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A Proton Moving Freely in a Constant Magnetic Field

  1. Jul 15, 2008 #1
    A proton moving freely in a circular path perpendicular to a constant magnetic field takes 1.50 µs to complete one revolution. Determine the magnitude of the magnetic field.


    I know that a proton has a charge of 1.6x10^-19 and a mass of 1.67x10^-27. Those are the only two relevant facts I know about a proton. The equation I was attempting to use was F=qvBsin(theta). I know that theta is 90 because the proton is perpendicular to the field. So sin(theta) will be one. Q is the charge of the proton stated above. I do not know v or B, v being velocity and B the magnitude of the force of the magnetic field. I tried numerous ways to solve for v but I can not figure out how. We do have a time but we have no radius of the circle so I do not know how to reap a distance from the given information. I know that I have not explicitly attempted a solution, but I am really really genuinely stuck on this one. Someone please help.
     
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  3. Jul 15, 2008 #2

    alphysicist

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    You've got that F=qvB from the magnetic force. What does F also have to equal since that magnetic force is causing it to move in a circular path?

    Also, for later in the problem, if a object is moving in a circular path of radius R with speed V, how is R and V related to the time it takes for the object to go around the circle once?
     
  4. Jul 15, 2008 #3
    I had tried to set F=0 because the proton has a constant speed and no acceleration. But that did not help because you can't solve 0=qvB for B which is what I need. R is related to the time it takes for a revolution by 2piR/V. However, I have no radius. If I am wrong about F being equal to zero, and I feel like I am; could you give me another nudge as to what F may be equal to. Possibly the charge of the proton since it is acting equally against that charge to hold it in circular rotation? No, I don't think so, just a guess.
     
  5. Jul 15, 2008 #4
    Were you referring to the fact that F would also have to equal m*(v^2/R)
     
  6. Jul 15, 2008 #5

    alphysicist

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    That's right; so set that equal to qvB to get your condition for circular motion in a magnetic field. Most books would solve what you get for r; what do you get?

    Once you have that, find out how R, V, and T are related, and then you can get the time in your equation.
     
  7. Jul 15, 2008 #6
    When solving qvB=m(v^2/R) for R, I got R=mv^2/qvB and from what I said earlier, I know that 2piR/V=T. So I can solve that for R as well and get R=TV/2pi. Then I can set my two equations equal and I have mv^2/qvB=TV/2pi. So now I successfully have time in my equation. Thanks so much for the help so far. However, I still have both B and v to solve for in my equation. All other variables I know.
     
  8. Jul 15, 2008 #7
    Oh wait! You can solve that because the v's eventually cancel out. Thanks al, you are the man. You helped me maintain my 100 homework average for summer physics 2!!
     
  9. Jul 15, 2008 #8

    alphysicist

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    Sure, glad to help!
     
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