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Force of one wire due to the other

  1. Jun 20, 2013 #1
    1. The problem statement, all variables and given/known data

    3.jpg

    D=1.0cm=0.01m
    I=1.0A


    2. Relevant equations
    B=(μ0I)/(2∏r)

    3. The attempt at a solution

    B1=(4∏x10^-7 * 1.0A)/(2∏ * 0.01m)= 2.0x10^-5

    B2=(4∏x10^-7 * 1.0A)/(2∏ * 0.01m)= 2.0x10^-5

    Not sure what to do with these two numbers.
     
  2. jcsd
  3. Jun 20, 2013 #2
    How does the force on a wire relate to the current through the wire and the B field?
     
  4. Jun 20, 2013 #3
    I would use the Right Hand Rule to find the relation. Since the current is going upward (thumb), the magnetic field B would be going into the page (fingers) and the force would be pointing to the left (direction the palm faces).
     
  5. Jun 20, 2013 #4
    The right hand rule will give the direction of the force but not the magnitude. You have found B, the magnetic field produced by a wire carying 1 amp at a distance of 10 cm. Now How do you find the magnitude of the force on the wire that is in this field?
     
  6. Jun 20, 2013 #5
    Would I use F=|q0|vBsinθ ?
     
  7. Jun 20, 2013 #6
    Wait, nevermind. That would be for a moving particle.
     
  8. Jun 20, 2013 #7
    F = BIL, remember this?
     
  9. Jun 20, 2013 #8
    F=ILBsinθ would be the equation, I think.
     
  10. Jun 20, 2013 #9
    Oh, saw your post after I posted. Oops. What would be L? I thought that L was the length of the wire, but that was not a given variable.
     
  11. Jun 20, 2013 #10
    If you have studied Vectors i would suggest you use the more general equation force that is i*L(cross)B
     
  12. Jun 20, 2013 #11
    F=ILBsinθ is technically correct and is used if the wire is not perpendicular to the B field which it is in this case. s far as what is L, look it up in your text. It will do you some good :-) Remember the force requested is per unit length.
     
  13. Jun 20, 2013 #12
    Oh ok, so...

    F/L=(1.0A)(2x10^-5)

    and sin90=1

    So F/L=2*10^-5
     
  14. Jun 20, 2013 #13
    would the units be Gauss/meter?
     
  15. Jun 20, 2013 #14
    What is the unit of force?
     
  16. Jun 20, 2013 #15
    oops, N/m!!
     
  17. Jun 20, 2013 #16
    Remember that this is a two step process. Step 1 is to determine the B
    field in the vicinity of the wire, and step 2 is to determine the force on the wire due to the B field generated by the other wire..
     
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