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### 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?
• I believe that at first, the particle would simply bend and continue traveling into the magnetic field. Eventually, it would complete a circle, but I guess that we are meant to assume that either the field was large enough to contain the circular path of the proton, or that the proton traveled far enough into the medium to enter a circular path.
• 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!
• because centripetal force=ma and a=v^2/r so F centripetal =(mv^2)/r
F magnetic=F centripetal
therefore (mv^2)/r=QvB
simplifying (mv)/r=QB
• i dont understand how the direction of the proton will still be perpendicular to the field lines after deflection. Please help..
• The force/acceleration is always perpendicular to the field and velocity. If the velocity is initially perpendicular to the field then the charge will move in a circle which is always perpendicular to the field.
• 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
• Hi, I checked it out on my TI-84 as well. They do have these constants build in.
Go to "APPS", then to "Scitools", then "2:UNIT CONVERTER". If you press f1 or f2 now, you'll end up in the constants menu!!

Good luck
• 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?
• From the combination of centripetal and magnetic force formulas (m*v^2/r, q*v*B*sin(a)) we can figure out the radius which is r=m*v/q*B. If the velocity increases then the radius increases too.
• 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.
• A proton is heavier than an electron, it's massive so in that case, it should have more inertia and it must resist the change in its state more than an electron so it should be deflected less and should form a circle with smaller radius but it doesn't. And we say, magnetic force is independent of mass but here, mass has an effect on radius and won't it require more force to deflect the particle with a larger angle??
• It does require more force to deflect a proton than it would require to deflect an electron the same amount. However, the force on both is completely independent of their masses and only relies on their velocities. (he does mention this at )
• How come the magnitude of velocity stay constant?
• Because the magnetic force is unable to exert any force in the direction of motion, which means it can't do any work on the charge, which means the charge can't gain any energy, which means it can't go any faster. You can't speed something up by pushing always in a direction that is perpendicular to the direction of its motion.
• So would the proton move in a circle forever?