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Magnetic Field formula Question

  1. May 7, 2006 #1
    I'm not sure why the answer to the following question would be zero, so far, I'm thinking because work is force times distance and in a circle, there is no distance after a complete orbit. But I have a strange doubt that it might be something else.

    A magnetic field of .1T forces a proton beam of 1.5 mA to move in a circle of radius .1 m. The plane of the circle is perpendicular to the magnetic field. Of the following which is the best estimate of the work done by the magnetic field on the protons during one complete orbit of the circle?

    What I have a real problem with is how to solve the second part:

    Of the following, which is the best estimate of the speed of a proton in the beam as it moves in the circle?

    Could someone at least give me a formula or something to start with?
  2. jcsd
  3. May 7, 2006 #2

    Andrew Mason

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    What is the cause of the magnetic field of the proton beam? What is the magnitude of the force if the protons move in a circle?

    What is the direction of the force relative to the direction of motion of hte protons if the protons are moving in a circle? Does such a force do any work? Use [itex]\vec{F} \cdot \vec{ds} = d\vec{W} [/itex]

  4. May 7, 2006 #3
    The question doesn't have the cause of the magnetic field of the proton or the magnitude of the force... Unless those questions were supposed to help me grasp a concept in which I still can't see...

    For the second part, I'm still clueless...
  5. May 8, 2006 #4

    Andrew Mason

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    This is a Lorentz force problem. The moving proton charge is seen as having a magnetic field which interacts with the static magnetic field according to the Lorentz force equation:

    [tex]\vec{F} = q\vec{v}\times \vec{B}[/tex]

    If it is moving in a circle, the centripetal force must equal the magnetic force on the proton. Work out the expression for v.

    Does v depend on the current? What is the direction of the force compared to the direction of the proton (hint: think 'cross product')

  6. May 8, 2006 #5
    Ok, for the first part:
    (-mv^2)/.1 = .1*.0015 * v
    v = -( .00015 * .1 )/m
    mass of proton is about 1.67*10^(-27)(I'll use 1.5*10^-27 because I can't use a calculator here)
    v = 10 * 10 ^ 22

    I think I did something wrong here...
  7. May 8, 2006 #6
    [tex] \frac{mv^2}{r} = Bqv [/tex]
    so ...
    [tex] v = \frac{Bqr}{m} [/tex]
    Now put the numbers in. If the answer is greater than c, then mass correction is probably your problem.
  8. Mar 10, 2009 #7
    The F is a centripetal force, and the d is tangent to the circle. Because the F perpendicular to the d, Then W = Fd cos Θ is equal to zero.
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