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Rotating square coil in constant magnetic field

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data

    There is one square coil of area A that rotates with speed [tex]\omega[/tex] around axis y. The magnetic field is uniform and points to the z direction. Show that the torque is [tex]\frac{A^2B^2N^2\omega sin\omega t}{R}[/tex].

    2. Relevant equations
    [tex]T = |\underline{m} \times \underline{B}|[/tex]
    [tex]\underline{m} = N I A[/tex]
    Thus [tex]T = NIABcos(\omega t)[/tex]

    Unsure about these:
    [tex]EMF = NIR[/tex]
    [tex]EMF = - N\frac{d \Phi}{d t}[/tex]

    3. The attempt at a solution

    Now I have read many similar questions on this forums, and I think I am close to fully understand the problem itself, but there is a piece missing, and I don't see from my initial equations where I am going wrong.

    I calculate I via the EMF, substitute it into T. But with this I end up with one N, whereas the supposed solution that I need to prove in the exercise is [tex]T = \frac{A^2B^2\omega sin^2\omega t}{R}[/tex].

    I am assuming one of the initial equations wrong, I suspect the EMF section, because I not understand it. Any help would be appreciated.
     
    Last edited: Apr 17, 2009
  2. jcsd
  3. Apr 17, 2009 #2
    Thinking of V that is probably wrong, the number of loops has nothing to do with the potential, so [tex]V = IR[/tex].

    So EMF is [tex]-N \cdot \frac{d \Phi}{d t}[/tex] because we take [tex]\Phi[/tex] at each loop, and just sum those up. I am wondering now why can we actually do that, since in a different alignment the flux in each loop must be different.
     
  4. Apr 18, 2009 #3

    Redbelly98

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    The flux is the same for each loop, so yes EMF = -N dΦ/dt
     
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