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Easy Magnet problem.

  1. Jan 12, 2005 #1
    I am having too hard of a time with such a simple magnet equation that I am starting to loose it.

    A particle with a mass of m carries a negative charge of - q. The particle is given an initial horizontal velocity that is due north and has a magnitude of v.

    What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction? Use g for the acceleration due to gravity.
     
  2. jcsd
  3. Jan 12, 2005 #2

    Curious3141

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    Let the charge be travelling a distance [itex]s[/itex] in time [itex]t[/itex]. Visualise this as a thin wire carrying current in a magnetic and gravitational field.

    The current carried by the charge is [tex]I = \frac{dq}{dt} = \frac{qv}{s}[/tex]

    A magnetic field directed perpendicular to the motion of charge will exert a force on the wire given by [itex]F = IsB[/itex]. This force has to balance the weight of the "wire" = [itex]mg[/itex].

    Hence [tex]mg = IsB = \frac{qv}{s}(s)(B) = qvB[/tex]

    Therefore [tex]B = \frac{mg}{qv}[/tex].

    That's the magnitude. For the direction, use the right hand rule, but remember that the current in the right hand rule is conventional current, so you need to reverse the direction for negatively charged carriers.
     
    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3
    Thank you, for putting it clearly.
     
  5. Jan 12, 2005 #4

    Curious3141

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