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Magnetic Forces Between Current Carrying Wires

  1. Aug 2, 2012 #1

    s-f

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    In each of the 10 cases below, 2 - 4 Very long straight wires are parallel to each other. Rank the strength of the total magnetic force on each of the labeled wires.

    So each case has 2-4 wires with varying currents and distances between; I have to calculate the magnetic force and then rank all of the wires by magnitude. For example, the 1st one looks like:

    Wire A: ----->-----3A
    Wire B: ----->-----2A

    Wire C: ----->-----6A

    Distance from Wire A to Wire B = d
    Distance from Wire B to Wire C = 2d (and from A to C is 3d obviously).

    Using Ampere's Law, I calculated Wire A's magnitude to be 12A/d, which is correct.

    I found Wire B = -12A/d which is wrong, the correct answer is 0. Apparently the force vectors cancel out but I thought they should be added. Can someone explain?
     
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  3. Aug 2, 2012 #2

    berkeman

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    Staff: Mentor

    Use the right-hand rule to get the direction of the magnetic field due to each wire. Wrap your right hand around Wire C, and your fingertips are pointing at you when you touch Wire B. Wrap your right hand around Wire A, and your fingertips are pointing away from you when you touch wire B.
     
  4. Aug 2, 2012 #3

    s-f

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    Ok, so then would I subtract the magnetic force between A/B from B/C to get the total magnetic force on B? As in:

    F/L = μ/2∏ (2A x 6A/2d) - (2A x 3A/d) = 0

    And then would C = -12A/d?

    Thanks
     
  5. Aug 2, 2012 #4

    berkeman

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    How did you get that for C? Remember, the force ratios with the current, and inversely with distance:

    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/wirfor.html

    Can you show your work that gave you that answer?
     
  6. Aug 2, 2012 #5

    berkeman

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    Also, I think they want you to rank the absolute value of the forces, so the negative sign for the sum may not matter.
     
  7. Aug 2, 2012 #6

    s-f

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    I thought the force of Wire C would be the same as A but with opposite sign/direction. If not, what is the force on Wire C? Can you please explain it using Ampere's Law?
     
  8. Aug 2, 2012 #7

    berkeman

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    As shown at the Hyperphysics page (and in your textbook), the force on second wire due to the current in a first wire is proportional to the current in the first wire, and inversely proportional to the separation distance.

    So add up the two forces from the currents in Wires A and B to get their effect on C. What is the current in Wire A? How far is it away from Wire C? What is the current in Wire B? How far is it away from Wire C? Add up those two quantities to get the relative force (the question appears to be asking for the relative magnitudes of the forces on each wire due to the others, so the other terms in the force equation cancel out, except for the currents and separation distances....)
     
  9. Aug 2, 2012 #8

    s-f

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    Using that I still ended up getting 12A/d. Not sure what I'm doing wrong here.
     
  10. Aug 2, 2012 #9

    berkeman

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    Show your work.
     
  11. Aug 2, 2012 #10

    s-f

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    (3A x 6A/3d) + (2A x 6A/2d) = 12A/d

    (that's for wire C)
     
  12. Aug 2, 2012 #11

    berkeman

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    Ah, my apologies, that looks correct. What do you get for Wire A?
     
  13. Aug 2, 2012 #12

    s-f

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    No I checked the answer key and the right answer is -12A/d.

    The answer for A was 12A/d which I did correctly. I understand Wire A, because you simply add the forces together. I don't understand why you subtract for B (answer is 0) and C (answer is -12A/d). I thought that if the currents in the wires were in the same direction, as all 3 are, you add the forces.
     
  14. Aug 3, 2012 #13

    berkeman

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    It may just be negative because the force is attractive for two wires with current flowing in the same direction. You can see that it is an attractive force by using the right-hand rule to get the B-field direction at the 2nd wire, and then using F = qv X B to see the direction of the force.
     
  15. Aug 3, 2012 #14

    s-f

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    I don't get how to apply the right hand rule to get the force one wire has on another, and then how to use that in Ampere's Law to calculate the magnetic force between the wires.

    For example, in this problem:

    Wire 1 ------>-------3A
    Wire 2 ------<--------2A
    Wire 3 ------>--------6A

    Now the forces are 0 on all three wires, but why?
     
  16. Aug 3, 2012 #15

    berkeman

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    Are the spacings equal? A quick look has me thinking non-zero forces on all 3 wires...
     
  17. Aug 3, 2012 #16

    s-f

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    No sorry the spacing isn't equal:

    The distance between Wire 1 and 2 is d. The distance from Wire 1 and 3 is 3d. And the distance from Wire 2 and 3 is 2d.

    So for Wire 3, I did:

    (6A x 2A)/2d + (6A x 3A)/3d = 12A/d

    But the answer is 0, so the force between Wires 2/3 must be subtracted from the force between Wires 1/3 like this:

    (6A x 2A)/2d - (6A x 3A)/3d = 0

    I don't know why the force is subtracted here (I think I must be using the right hand rule wrong if I'm adding rather than subtracting forces).
     
  18. Aug 3, 2012 #17

    berkeman

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    Yeah, they need to subtract.

    For the first term: (6A x 2A)/2d wrap your right hand around Wire 2 with your thumb pointing left in the direction of the current. Your fingers touch Wire 3 with your fingertips pointing toward you.

    For the second term: (6A x 3A)/3d wrap your right hand around Wire 1 with your thumb pointing right in the direction of the current. Your fingers touch Wire 3 with your fingertips pointing away from you.

    So for however you set up your x-y-z coordinates, the vectors you get for those two terms will point opposite directions, and will subtract.

    Makes sense now?
     
  19. Aug 3, 2012 #18

    s-f

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    Yes that explains it perfectly, thanks!
     
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