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Homework Help: Magnetic field between 2 parallel wires

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Sorry. I know this question is on the forum somewhere, but I still don't get it... Thus.

    For parallel wires, at the mid-point of the wires, what is the strength of the magnetic field? Do I use the formula [itex]B=\frac{μi_{1}i_{2}L}{2∏r}[/itex], or is it the sum of
    [itex]B=\frac{μi}{2∏r}[/itex], for both wires? I'm really confused:(

    For example,
    Two parralel wires are 8 cm apart. The magnetic field halfway between them is 300 uT. What equal currents must be in the wires?

    I would use equation no. 1, but it is actually 2x equation (2). Thus I don't really get, how to calculate the magnetic field between 2 parallel current carrying wires? And when to use equation (1)?

    Thank you.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 9, 2012 #2


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    For one wire you would obviously use eqn 2. For two wires it will be just the sum of the fields due to each wire, so just use eqn 2 for each and add them up. (Note that the currents must be in opposite directions or the fields would cancel.)
    I don't recognise eqn 1. It doesn't make sense dimensionally. Where did you find it?
  4. Nov 9, 2012 #3
    Oh, okay, thanks!

    Uh the 2nd formula is in quite a lot of places actually, though I might have misunderstood it... Places like http://www.cartage.org.lb/en/themes/sciences/physics/electromagnetism/Magnetostatics/MagneticField/Forcesoncurrents/parallelwires/parallelwires.htm [Broken]

    and the attached is the Halliday Textbook... In such a case, what do they mean by the 1st equation? Is it the force acting on one wire or something? Thanks!

    Attached Files:

    Last edited by a moderator: May 6, 2017
  5. Nov 9, 2012 #4


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    Yes, it's the force the wires experience from each other, not the field.
    Field at 1st wire = B1 = μ i2/d
    Current in 1st wire = i1
    Force per unit length acting on first wire = B1 i1 = μ i1i2/d
    Force on 1st wire of length L = μ i1i2L/d
  6. Nov 9, 2012 #5
    Thank yu very much! Now I get it:D
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