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Torque due to electromagnetic induction

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A circular coil of radius r carries a current I. A magnetic induction B acts at right angles to a diameter of the coil. Show that the current experiences a torque T about the diameter given by T=Iπ[itex]r^{2}[/itex]Bsinω, where ω is the angle between the normal to the plane of the coil and the induction B.

    2. Relevant equations

    F=∫IdlxB

    3. The attempt at a solution

    T=Fr/2=Ir/2∫dlxB
    Taking the modulus of both sides, T=Ir/2∫Bdlcos(ω)=IrBlcos(ω)/2=Iπ[itex]r^{2}[/itex]Bcosω
    I suspect the step where I take the modulus of the inside of the integral is wrong, but I'm not sure why.

    More generally, I'm unsure how to calculate a cross product such as dlxB around a circle. The modulus of this is Bdlsin(θ), but what is the angle θ?

    Thanks in advance for any help! :-)
     
  2. jcsd
  3. Mar 28, 2013 #2

    rude man

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    It's easier if you let ω = π/2 at first. So draw the circular loop in the x-y plane with center at the origin and radius = R. Draw a diameter along the x axis. What direction is the B field in? Call θ the angle between the x axis and a point on the loop in the 1st quadrant.

    Then use d F= I d l x B to get the force on the loop in the direction you're looking for. Remember, they're asking for torque about the diameter, so you're looking for the component of d F = dF in the appropriate direction.

    Now you have dF about the diameter as a function of θ. Now, given dF, what is torque d τ about the diameter? Watch out , this is tricky - the torque is about the diameter, not the origin, so don't just use dτ = R * dF.

    Then integrate dτ from θ = 0 to π/2 and multiply by 4 since there are 4 quadrants.

    Finally, include the effect of ω not being π/2 but any angle 0 < ω < π/2. This is a tilt angle about the diameter.
     
  4. Mar 28, 2013 #3
    Thanks so much for your help. I'll try the method you suggested now. :thumbup:
     
  5. Mar 29, 2013 #4
    Hi. I'm trying to use the method above but I'm having some problems.
    dT=rcosθdF, and T=4∫dT between 0 and π/2. But I don't know how to write dF as a function of dθ.
    How do I use the expression dF=I dl x B to find dF?

    Thanks so much!
     
  6. Mar 29, 2013 #5

    rude man

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    Don't start figuring torque. Start with force.

    Did you draw the diagram I suggested? Let's take two points on the loop, one at (R,0) the other at (0,R). What is dl x B as a function of R and dθ for those two locations? Remember, B points in the j direction if you let ω = π/2 for the time being.
     
  7. Mar 29, 2013 #6
    This is the point I don't understand. Since dl and B are perpendicular, dl x B = B dl, but that's not a function of theta...
     
  8. Mar 29, 2013 #7

    rude man

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    d l and B are not perpendicular at (R,0) which would correspond to θ = 0 if you drew my diagram.
     
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