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Part of 110 pointsWhat speed would proton need to achieve in order to circle Earth 1030 kut above the Ia ghetic equator; wfere the Earth s Inag- Iletic field is dir...

Question

Part of 110 pointsWhat speed would proton need to achieve in order to circle Earth 1030 kut above the Ia ghetic equator; wfere the Earth s Inag- Iletic field is directed OnL line between Inag- Iletic nlorth and south and has a1 intensity of 3.83 X 10 T? The mass of proton is 1.673 x 10-27 kg: Answer in units Of In/8

part of 1 10 points What speed would proton need to achieve in order to circle Earth 1030 kut above the Ia ghetic equator; wfere the Earth s Inag- Iletic field is directed OnL line between Inag- Iletic nlorth and south and has a1 intensity of 3.83 X 10 T? The mass of proton is 1.673 x 10-27 kg: Answer in units Of In/8



Answers

MIXED REVIEW What speed would a proton need to achieve in order to circle Earth $1000.0 \mathrm{km}$ above the magnetic equator? Assume that Earth's magnetic field is everywhere perpendicular to the path of the proton and that Earth's magnetic field has an intensity of $4.00 \times 10^{-8} \mathrm{T}$. (Hint: The magnetic force exerted on the proton is equal to the centripetal force, and the speed needed by the proton is its tangential speed. Remember that the radius of the circular orbit should also include the radius of Earth. Ignore relativistic effects.)

This problem was the concept of the motion of a church in a magnetic things. So in a uniform magnetic field the radius covered by the circular part of the church is given by the situation. So from the situation we can write the speed is a cuBA into are into the magnitude of the Church times the magnetic field strength upon the mass of the church. Okay. And this value are can be replaced with that. It is of the earth plus the altitude edge. So the speed vehicles r plus edge into the charge on the proton times the magnetic field upon the mass of the proton. I substitute the value to calculate the speech, the vehicles. The ideas of the author's 6.38 into tender six m. That's the altitude at just 1000 km. Or we can write 1000 into. Then there's three m into the charge on the platoon that is 1.6 into And there's -19 columns into the strength of the magnetic field and that is full And two tenders -8 Tesla Upon the mass of the problem that is 1.67 and two tenders managed 27 kg. So we get the speed of the opportunities. 2.83 Into Tender, seven m or second. No, we need to find the direction. Okay. The magnetic field is oriented along the north direction. So this is the magnetic field and along this direction is not so long. This election is not. Okay. And the 4th on the Platoon. Uh the force on the tone will be along the world grid downward direction. So let's see the forces along the downward direction along this direction. That is the magnetic force on the purple. So by using the right hand, come through, uh the force uh will be download when the velocity of the proton is a long the direction. Okay? So which means the proton is moving to us, the west so they can buy it the velocity of the protagonists 2.83 into And the seven NATO's, or second along us.

Okay, So a proton moves east eastward in the plane of theirs magnetic equator, so that assistance from the ground remains constant. What is the speed of the proton of the Earth's magnetic field? Points north and has a magnitude of 5.0 times 10 to the negative five Tesla. So the proton is moving east. So this is our proton moving east and the magnetic field. It's pointing up, so the gravitational force is going to pull it down. So since the proton is moving perpendicular to the magnetic field, we know that the magnetic force is equal to Q B B, and this is going to be equal to the gravitational force that's keeping get suspended in the air. So it wants us to find the velocity. So we confined velocity by rearranging this equation. The equals M G over Q B. So the mass of a proton is one point 673 times 10 to the negative. 27 kilograms G is 9.81 meters per second. Squared Q is the charge of a proton, which is 1.60 times 10 to the negative. 19 Cool owns and be is given as 5.0 times 10 to the negative five. Tesla. Plugging these values into this equation well, give us a velocity equal to 2.1 times 10 to the negative third in units of meters per second.

Alright, problem. Eight years We're told that we have got a proton. That's this little point right here with peace that's traveling street down above the equator straight down towards the ground with a speed of 375 minutes per second. And we're told that the magnetic field and it's at altitude is a magnitude of 4.5 times 10 to the negative five. Tesla's so were asked to find what the acceleration of this particle is. Anderson of the acceleration. Of course, we need to know the force on the particle and so to find the force will write down our expression. Fourth Forces forces equal to make to charge times, velocity times, Meg Field Times Sign of faith. So which of these things do we have? All we know that it's a proton. So I've got charge. We know its speed and we know the magnitude of the force. What we don't know is what fetus? What is data? So we weight ourselves that what it was the earth look like. Here's the North Pole years, the South Pole and we have magnetic field lines that looked like this running like this. They go all the way around. So at the equator, along the equator, on the magnet Phelan's running yourself now are Proton is actually traveling in the vertical direction. It's actually not going north, south, east or west, but it's heading straight towards the ground. So we know that it's actually traveling perpendicular to the magnetic field. So we're looking at something like that. So since that particular, that means that isn't any degrees and sign if they of 90 degrees is what? So we can just go ahead and just do the calculation. So for us is equal to que times v times be so. This is a 1.6, so two times 10 to the negative 19 Coolum. Our philosophy is thes e 375 minutes per second. It's given the problem and our 4.5 times 10 to the data five Tesla's is are making the field looks just like that. All right, so we punched all this into our calculator, being sure that we include all the times 10 to the negative 19 and whatnot they get. So what I get for the answer to this is that the force on the proton is actually 2.43 times 10 to the negative 20 one needn't. Okay, so acceleration is equal to force divided by the mass. So the force two point for three times 10 to the negative 21 Newton's divided by the mass of a proton. So we looked that up. That's something that is just recorded. So that's the proton. We've got 1.67 times, 10 to the negative, 27 kilograms. We punch this into Recruiter, in which we gets bigot. So I have the answer to be 1.5 times 10 2 to 6 meters for a second squared, and that should be the end of a problem for us.

Hi in this given problem there is our space having X. Axis Xeroxes. And while access I shown in this figure my access is into the plane of paper like this. Then there is a magnetic field in this space which is a long their access positive. That access like this. In this magnetic field the proton is entering with our velocity V. Are making an angle of 30 degree with the X. Axis here. This velocity of the proton has given us 1.5 Into 1080 per six m/s. And the magnitude of this magnetic period along Z axis this is B Is equal to 0.50. That's not. And this angle Theta is 30°.. So the two components of this velocity, the component along X axis. Weeks. That will be because 30 degree and a core component alongside taxes we set that will be we sign 30 degree. Now in the first part of the problem we have to discuss about the path followed by this proton within this magnetic field. So as under the influence of this magnetic field, the component the vertical. The horizontal component of this charged particle will constitute a circular motion. But under the influence of this visit it will move vertically upward also. And in each circular motion a complete circular motion. The charged particle means the protein will keep on moving vertical upward vertically upward. Also, hence in this way this charge particle constitutes a helical path in the magnetic field so we can write proton will move in a circular path due to the component of velocity VX. Which is given as we cost 30°. do you do the component visit which is equal to be signed 30 degrees. It will move vertically upward also. So ultimately it will cover a helical part which is the answer for them. First part of this problem. Now in the second part of the problem, we have to find the radius of this helical path which will be given by em into the X velocity responsible for this circular motion divided by B into E. The charge for a proton. So here this is M. We cost 30 degree divided by being too E. Finally plugging in all known values for the mass of the proton. This is 1.67 into 10 for -27 kg for the velocity, this is 1.5 Into 10. We should part six m/s into co sign of 30 degree, divided by be the magnetic field, which is having A magnitude of 0.50. Tesla multiplied by the charge over a proton, which is 1.6 into 10 for -19. So finally, this radius comes out to be 2.7 into 10 days to par -2 m. Or we can say this is 2.7 centimetre, answer for that. The second part of this problem. Then In the 3rd part of the problem, don't we have to find revolutions per second made by this proton. In the mathematical means we have to find its frequency and the expression for the frequency is B. E. The product of magnetically with the charge Divided by to buy em. So again, plugging in these non values, this is 0.50 Tesla for magnetic field 1.16- 10 for -19. Cool. Um, charge order brought on two into 3.14. Multiplied by mass of the proton. 1.67: 10 for -27 kg. So finally this frequency here comes out to be 7.6 into 10 46 hers, which becomes answer for them. 3rd part of the problem, Then In the 4th part of the problem, we have to find speed of the proton along the field line means we zet and that is we sign 30° means this is 1.5 Into 10. Parts six into force in 30° disease. 0.5 m/s. or we can say visit is equal to 7.5 into And which are five m For a second answer for the 4th part of this problem then, mhm. In the 5th part of the problem, they have to find the gap between the two boys of this helical path means here. If you look into this diagram, The gap between two consecutive helical part, circular circles of the helical part will be D. And that D. Will be equal to distance, is equal to speed into time speed, which is vertically upward means visit into time means time period and we know time period is equal to one upon F means frequency. So finally this D. Here will be given by Or visit. That was 7.5 Into 10. They should part five meet up for second and for frequency this was 7.67 to 10 days, part six revolutions for a second. So finally this gap here will be given by 9.75 into 10 par -2 m. Or we can see This is 9.75 centimetre, which is an answer for that 5th and the last part of this problem. Thank you.


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