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Maximum torque on a wire

  1. Nov 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the maximum torque on a square shaped wire provided N=120; l=4cm;R=10##\Omega##; B=0.2 T ; ##\omega=200 revs/sec##



    2. Relevant equations
    ##\tau=IaBsin(\theta)##
    ##\epsilon=\frac{d}{dt}\Phi##

    3. The attempt at a solution
    ##i=\frac{\epsilon}{R}=-\frac{\frac{d}{dt} (BAcos\omega t)}{R}##
    ##i_{max}=\frac{NBA\omega}{R}##
    ##\tau=il^2 B sin(\omega t)##
    ##\tau_{max}=il^2B = 2.6x10^-5 Nm##

    This solution depends on the fact that my area vector is initially parallel to the magnetic field.
     
  2. jcsd
  3. Nov 21, 2014 #2

    andrevdh

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    Shouldn't it be since
    the torque on the coil is τ = BANi and i = BANω/R we have that
    ∴ τ = (BAN)2ω/R
    = 29,5x10-3 Nm
    ?
     
  4. Nov 21, 2014 #3
    I don't see why the term ##\omega/R## is in there (the equation for current). Could you please explain? thanks
     
  5. Nov 21, 2014 #4

    haruspex

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    This is not valid. It is not the case that ##\tau(t)=i_{max}l^2 B sin(\omega t)##. You cannot take the max value until you have the general equation for ##\tau(t)##.
     
  6. Nov 24, 2014 #5

    andrevdh

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    But you also have it ??? ..... i = emf/R = NBAω {sin(θ)} /R
    That is the emf is the time derivative of the magnetic flux is NBAω sin(θ)
    I think what haruspex is saying is that there is another time varying term
    for the torque. So you have a sin(θ)cos(θ) term in the general equation for the torque?
     
    Last edited: Nov 24, 2014
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