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Homework Help: Force and torque on a wire carrying a current

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A wire along the x-axis carries current I in the negative x-direction through the magnetic field
    [tex]\vec{}B[/tex] = B0*(x/L) [tex]\hat{}k[/tex] 0[tex]\leq[/tex]x[tex]\leq[/tex]L
    = 0 elsewhere

    Part a was to draw a graph of B versus x over the interval -L[tex]\leq[/tex]x[tex]\leq[/tex]L, which I did.

    b. Find an expression for the net force F_{\rm net} on the wire. Express your answer in terms of the variables I, L, and B0

    c. Find an expression for the net torque on the wire about the point x = 0.
    Express your answer in terms of the variables I, L, and B0

    2. Relevant equations

    Fnet=I*L*B*sin[tex]\alpha[/tex]
    [tex]\tau[/tex]net=I*L2*B*sin[tex]\alpha[/tex]

    3. The attempt at a solution

    I thought that because I is carried in the -[tex]\hat{}i[/tex] direction and B points in the [tex]\hat{}k[/tex] direction that [tex]\alpha[/tex]=90 degrees, meaning I would be multiplying by 1. However, for both parts, when I submit I*L*B0 and I*L2*B0, respectively, the website says "Your answer either contains an incorrect numerical multiplier or is missing one."

    For part b I've tried submitting -I*L*B0 because actually calculating I[tex]\hat{}L[/tex] X [tex]\hat{}B[/tex] yields a negative answer, but that didn't work either.

    I've also tried to use x in my answer, and the website says the correct answer doesn't depend on it.

    Any help would be much appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2010 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    This is only true when the magnetic field is constant over the length of the wire. More generally, you will have to divide the wire into pieces [itex]d\textbf{l}[/itex] so small that the magnetic field is effectively constant/uniform over the length of the piece, calculate the force on each piece:

    [tex]d\textbf{F}=Id\textbf{l}\times\textbf{B}[/tex]

    and then add up (integrate) all these little forces:

    [tex]\textbf{F}=\int Id\textbf{l}\times\textbf{B}[/tex]

    You must do something similar for the torque.
     
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