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Currents and Magnetic Field

  1. Apr 3, 2005 #1
    Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 42.0 cm , as shown in the figure. The currents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I.
    ******^
    ****** |
    ------10 A------I
    -------|--------|-------- 8 A --->
    ______|_______|
    ______|_______|
    -------|------- |--------- 20 A ---->

    10 A point upward; 8A, 20 A to right

    Find the magnitude and direction of the current I that will make the magnetic field at the center of the square equal to zero.

    I know that I must find the magnetic field at the center of the square due to the wires whose currents are known. Then, the current whose contribution to the magnetic field will exactly cancel the contribution of the other three wires must be found.

    Ampere's law must be used, I think, to find B at the center, but how do I apply it? There are more than 1 current, so how do I use it for the 3?

    Thank you.
     
    Last edited: Apr 3, 2005
  2. jcsd
  3. Apr 3, 2005 #2

    Doc Al

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    Staff: Mentor

    Right.
    Find the field from each current-carrying wire separately and add them up. Either look up the formula for the magnetic field due to a long current-carrying wire, or derive it from Ampere's law (it's easy).
     
  4. Apr 3, 2005 #3
    So I would use B = (mu_0*I)/(2*pi*r) for each different current? What would I plug in for r?
     
  5. Apr 3, 2005 #4

    Doc Al

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    Right.
    The distance from each wire to the point where you are calculating the field--which is at the center of the square.
     
  6. Apr 3, 2005 #5
    What about each current's direction? Must I take that into account when I am adding up the three?

    This is what I did to get B at center:

    Using mu_0 = 4*pi*10^-7, r = .21 m, I_1 = 10 A, I_2 = 8 A, I_3 = 20 A in formula,

    I found B_1 = 9.52*10^-6 T, B_2 = 7.62*10^-6 T, B_3 = 1.90*10^-5T and added these to get B at center, 3.62*10^-5 T, which is wrong.

    Point me on the correct path.
     
    Last edited: Apr 3, 2005
  7. Apr 3, 2005 #6

    Doc Al

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    Absolutely! Use the right hand rule to find the direction of the magnetic field surrounding a current-carrying wire. (For example: If the current points along the +y axis, the field to the right of the wire would point into the page or along the the -z axis.)

    Your numbers for each B field contribution look correct. I can't tell from your diagram which way the currents point, so I can't tell what sign each field contribution must have.
     
  8. Apr 3, 2005 #7
    So if it's pointing to -z, would that make that a -current?


    In my diagram 10 A is pointing vertically up (+y?) on the left vertical wire while the 8 A is pointing to the right (+x?) on the upper horizontal wire and 20 A is also pointing to the right on the lower horizontal wire. I is the right vertical wire's current with unknown direction.


    The wires are sort of arranged like a big number sign to form a square #.
     
    Last edited: Apr 3, 2005
  9. Apr 4, 2005 #8
    For the right-hand rule, is it where your thumb points in the current's direction and the fingers curling around the wire point in the magnetic field's direction? How do I express the direction for each current in the equation (whether its negative or positive) to find B at center to get I?
     
  10. Apr 4, 2005 #9

    Doc Al

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    Correct.

    For the three currents that you know, find the sign of the B field that each contributes. For your purpose, the sign is arbitrary: I would choose into the page to be negative, out of the page to be positive.

    Thus: The magnetic field at the center due to the 10A current is negative, due to the 8A current is negative, but the field due to the 20A points out of the page, so it's positive. Add them up with the proper sign. Then you can figure out what additional B field is needed to cancel it out. From that you can find the direction and magnitude of the current needed in that fourth wire.
     
  11. Apr 4, 2005 #10
    I found I (mag.) to be 2 A. Will the direction of I be downward?
     
  12. Apr 4, 2005 #11

    Doc Al

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    Which way must its field point to cancel out the others?
     
  13. Apr 4, 2005 #12
    The magnetic field for I must be negative (-1.90*10^-6) to cancel out the others, so since we are looking at the vertical wire, it must be upward instead?
     
  14. Apr 4, 2005 #13

    Doc Al

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    If the magnetic field must be negative (which means pointing into the page, I presume) to the left of the vertical wire, then the right hand rule tells us that the current must be downward.
     
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