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Velocity Components and Magnetic Force

  1. Mar 3, 2007 #1
    1. The problem statement, all variables and given/known data

    A particle with charge 9.42×10−8 C is moving in a region where there is a uniform magnetic field with a magnitude of 0.440 T in the +x-direction. At a particular instant of time the velocity of the particle has components -v_x = 1.70×10^4 m/s, -v_y = 3.19×10^4 m/s, v_z = 5.83×10^4 m/s.

    a. What is the y-component of the force on the particle at this time?

    b. What is the z-component of the force on the particle at this time?

    2. Relevant equations

    F = q*v*B*sin theta

    3. The attempt at a solution

    In drawing the y and z components,

    The y-component is pointing to the south, so the force’s y-component is pointing upward (positive?)?

    The z-component is pointing upward out of the plane, so the force’s z-component is pointing north (positive?)?

    As for calculating the individual force components, what else must be done aside from applying the F = q*v*B definition? (Just using F_y = v_y*B*q or F_z = v_z*B*q is incorrect, right?)

    Thank you for any guidance.
  2. jcsd
  3. Mar 5, 2007 #2
    For a full mathematical treatment of this type of problem it is necessary to construct a vector for velocity and magnetic flux density and find their vector cross product. But I don't think going into that would help you.

    One thing to emphasise is that the x y and z directions are the same for forces, velocities and any other vector. Some of your statements suggest that you think that for example F_z and v_z are not in the same direction. They most definitely are. Maybe I have misunderstood you but it is an important point.

    One way to tackle this is to imagine there are three identical charges of 9.42x10^-8 C. One is moving in the x direction at 1.7x10^4, one in the y direction at 3.19x10^4 and one in the z at 5.83x10^4. For each particle work out the force on it. The x moving particle will have no force since v and B are parallel. The y moving particle will have a force in the z direction by Fleming's LH motor rule and the z moving particle will have a y direction force again by FLHMR. Then you simply need to understand that these 3 forces( including the zero force) all apply to one particle moving with the original vector components.

    (You very nearly had the right answer in the beginning)
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