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Magnetic force cross product

  1. Oct 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Considera 1.0 C charge moving with a velocity of v = -2.0i + 2.0j - 1.0k in a magnetic field of B = -4.0i + 1.0j – 3.0k.
    What force is this charge experiencing?
    What is the angle between the velocity and magnetic field vectors?

    2. Relevant equations
    F = q(E + v x B) sin(theta)


    3. The attempt at a solution

    Don't know which equations to use///
     
  2. jcsd
  3. Oct 10, 2014 #2

    jtbell

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    Staff: Mentor

    What you have here is a mish-mash of two separate equations. The equation $$\vec F = q (\vec E + \vec v \times \vec B)$$ gives you the force vector if you know the electric field vector, the velocity vector and the magnetic field vector. The equation $$F = qvB \sin \theta$$ gives you the magnitude of the force vector when you know the speed (magnitude of the velocity vector), the magnitude of the magnetic field vector, and the angle between the velocity and the magnetic field vectors.

    In this problem you're given the velocity vector and the magnetic field vector. Which equation does this suggest you should use?
     
  4. Oct 10, 2014 #3
    I am assuming that it would be F = qvBsin(theta) for the simple fact that I don't know the electric field vector... would that be a correct assumption?
     
  5. Oct 10, 2014 #4

    jtbell

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    If the problem statement says nothing about the electric field, isn't it reasonable to assume that there is no electric field, and set it equal to zero? :)
     
  6. Oct 10, 2014 #5

    RUber

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    No.
     
  7. Oct 10, 2014 #6

    jtbell

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    Why not?
     
  8. Oct 10, 2014 #7

    RUber

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    Sorry, I missed the fact that the magnetic field was constant.
     
  9. Oct 10, 2014 #8

    jtbell

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    Note that the second equation gives you the magnitude of the magnetic force only, without regard to the presence of an electric field. The first equation gives you the sum of the electric and magnetic forces. If you want only the magnetic force, you set ##\vec E = 0##. If you want only the electric force, you set ##\vec B = 0##.

    It's possible to make the second equation more complicated, to allow for the possible presence of both fields. In that case it would still reduce to what you have here if you set E = 0.
     
  10. Oct 10, 2014 #9

    jtbell

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    Aha, you were thinking of Maxwell's equations: a changing ##\vec B## is associated with an ##\vec E##. Most textbooks and courses cover that long after the basic stuff about the Lorentz force ##\vec F = q(\vec E + \vec v \times \vec B)## and how to calculate the cross product, so I wasn't expecting you'd be coming at it from that direction.
     
    Last edited: Oct 10, 2014
  11. Oct 10, 2014 #10

    RUber

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    That's right.
    Using ##\vec v \cdot \vec B = |v||B|\cos\theta ## the second equation requires fewer vector operations. But that would not resolve the direction of the force.
     
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