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What is the current in the loop

  1. Dec 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A piece of wire slides, without friction, along two similiar wires. The resistance per unit length of all wires is 0.350 ohms/m, and a constant magnetic field, B= 1.90T, points up, perpendicular to the loop. The distance between the parallel, fixed wires is 10.0cm. At the instant the sliding wires moves with velocity , v= 1.3 m/s, the loop length is L= 30.0cm. what is the current in the loop

    2. Relevant equations
    E = -d0 / dt
    I=E/R
    3. The attempt at a solution
    0.350 ohms/m
    B= 1.90 T
    d = 0.1 m
    V = 1.3m/s
    L= 0.3 m
    E = (1.90T)(0.13m/s) = 0.247V

    Resistance in loop = (o.3m)90.350ohms/m) = 0.105 ohms

    E= 2.47x10^-1 V
    R= 1.05x10^-1 ohms

    I = 2.47x10^-1 V / 1.05x10^-1 ohms = 2.35A

    My answer is 2.35 A but was told that there is some change in quantity that I missed? I feel like that answer is correct, can anyone show me where I might have skipped a step or got my units wrong? thanks.
     
  2. jcsd
  3. Dec 1, 2015 #2

    gneill

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    Staff: Mentor

    It looks like you haven't accounted for all four sides comprising the loop when you calculated the resistance.
     
  4. Dec 1, 2015 #3
    So I would have R = (0.3m)(0.350 ohm/m)(4) = 0.42 ohm?
     
  5. Dec 1, 2015 #4

    gneill

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    Staff: Mentor

    No, check the problem description to find the dimensions of the loop. It's a rectangle, not a square.
     
  6. Dec 1, 2015 #5
    width = 10.0cm (0.1m) length = 30cm (0.3m)

    Rectangle area = 0.03m^2?

    ...i shouldn't be this confused haha
     
  7. Dec 1, 2015 #6

    gneill

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    Staff: Mentor

    You want the perimeter.
     
  8. Dec 1, 2015 #7
    width = 2(0.1m) length = 2(0.3m)
    = 0.8 m haha! ?
     
  9. Dec 1, 2015 #8

    gneill

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    Staff: Mentor

    Yes, 0.8 m total length for the perimeter. So what's the resistance of the loop?
     
  10. Dec 1, 2015 #9
    R = (0.8m)(0.350ohm/m) = 0.28 ohm....?
     
  11. Dec 1, 2015 #10

    gneill

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    Staff: Mentor

    Yes. Use that in your current calculation.
     
  12. Dec 1, 2015 #11
    I= E/R

    I= 2.47x10^-1 V / 2.80x10^-1 ohm
    I= 0.88 A
     
  13. Dec 1, 2015 #12
    does that look better?
     
  14. Dec 1, 2015 #13

    gneill

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    Staff: Mentor

    Yes, that looks better.
     
  15. Dec 1, 2015 #14
    Thank You!
     
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