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Crossed Electric and Magnetic fields problem

  1. May 1, 2005 #1
    "A magnetic field has a magnitude of 1.2 x 10-3 T, and an electric field has a magnitude of 5.1 x 10^3 N/C. Both fields point in the same direction. A positive 1.8 µC charge moves at a speed of 3.2 x 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge."

    Using the crossed magnetic and electric field equation, wouldn't the net force be F=qE+qvB? I resolved it to 0.01609N, which isn't the correct answer. Any ideas?

    ~Futron
     
  2. jcsd
  3. May 1, 2005 #2

    OlderDan

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    Did you consider the direction of the forces? I did not check your answer, but that is a likely source of error.
     
  4. May 1, 2005 #3
    Wouldn't Fnet be the sum if they are both pointed the same direction?

    ~Futron
     
  5. May 1, 2005 #4

    OlderDan

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    The fields are in the same direction. Electric force is parallel to the electric field. What about the magnetic force?
     
  6. May 1, 2005 #5
    It is perpendicular relative to the magnetic field. How would I then calculate that into the Fnet equation? Thanks.

    ~Futron
     
  7. May 1, 2005 #6

    OlderDan

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    Not only is it perpendicular to the magnetic field, it is perpendicular to the direction of the velocity of the charge. Of imporatance here is that since the electric force is parallel to the electric field, and the magnetic force is perpendicuar to that, the two forces involved are perpendicular. They must be added as vectors.
     
  8. May 2, 2005 #7
    Got it. Thanks for your help!

    ~Futron
     
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