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Homework Help: Torque from straight wire segment

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data
    A straight wire segment of length L makes an angle of 21.0 degrees with respect to the x axis. The wire carries a current of 4.10 A in the direction shown in the figure. There is a magnetic field in the vicinity of the wire which points in the negative z direction with a magnitude given by B=B0(x/L)^3, where B0=3.80 T and L=4.91 m.

    What is the force in the y direction from the magnetic field?

    If the wire can rotate freely about the z axis what is the magnitude of the torque?
    2. Relevant equations

    dF=Idl x B

    Torque= r x F

    3. The attempt at a solution

    I found the magnetic force using the first equation

    dF=IdL x B and since B is with respect to x i simplified the cross product to dx

    dF=Icos(theta)dx B

    And plugged in the equation for B and integrated from 0 to L*cos(theta)

    Now for the torque I thought back and if you have the wire laying across the x axis or at least the xz plane and applied a force in the positive y direction you would be rotating about the z axis so i did r x F with the F from the previous answer and r= L*cos(theta). Is that incorrect?
  2. jcsd
  3. Oct 17, 2013 #2

    rude man

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    You didn't show the drawing so I'll assume the wire runs from, and pivots at, the origin.

    You haven't shown enough math detail for me to trace your steps so I'll just give you my way of approaching the problem.

    If you write out the vectors in vector notation it makes things easier (at least for me):

    dF = I dl x B
    dl = dx i + dy j
    B = -B0 (x/L)^3 k
    dτ = dl x F,

    all vectors in bold.

    Can you write dy = (constant) dx to eliminate dy?

    Then you can formally use vector math from then on. All integrations will be with respect to x only.
  4. Oct 17, 2013 #3
    I tried it your way(dont know how to make the vectors bold):


    I used dl= dx i + tan(theta) dx j
    So when I took the cross product i got dF=-(I*B0*tan(theta)/(4*L^3))*x^3 dx i + (I*B0/(4*L^3))*x^3 dx j

    And Integrated that from 0 to L*cos(theta) so
    F= -(I*B0*tan(theta)*L*cos^4(theta)/4) i + (I*B0*L*cos^4(theta)/4) j

    this is the same answer that I got (though I only calculated the y component)
    To simplify writing it the Fi(magnitude)=5.576 N for the i component and Fj=14.53 N for the j component
    Now for the torque:
    dl= dx i + tan(theta) dx j

    F=Fi i + Fj j

    So crossing those two you end up with just the k vector

    (Fj dx + Fi tan(theta) dx) k
    Went ahead and used the fact that Fi is negative here

    So again integrate (Fj+Fi*tan(theta) dx from 0 to L*cos(theta)
    = Fj+Fi*tan(theta)*L*cos(theta)= L*(Fj*cos(theta)+Fi*sin(theta))

    Numerically I got 76.41 N*m, still not the right answer
    Last edited: Oct 17, 2013
  5. Oct 17, 2013 #4

    rude man

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    Select "Go Advanced" & use the "B" on the toolbar.

    I need to leave for about 1 hr but will try to follow your work which basically looks right.
  6. Oct 17, 2013 #5

    rude man

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    Your F is the same as mine.
    Let a = tan(dy/dx) = tan(theta) where theta = 21 deg.

    Define F = Fx i + Fy j. Note that Fx < 0, we both got that.

    I question how you obtained torque τ. Did you perform

    dτ = (dx i + a dx j) x (Fy j + Fx i) correctly? Did you wind up with

    dτ = (Fydx - aFxdy) k ?

    You had a sin(theta) term in there & I didn't.
  7. Oct 17, 2013 #6
    Thanks for your help rude man, I ended up crossing r x dF to get the answer, not dr x F
  8. Oct 17, 2013 #7

    rude man

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    You are right. I wish I could say that was a typo on my part but it wasn't.

    Differential amount of force along the wire is dF = I dl x B and
    dτ = l x dF.

    Good catch!
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