that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

(II) An electron enters a uniform magnetic field B=0.28 T at a 45° angle to B (→ above B). Determine the radius r and pitch p (distance between loops) of the electron’s helical path assuming its speed is 2.2 x 10⁶ m/s . See Fig. 27–48.

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Key Concepts

Lorentz Force

Radius of Circular Motion

Helical Motion

Watch next

(II) An electron enters a uniform magnetic field B=0.28 T at a 45° angle to B (→ above B). Determine the radius r and pitch p (distance between loops) of the electron’s helical path assuming its speed is 2.2 x 10⁶ m/s . See Fig. 27–48.<IMAGE>

Key Concepts

Lorentz Force

Radius of Circular Motion

Helical Motion

Watch next

(II) An electron enters a uniform magnetic field B=0.28 T at a 45° angle to B (→ above B). Determine the radius r and pitch p (distance between loops) of the electron’s helical path assuming its speed is 2.2 x 10⁶ m/s . See Fig. 27–48.

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