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What are the magnitude and the direction of force, Fbc, that B exerts on segment bc o

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    A coil of wire consisting of 40 rectangular loops, with width 16 cm and height 30 cm, is placed in a constant magnetic field given by
    B = 0.055Tx + 0.210T.z
    The coil is hinged to a fixed thin rod along the y-axis (along segment da in the figure) and is originally located in the xy-plane. A current of 0.150 A runs through the wire.


    http://www.webassign.net/bauerphys1/27-p-051.gif


    What are the magnitude and the direction of force, Fbc, that B exerts on segment bc of the coil?
    magnitude Fbc =
    direction °



    What are the magnitude and the direction of the torque, t, that B exerts on the coil?
    magnitude t =



    3. The attempt at a solution

    i am just lost

    i have the formulas but i'm not sure how to find the magnitube for B
    hints pleasE?
     
  2. jcsd
  3. Jul 24, 2011 #2

    gneill

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    Staff: Mentor

    Re: What are the magnitude and the direction of force, Fbc, that B exerts on segment

    What's the formula for the force of a magnetic field on a current carrying wire of a given length?
     
  4. Jul 24, 2011 #3
    Re: What are the magnitude and the direction of force, Fbc, that B exerts on segment

    F = ILB
    yeah but B is in x and z direction so do i just do

    0.15 A * 0.3 M * (0.055T x + 0.210T z)

    and then do i square x and z then rad it? i tried that but it's not right
     
  5. Jul 24, 2011 #4

    gneill

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    Staff: Mentor

    Re: What are the magnitude and the direction of force, Fbc, that B exerts on segment

    It's a vector equation:
    [tex] \vec{F} = I \; \vec{L} \times \vec{B} [/tex]
    In this case you have 40 wires of length 30cm all carrying the same current I, so multiply the result by 40.
    [tex] \vec{F} = 40I \; \vec{L} \times \vec{B} [/tex]
    You should be able to write vectors for both L and B and perform the cross product (do it manually, it's probably easier). The result will be your force vector.
     
  6. Jul 24, 2011 #5
    Re: What are the magnitude and the direction of force, Fbc, that B exerts on segment

    so

    40 * .15 * 0.3 X (0.055x + 0.210z)


    using the cross product i get (0.3 * .21, 0 , -.3*.055)
    if i add them i get 0.0465

    and 40 * .15 * .0465 = .279 N

    is this right?
     
  7. Jul 24, 2011 #6

    gneill

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    Staff: Mentor

    Re: What are the magnitude and the direction of force, Fbc, that B exerts on segment

    The cross product yields a vector, so you can't just add the components. You have to take the magnitude of the vector to find the magnitude of the force. First write out all three components of the force vector.
     
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