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Simple magnetic fields problem

  1. Feb 24, 2013 #1
    One component of a magnetic field has a magnitude of 0.0266 T and points along the +x axis, while the other component has a magnitude of 0.0648 T and points along the -y axis. A particle carrying a charge of +2.96 × 10-5 C is moving along the +z axis at a speed of 4.71 × 103 m/s. (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the +x axis.


    I got the correct answer for part A, which is simply 9.77x10-3 N.
    Fx ~ 0.003708 N
    Fy ~ 0.009034 N
    F = sqrt(Fx^2 + Fy^2)
    F = 9.77x10-3 N

    However, I am not getting the correct answer for B.

    I thought it was simply:
    theta = InverseTangent(Fy/Fx) = 67.68 deg but this wasn't correct.

    Then I thought I had to add 90 to get theta = 67.68 + 90 = 157.68 deg since F is perpendicular to both field (B) and velocity (V), but this was not correct either. What am I doing incorrectly? Thank you!
     
    Last edited: Feb 24, 2013
  2. jcsd
  3. Feb 25, 2013 #2
    According to the vector product (hope you know this version) the force on the positive charge
    is perpendicular to both the velocity vector and the magnetic field vector:

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

    , that is you turn the velocity vector towards the magnetic field vector. The force vector points in the direction in which a righthand screw advances due to this turning. This means [itex]\vec{F}[/itex] will be perpendicular to the plane containing the two vectors [itex]\vec{v}[/itex] and [itex]\vec{B}[/itex]. Or you may know the right-hand rule for determining the direction of [itex]\vec{F}[/itex].
     
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