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Ampere's law and induced current

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data
    ?temp_hash=248bebc91e160ebfb2d6ecd9b30a5d9e.png

    2. Relevant equations
    ∫BdL = μI

    3. The attempt at a solution
    a) the magnetic field is a circle around the wire so the length of the path is the circumference of the circle which is 2πr

    ∫BdL = BL = μI
    B(2πr) = μI
    B = μI/2πr
    in this problem it is into the page (right hand rule)

    b) The force on the top part of the rectangle will be down and the bottom part it will be up but they will cancel because the directions are opposite and B will be the same at each point.
    The net force on the rectangle will be the sum of the force on the left part and the right part. The force on the left part will be to the right (F1) and the force on the right part will be to the left (F2)
    so F = F1 - F2
    F = ILB1- ILB2
    the length is the same so
    F = IL(B1-B2) = IL(μI/2πr1 - μI/2πr2) = I^2Lμ/2π(1/r1 - 1/r2)
    r1 = .02m r2 = .05 m I = 6 A L = .1 m
    F = (36)(.1)μ/2π(1/.02 - 1/.05) = (3.6)(30)μ/2π (to the right)

    c) I think the net torque is 0 but I am not sure. I think it is because they are on the same plane so sinθ = 0 ???
     

    Attached Files:

  2. jcsd
  3. Apr 24, 2015 #2

    rude man

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    Good on parts a and b. On part c they did not specify the axis about which torque is to be determined. I would guess they meant the wire.
    So torque is developed in one angular direction by the outside section of the loop and in the other by the inside section. What are those torques?
    (Remember, torque is a vector so your answer should include the direction of net torque, e.g. by use of the appropriate unit vector.)
     
  4. Apr 24, 2015 #3
    Torque is rFsinθ where r is the distance from the axis, F is the force and θ is the angle between them. The top and bottom sections of the loops torques will again cancel.
    for the left section the F and r are parallel so sinθ is 0 and the torque is 0. for the right section F is to the left and is again parallel with r so the torque would be 0?
     
  5. Apr 24, 2015 #4

    rude man

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    You are correct. I got confused on the direction of F. o:)
     
    Last edited: Apr 24, 2015
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