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Torque on a Coil of wire

  1. Jun 30, 2009 #1
    1. The problem statement, all variables and given/known data
    a circular coil of wire has a diameter of 2 m and contains 10 loops. The current in each loop is 3 amps, and the coil is placed in a 2 tesla magnetic field. Determine the maximum torque exerted on the coil by the magnetic field.

    2. Relevant equations
    F = IL x B

    torque = y x B

    3. The attempt at a solution

    I am trying to work this problem out from F = IL x B. I get stuck after I get to integrating the total torque on the circular coil. can you help???
  2. jcsd
  3. Jun 30, 2009 #2


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    There's a general equation relating torque to the number of loops, current, coil area, and magnetic field strength. It's T=NIABsin theta; with this formula, your question should be trivial.
  4. Jun 30, 2009 #3
    i'm not in the habit of memorizing formulas and plugging in values. I'd rather derive starting from the basics.... please don't throw out no brainer formulas... it's a waste of time.
  5. Jul 2, 2009 #4
  6. Jul 2, 2009 #5
    Don't get mired thinking in cartesians! You just need to integrate one equation going from 0 to 2pi. Start by thinking about a small element of the loop....
  7. Jul 2, 2009 #6
    yes, the 1 equation would be

    [tex] \sum \vec F_{in} = \int d \vec F_{in} = \int -IB_0 R cos \theta d\theta \cdot \hat{k}[/tex]
    [tex] \sum \vec F_{out} = \int d \vec F_{out} = \int I B_0 R cos \theta d\theta \cdot \hat{k}[/tex]

    what do I do next????? and where do these forces apply so that I can properly apply the torque equation.

    [tex] \vec {\tau} = \vec R \times \vec F_{perpendicular}[/tex]
    Last edited: Jul 2, 2009
  8. Jul 2, 2009 #7


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    Since we're taking an integral, you should have dL=Rdθ rather than "L" inside the integrand. Since we're trying to calculate torque, calculating the total force wouldn't be particularly useful.

    Think about it this way:

    (1) The coil must rotate around an axis that passes through its center.
    (2) What is the force on an infinitesimally small piece of coil? Express this in terms of θ (angular displacement from the axis), I, B, and dL=Rdθ.
    (3) What's the moment arm for this small piece of coil? What's the torque?
    (4) Integrate over the entire coil and you're done.

    For an extra challenge, try deriving the formula I gave you, T=NIABsinθ, for an arbitrary shape. That's quite challenging!
  9. Jul 2, 2009 #8
    oooops! mistake! I've corrected it
  10. Jul 2, 2009 #9
    for each radius vector that extends out to the circle, there is a different magnitude of the Force component. How would you set it up???
  11. Jul 2, 2009 #10


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    Consider just one tiny piece of coil of length dL=Rdθ. Since this piece is infinitesimally small, it's also infinitely straight, so the force on it would be F=IBdLsinθ. So, what's the moment arm and what's the torque on this small piece of coil?
  12. Jul 3, 2009 #11

    torque would be r x IBdLsin@. But dL sin@ is not the magneticforce on the coil. IBr sin@ is the magnetic force
  13. Jul 5, 2009 #12
    [tex] \vec F_{in} = -I L B_x cos\phi \cdot k[/tex]
    [tex] \vec F_{out} = I L B_x cos\theta \cdot k[/tex]

    [tex] dL = R d\phi [/tex]
    [tex] dL = R d\theta [/tex]
    [tex] d\vec F_{in} = -I R B_x cos\phi d\phi \cdot k[/tex]
    [tex] d\vec F_{out} = I R B_x cos\theta d\theta \cdot k[/tex]

    [tex] \sum \vec F_{in} = -I R B_x \int_{-\pi / 2}^{\pi / 2} cos\phi d\phi \cdot k[/tex]
    [tex] \sum \vec F_{out} = I R B_x \int_{-\pi / 2}^{\pi / 2} cos\theta d\theta \cdot k[/tex]

    [tex] \sum \vec F_{in} = -2I R B_x \cdot k[/tex]
    [tex] \sum \vec F_{out} = 2I R B_x \cdot k[/tex]

    [tex] \vec \tau_{1} = \vec R_{in} \times \sum \vec F_{in}[/tex]

    [tex] \vec \tau_{2} = \vec R_{out} \times \sum \vec F_{out}[/tex]


    [tex] \vec R_{in} = <Rcos\phi , -Rsin\phi , 0> from -\pi / 2 \rightarrow 0 [/tex]
    [tex] \sum \vec F_{in} = <0 , 0 , -2 I R B_x> [/tex]

    [tex] \vec R_{in} = <Rcos\phi , Rsin\phi , 0> from 0 \rightarrow \pi / 2 [/tex]
    [tex] \sum \vec F_{in} = <0 , 0 , -2 I R B_x> [/tex]

  14. Jul 6, 2009 #13
  15. Jul 6, 2009 #14


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    I might as well give you the answer, since this question has been up for this long. I said before that the force on an infinitesimally small piece of coil would be dF=IBdLsinθ=IBRsinθdθ. The moment arm would be Rsinθ (if you can't see this, draw a diagram). Torque=force * moment arm=IBR2sin2(θ)dθ Integrate that from 0 to 2π and you get the answer: IBπR2. With N coils of wire, total torque would be T=NIBπR2
  16. Jul 6, 2009 #15
    there needs to be a pi R squared in the final equation. and what is the integral of sin squared of theta????
  17. Jul 6, 2009 #16


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    But there is a pi R squared; look closely: T=NIBπR2. That's NIB times pi*R2.

    As for the integral of sin squared theta, remember that cos 2θ=1-2sin2θ, so sin2θ=(1-cos 2θ)/2
  18. Jul 6, 2009 #17
    would you look through my steps and tell me what I'm doing wrong. I can get to the final equation if I don't take a step by step detailed approach....

    Must I wait to take the integral of Cos until after I cross it with the radius to find the torque???
  19. Jul 7, 2009 #18
    I think I got it...but I ended up getting 2 pi radius squared in the final equation instead of Pi radius squared.

  20. Jul 7, 2009 #19
    Also, I took the integral of [tex] cos^2 \theta [/tex] instead of [tex] sin^2 \theta [/tex]
  21. Jul 7, 2009 #20


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    How is the integral of (1-cos2θ)dθ/2 from 0 to 2pi 2piR^2? The integral of cos2θdθ from 0 to 2pi is 0 because we're integrating over two cycles. The integral of dθ/2 from 0 to 2pi is pi.
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