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Grade 12 Electromagnetic Problem

  1. Dec 12, 2004 #1
    A wire, whose linear mass density is 150g/m, carries a current of 40A (supplied by a flexible negligible weight). This wire lies parallel to, and on top of, another horizontal wire on a table. What current must flow through the bottom wire in order to repel and support the top wire at a height of 4.0 cm above it? The top wire is held in place by frictionless guide plates.


  2. jcsd
  3. Dec 14, 2004 #2
    The field due to a wire is

    [tex]B=\frac{\mu_{0}I}{2\pi r}[/tex]

    and the direction can be determined by use of the right hand rule. The force associated with these fields is given by the lorentz force law


    when the field and the motion are perpendicular. Lets look at this another way... lets say that the lorentz force law read

    [tex]F_{mag}=\frac{qLB}{\Delta t}[/tex]

    we can then move that [tex]\Delta t[/tex] over to [tex]q[/tex] and get somthing that looks like this

    [tex]F_{mag}=ILB=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L[/tex]

    Utilizing Newton's Second law, in order for one wire to repel and support the other the forces must be equal.


    so we set the forces equal. ([tex]\mu[/tex] and [tex]\mu_{0}[/tex] are not related in any kind of way,[tex]\mu[/tex] is the linear mass density)

    [tex]\mu gL=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}L[/tex]

    and we get a result that is only dependant on the second current

    [tex]I_{2}=\frac{\mu 2\pi rg}{\mu_{0}I_{1}}[/tex]

    Before you start putting numbers in make sure you understand what I have said here. And remember to give the current the proper polarity as indicated by the right hand rule.
    Last edited: Dec 14, 2004
  4. Dec 14, 2004 #3
    Thanks for the help
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