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Finding EMF induced

  1. Jul 30, 2011 #1
    How would one go about finding the EMF induced when a two-turn coil of wire is straightened into a one-turn coil?
     
  2. jcsd
  3. Jul 30, 2011 #2
    [itex] \varepsilon = −\frac{\partial\Phi_B}{\partial t} [/itex]

    I'm assuming a uniform magnetic field at an angle [itex] \theta [/itex] with the "center" normal axis of the coil. In this case the change in the magnetic flux upon changing the coil from being a doubly wounded coil to a single winded coil (assuming same circular shapes and same normal axis in both cases), would result in a change in area.

    The change in area would be given by: [itex] \Delta A = \pi{(2\cdot r)}^2 - 2\cdot \pi r^2 = \cdots [/itex].

    Now, since the magnetic field is assumed uniform and let's also assume time independent, the change in magnetic flux during the transformation of the coil, is given by.

    [itex] \varepsilon = −\frac{\partial\Phi_B}{\partial t} = −\frac{\partial A\cdot B\cos \theta}{\partial t} = −B\cos \theta\cdot\frac{\partial A}{\partial t} [/itex]

    Since it is rather difficult to calculate [itex] \frac{\partial A}{\partial t} [/itex], I would just satisfy myself with noting how long it takes to transform the coil [itex] \Delta t [/itex], and then just use the average of [itex] \frac{\partial A}{\partial t} [/itex]. Thus

    [itex] \langle{\varepsilon}\rangle = −B\cos \theta\cdot\frac{\Delta A}{\Delta t} [/itex]
     
    Last edited: Jul 30, 2011
  4. Jul 30, 2011 #3
    wow...I initially thought that Bcranger was simply asking how the final coil compared to the initial one as far as resulting EMF for same conditions...

    ...after reading Isak's post, I seem to understand that we are talking about what happens to the EMF in the coil DURING the process of unwinding it from 2 down to 1 turn coil. is that right?

    Are we assuming that nothing is moving in the first place? You know, emf=0 at the beginning and at the end? or what? This is important since if things are moving then the initial area needs to be multiplied by the number of turns for a proper initial state...

    assuming nothing moving...

    if the coils is initially circular and intended to end up as another circle with twice the perimeter; then, yes, as Isak indicates, the initial cross sectional area is pi.r2 and the final cross sectional are is pi.(2r)2=4pi.r2

    ...but, to be proper, you are not going straight from one to the other one...you first need to go through zero cross sectional area as you untwist one of the turns...unless you are thinking of simply stretching your coil and leaving a 'kink' (spelling?) in the wire.

    so, you are going from 1 to 0 to 4 pi.r2 cross section area in the process

    for the case where the magnetic field is also moving and hence inducing emf at all times, then, you also need to count the other turn and so

    you are going from 2 to 1 to 4 pi.r2 cross section area in the process

    my 2 cents
     
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