Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.

    Thanks,

    ...Dan
     
  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

    [tex]F_{mag}=qvB[/tex]

    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.

    [tex]0=F_{mag}-F_{grav}[/tex]

    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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook