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The net force on a moving positive charge

  1. Feb 15, 2007 #1
    The net force on a moving positive charge....

    1. The problem statement, all variables and given/known data
    A magnetic field has a magnitude of 1.2 x10^-3 T, and an electric field has a magnitude of 5.4 x10^3 N/C. Both fields point in the same direction. A positive 1.8 µC charge moves at a speed of 2.9x10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

    2. Relevant equations
    Force = qE
    Force = qvBsin(theta)

    Electric force + Magnetic force = Net force

    3. The attempt at a solution
    qE + qvBsin(theta) = net force

    (1.8e-6 C * 5.4e^3 N/C ) + ( 1.8e-6C * 2.9e^6m/s *1.2e^-3T*sin(90) =

    .0972N + .006264N = .1035N of net force
    Last edited: Feb 15, 2007
  2. jcsd
  3. Feb 15, 2007 #2


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    Indeed, the electric force has a magnitude of F1=qE and the lorenz force F2=qvBsinθ=qvB because θ=π/2.
    What you`re doing wrong is the "qE + qvBsin(theta) = net force" part.
    That`s only true if you right it with vectors.
    You want the magnitude of the net force. Draw a diagram with the forces acting on the charge, and you`ll easily get the answer
  4. Feb 15, 2007 #3
    im not sure i follow...i made a diagram of the forces acting on the charge (don't laugh):
    green = magnetic field
    blue = electric field
    grey line = direction of velocity of + charge
    black dot = + charge

    the way i'm seeing it is the force from the magnetic field is pointing out of the page and the force from the electric field is pointed upward. the net force would be a combination of the two, so why can't i add them like vectors?
  5. Feb 15, 2007 #4
    Thats not what you did, you just added them together
  6. Feb 15, 2007 #5
    ok to correct myself, i made a diagram of what i believe the magnetic field and the electric field to look like. i realize the force of the magnetic field will be pointing out of the screen towards me; the force of the electric field will be directed parallel along the field. does that clear things up?
  7. Feb 16, 2007 #6


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    Your diagram is correct, no doubt about that. The forces` direction is as you say as well.
    But this
    "1.8e-6 C * 5.4e^3 N/C ) + ( 1.8e-6C * 2.9e^6m/s *1.2e^-3T*sin(90) =

    .0972N + .006264N = .1035N of net force"
    is absolutely wrong. You cant add together the two forces because they havent the same direction, they are perpendicular to each other .....
    So what are we doing in this occasions?
  8. Feb 16, 2007 #7
    ok, maybe i'm wrong again here but i think i need to find the force that acts in the middle of the 2 forces, since they are perpendicular. im sure i would use some trigonometric function to accomplish this..
  9. Feb 16, 2007 #8
    "add them like vectors" with the pythagorean theorem
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