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Magnetic field torque

  1. Jun 20, 2008 #1
    1. The problem statement, all variables and given/known data

    a coil 2 cm in diameter has 300 turns. what is the maximum torque on this coil when it carries a current of 10mA and is placed ina constant magnetic field of 0.05 Teslas?

    2. Relevant equations

    magnetic force on a current F = IL X B where X indicates cross product, I is current, L is length of coil, B is magnetic field

    torque tau = rFsin(theta) where r is radius, F is magnetic force on current

    3. The attempt at a solution

    basically subbed in IL X B for F in rFsin(theta)
    and since r is radius, it is two times the diameter, 2*0.02 meters

    so torque tau = (2*0.02)[(10*10^-3)(300)(0.05)sin(90)]sin(90) = 0.06 N/m

    and since we won't max torque, force must be applied perpendicular, thus theta = 90 degrees.

    i am unsure of the length L though, the problem states diameter is 2cm and contains 300 turns, how much is 300 turns in meters?

    is my approach correct, though?

  2. jcsd
  3. Jun 20, 2008 #2


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    There's really quite a lot of errors in there. To start with radius isn't 2*diameter. You're also using inappropriate formulas all over the place. What you want to find the force is the magnetic moment of the coil cross the applied field. What's a formula for the magnetic moment of a coil? You certainly can't convert 300 turns to meters, come on. You can use it to find the total current circulating. Start all over again with a clear head, ok?
  4. Jun 20, 2008 #3
    i agree, it did seem a little too straightforward with my initial approach

    magnetic dipole moment, mu = NIA where N is number of turns in coil, I is current, A is area( is it area of wire it self(a circle; A = pi(r^2) ?) --> I = 10*10^-3 ampere, N = 300, A = pi(r^2) = pi((2*0.02)^2) = 0.005 m^2

    and torque, tau = mu X B where B is magnetic field = 0.05Teslas

    once i determine area A, and then mu, will i then find torque?

    so tau = mu X B = mu(B)sin(theta) = NIABsin(theta) = ((300)(10*10^-3)(0.005)(0.05))sin(theta) where theta = 90 degrees
    ---> ((300)(10*10^-3)(0.005)(0.05))sin(90) = 7.5*10^-4 N/m

    correct now? not too sure about area, A though.
  5. Jun 20, 2008 #4


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    You are making progress. Sure, area=pi*r^2. But you still have that weird radius=2*diameter thing going on. Can you do something about that? It's not right, is it? Yes, I think the question means you to take the coil to be a circle. Otherwise, they wouldn't give a 'diameter' right?
  6. Jun 20, 2008 #5
    oh oh oh, i was getting confused with diameter and radius, diameter = 2 * radius, not the other other way round, so radius r = diameter/2 = 0.02/2 = 0.01 m

    in that case

    tau = mu X B = mu(B)sin(theta) = NIABsin(theta) = ((300)(10*10^-3)(3.14*10^-4)(0.05))sin(90) = 4.71*10^-5 newton meters
  7. Jun 20, 2008 #6


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    If you have all of the right numbers in the right place then it should be good. Your formula is correct. Do I have to check your math, or do you trust yourself? Say, 'yes, I trust myself'.
  8. Jun 20, 2008 #7
    yes, I trust myself, was mainly curious about the equation, but thanks for your help again, i need to learn to use the information given, properly
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