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Electric Motor Question

  1. Mar 9, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A coil is formed by wrapping 80 cm of thin wire round a non-conducting square of size 2 × 2 cm which is placed in a horizontal magnetic field of strength 20mT.

    A)Sketch the orientation of the coil that will result in the maximum torque acting, and the value of this torque when the coil is carrying a current of 200mA.
    B) Explain how the above system can be modified to produce a simple electric motor where the coil rotates in the magnetic field.

    3. The attempt at a solution
    A)We have derived in class that the torque on a coil suspended in a B field is ##\tau = AIB\sin \theta##. So this will be maximal when ##\theta = \pi/2## which means ##\tau = (4 \times 10^{-4} m^2 \cdot (200 \times 10^{-3} A) \cdot (20 \times 10^{-3})T \approx 1.6 \times 10^{-6}Nm##

    Is this correct? I haven't used the length of the wire here but I guess I assumed it covered the whole sqaure. If it is 80cm, then each full turn will give 4cm of the wire used which means to wrap the whole wire around the sqaure I need 20 turns. However, I don't know the thickness of the wire so I am unable to tell how much of the 2x2 square those 20 turns take up. Is that 80cm a distractor?

    B) I get the general idea about having to switch the direction of current every 180 degrees so as to reverse the magnetic force acting on the coil. But I have a question. Say the B field is horizontal to the left. Put the square parallel to the field such that a current runs clockwise. Then the magnetic force tends to tilt the sqaure upwards until it is perpendicular to the B field. The torque is then zero and the square will fall back down. How then is it possible to get the square to exceed 90 + degrees? I thought it could simply be inertia which makes it just exceed the 90 mark, but then if this is the case then since the current in the square is now flipped, this tends to push the sqaure back towards the 90 mark. So how is it even possible to get the sqaure to 180 degrees? (I hope this makes sense, if not I will attach a sketch).

    Many thanks.
     
  2. jcsd
  3. Mar 9, 2013 #2

    rude man

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    No. You need to take into consideration the number of turns in the coil, which you have correctly computed to be 20.
    Better attach a sketch ...

    :smile:
     
  4. Mar 9, 2013 #3

    CAF123

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    Hi rude man,
    I thought as much, but I don't know the thickness of the wire. So assuming I start winding the wire round right at the top of the square then to get the whole wire around the square I require 20 turns. But I believe in order to find the area of the coil I need to find ##A = 2 \times x##, ##0 < x \leq 2##.

    Ok, I will do.

    Edit: See attached sketch
     

    Attached Files:

    Last edited: Mar 9, 2013
  5. Mar 9, 2013 #4

    rude man

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    Don't worry about the thickness of the wire. Pretend it's zero.
     
  6. Mar 9, 2013 #5

    CAF123

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    I presume we are also meant to assume that the winding of the coil does not overlap. If we start at the top and wind all the way round once, then that will be 4 cm of the wire wrapped around. How do we know how much area this would cover (clearly 2cm along, but how much down?)
     
  7. Mar 9, 2013 #6

    rude man

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    See my previous post.

    You have a coil of dimensions 2 cm x 2 cm. That is a perimeter of 8 cm. So I was wrong too - you will have 80/8 = 10 infinitely thin turns of wire. So N = 10 and A = 4 cm^2. Change that to m^2 if I were you.
     
  8. Mar 9, 2013 #7

    CAF123

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    If the coil is a planar square, then will it not be 20 turns? (I can see that if the wire was wrapped around a central core for example then it would be 10).
     
  9. Mar 9, 2013 #8

    rude man

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    The wire is wrapped about the periphery. That makes it 10 turns.

    Assuming you get past this nuance, have you decided on a max. torque orientation yet?
     
    Last edited: Mar 9, 2013
  10. Mar 9, 2013 #9

    CWatters

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  11. Mar 10, 2013 #10

    CAF123

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    How can you tell it is wrapped around the perimeter? I was assuming it was wrapped lengthways. The maximal torque is when the the coil and B field are parallel.

    So ##\tau = 1.6 \times 10^{-5} N m##. Do you have any ideas about my problem regarding the electric motor? I get the idea behind it, but I had a question about it. (see my post 2 for the sketch)
     
  12. Mar 10, 2013 #11

    rude man

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    You need to be specific about the orientation of the coil. It's best to talk of the normal to the coil's area. So where does the normal vector point with respect to the B field ?
    Take it one step at a time to determine the torque:

    1. what is the force on one side of the coil?
    i = 0.2A
    l = 2cm = 0.02m
    B = 0.02T
    10 turns
    F = i l x B = ?

    2. Torque = F x d, what is d here?

    3. Repeat for the other side.
     
  13. Mar 10, 2013 #12

    CAF123

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    If the coil is wrapped around the perimeter of the square then when the square is in the plane of the B field the torque is maximal (i.e the normal vector of the square points in the upwards direction, perpendicular to the B field.)


    Yes, so since B and L are perpendicular, F = BIL(N) = 0.0008N
    There will be a torque with respect to the centre of the square, so d is 1cm. This means ##\tau## = d(BILN) = 8 x 10-6N m
    On the other side, the same but the force will point in the opposite direction.
    So total torque: ##\tau## = F(1cm) + F(1cm) = 2(0.01)F = 1.6 x 10-5N m
     
  14. Mar 10, 2013 #13

    rude man

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    Excellent! good work.

    Now, as to the motor: what happens when the normal to the coil rotates until it's in line with the B field & overshoots a small amount due to inertia? What is the direction of the torque now? Same as before or opposite?
     
    Last edited: Mar 10, 2013
  15. Mar 10, 2013 #14

    CAF123

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    Thanks, any ideas about my problem regarding the second part of the problem? (I posted the sketch in post 2).
     
  16. Mar 10, 2013 #15

    rude man

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    See my edited last post.
     
  17. Mar 10, 2013 #16

    CAF123

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    It will be in the opposite direction.
     
  18. Mar 10, 2013 #17

    rude man

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    Right. So what would be a way to continue making it go in the same direction?
     
  19. Mar 10, 2013 #18

    CAF123

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    Reverse the direction of the B field or the current every 90 degrees.
     
  20. Mar 10, 2013 #19

    CWatters

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    http://en.wikipedia.org/wiki/Commutator_(electric [Broken])
     
    Last edited by a moderator: May 6, 2017
  21. Mar 10, 2013 #20

    rude man

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    Yes and no.

    Yes, 90 deg after the max. torque position you would want to do that. But what about the other positions of the normal? Should you switch again 90 deg after that?

    Consider the direction of torque at every orientation of the coil and you will change your mind about reversing the current or field every 90 deg.
     
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