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Work on a proton in uniform magnetic field

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A magnetic field of 0.1 T forces a proton beam
    of 1.5 mA to move in a circle of radius 0.1 m.
    The plane of the circle is perpendicular to the
    magnetic field.
    Of the following, which is the best estimate
    of the work W done by the magnetic field on
    the protons during one complete orbit of the
    circle?
    1. W = 0 J
    2. W = 10^22 J
    3. W = 10^−22 J
    4. W = 10^20 J
    5. W = 10^−5 J


    2. Relevant equations
    F = qVB = ILB
    F = mv^2 / r
    C = 2 pi*r

    3. The attempt at a solution
    Since work is energy, I used the formula for kinetic energy...

    W = 1/2 mv^2

    Fcentripetal = Fmagnetic
    mv^2 /r = qvB
    v = rqB / m

    W = 1/2 m (rqB / m)^2
    W = 1/2 (1.6725 x 10^-27) * (0.1 * 1.602 x 10^-19 * 0.1 / 1.6725 x 10^-27) ^2
    W = 7.67235874 x 10^-16 J

    This isn't close to any of the answer choices. Can someone help me?
     
    Last edited: Feb 6, 2009
  2. jcsd
  3. Feb 6, 2009 #2

    LowlyPion

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    Homework Helper

    The Force supplied by the B field is

    F = qV X B (the cross product)

    What is the direction of this force relative to the direction of the motion of the particle?

    Isn't Work

    W = F ⋅ D (the dot product)
     
  4. Feb 6, 2009 #3
    I tried a multitude of ways, including that, and the order was between -13 and -17. The closest answer to that is choice 3, but that's wrong.

    F = qvB
    W = qvB * 2 pi r

    (in part 2, I solved for v and found that it's approximately 10^6)

    W = (1.602 x 10-19) (10^6) (0.1) * 2 pi * 0.1
    W = 1.00656629 x 10^-14 J

    Again, nothing close.
     
  5. Feb 6, 2009 #4
    OR, does anyone think choice 5 is a typo, and it meant to be 10^-15?

    My online system sometimes has problems/answers like that unfortunately...
     
  6. Feb 6, 2009 #5

    LowlyPion

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    What angle is the B field and the Force making with the direction of motion?
     
  7. Feb 6, 2009 #6
    Everything is at a right angle.
     
  8. Feb 6, 2009 #7

    LowlyPion

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    Homework Helper

    So how much work does a force at right angle do again?

    Work is F⋅D which is the dot product.

    So what is the dot product of the Force with the direction of motion?
     
  9. Feb 6, 2009 #8
    Ohhh, so it's zero. Gah, I was supposed to remember that detail from that chapter we learned months ago.

    Thanks again for all your help, LowlyPion.
     
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