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Magnetic Force direction

  1. Mar 28, 2005 #1
    Hi, I have a question: If there is a negative charge moving w/a velocity that is 50 degrees counter-clockwise from the +x-axis and a B field in the +z direction, I get that the magnitude of the Magnetic Force=(abs. val of q)(abs. val of v)(abs. val of B)sin(theta), where theta=angle b/t v and B, which is 90 degrees b/c v is perpendicular to B. But, how do you get the direction of the Magnetic Force? I know that F should be pointing down in the fourth quadrant by the right hand rule. The answer manual says that F is 40 degrees clockwise from the +x-axis, but why? Is the reason b/c v and F must be perpendicular?

    Thanks,
    Elizabeth
     
  2. jcsd
  3. Mar 28, 2005 #2

    Andrew Mason

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    The Lorentz force is given by:

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

    The convention for the cross product is given by the right hand rule (clockwise from v to B, v x B goes into the page). But you have to watch the sign for q. You can't use the absolute value of q. If it is a negative charge, you have to include the sign, so the force is in the opposite direction of v X B.

    AM
     
  4. Mar 28, 2005 #3
    Thanks...ok, but so then why is the F 40 degrees CW from the +x-axis? If instead of pointing in the 4th quadrant, it points in the 2nd quadrant now b/c q is negative, then shouldn't the answer be 40 degrees CW from the -x-axis???

    Thanks.
     
  5. Mar 28, 2005 #4

    Andrew Mason

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    If it was travelling along the x axis, v x B would be in the direction of the y axis. Since it is moving at 50 degrees to the x axis toward the y axis, v x B would be 50 degrees to the y axis toward the -x axis. Multiplying by -q, you add 180 degrees to that and you get 50 degrees from the - y axis toward the x axis, or -40 degrees from the x axis (toward the y axis).

    AM
     
  6. Mar 29, 2005 #5
    Ok, so let me just get this straight. Let's say the charge is moving along the +x-axis, so directly to the right and B is out in the +z-direction, then vXB gives an F down, but b/c the q is negative, F really goes up along the +y-axis. Now, b/c v is really 50 degrees CCW from the +x-axis, you shift everything 50 degrees CCW, too, so then you have F 50 degrees CCW from the +y-axis. But then how do you get 40 degree CW from the +x-axis? Isn't the negative value of the q already accounted for when you shift everything 50 degrees CCW?

    Sorry for so many questions, but I appreciate your help!

    Thanks,
    Elizabeth
     
  7. Mar 29, 2005 #6

    Andrew Mason

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    The problem here may be the direction of the axes. Normally, one has the x axis going to the right, y axis up, and z axis going into the page. Or the x axis to the right, the z axis up, and the y axis coming toward you out of the page.

    vxB is not the force. vXB gives you the force/unit charge or the field E. If v is along the x axis, and z goes into the page, E is up (y axis). If you shift v CCW 50 degrees (which is what the problem gives you), you shift E 50 degrees CCW from the y axis. The force, however, is in the opposite direction due to negative charge: which is 50 degrees CCW from the -y axis or -40 deg. (ie. CW) from the x axis.

    AM
     
  8. Mar 29, 2005 #7
    so, would it not work if I assume that the +z-axis is OUT of the page?
     
  9. Mar 29, 2005 #8

    Andrew Mason

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    If the question assumed that the z axis went out of the page, you would get the right answer. But it appears to me that the question assumes that the z axis goes into the page. It is a poor question because it does not explain the directions clearly. One can avoid ambiguity by using North South, East West, Up and Down.

    AM
     
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