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## Physics library

### Course: Physics library>Unit 13

Lesson 1: Magnets and Magnetic Force

# Magnetic force on a proton example (part 2)

Sal determines the radius of the circle traveled by the proton in the previous example by using the formula for centripetal acceleration. Created by Sal Khan.

## Want to join the conversation?

• the proton should have come out after executing half the circle.because by then it would have got out of the magnetic field.rite? • Sal did an awesome job on this video! However I do understand the centripetal acceleration formula but I do not understand why at he put an M in front of the MA formula? I do know that he may have broken up the force in MA but where did the A go? Thank you! Also please explain why he made it equal the force after denoting the MA, thank you! • i dont understand how the direction of the proton will still be perpendicular to the field lines after deflection. Please help.. • On 4.43,the calculator use to find the mass of the proton,is that only in TI-85,because I have TI-84 & I can't figure how to find that on my calculator
(1 vote) • Is this the same method they've been using to smash particles in search of the Higgs Boson? I know it's a little sidebar, but, I think it will help me grasp it a little more if this is the same process. Thanks. • My Physics reference book states that the direction a proton moves in is anticlockwise, while the electron moves clockwise, which is opposite from what Sal said. Can anyone explain to me why, or my book has a mistake?
(1 vote) • It depends which way the magnetic field is going.
If you have a magnetic field into the page, a proton coming from the left will circulate counterclockwise. You can determine that with the right hand rule
An electron would go the opposite direction.
If you reverse the magnetic field, both particles will reverse, too.
Sal's field is coming out of the page.
• A question: what happens to the radius if the velocity increases? Does it stay the same? If velocity increases the force increases also. In the first yellow line both increases would cancel out and the radius in unaffected? • At , why is the centripetal force equal to only the MAGNITUDE of the magnetic force and not the vector force? I thought that the magnetic force WAS the centripetal force in this case, just like how gravity or tension or friction are sometimes the centripetal force. • Good question! you are correct in noting that magnetic force should be a vector. In general, the force on a charge in a magnetic field is F = q v X B (Asterisks denote vectors here, and the X denotes a cross product.)
However, Sal decided to ignore the cross product, since v and B are perpendicular (B comes out of the screen, so anything in the plane of the screen is perpendicular to it), and we can thus simplify the magnitude of magnetic force to be q v B. Moreover, he uses only the magnitude of this force because we hardly ever use the vector form of centripetal force. If you look back at the circular motion videos, you'll notice that unit vectors hardly ever come up.
(1 vote)
• At , Sal says that the radius is 1.25 m. But since the proton is moving at 1/5th the speed of light, shouldn't we take relativity into account? The mass of the proton actually increases as per the formula
M=Mp*(1-(v^2/c^2))^-0.5
And the corrected radius is actually 1.28m. Isn't it??
(1 vote) • I didn't check your math but it looks right. He probably should have used a lower velocity. As you probably know, relativistic answers are always more correct than Newtonian ones, but the difference is usually too small to worry about and it makes it a lot harder to teach and learn basic physics. All those problems where we ignore friction are also "wrong", right? 