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Deflection of suspended coil (fluxmeter)

  1. Jul 14, 2013 #1
    1. The problem statement, all variables and given/known data
    A fluxmeter is a device sometimes used for measuring magnetic fields. It is a suspended coil instrument in which the suspension has virtually no restoring twist. The instrument is used in series with a search coil which is removed from or rotated in the field to be measured. The fluxmeter coil and the search coil form a contiuous Circuit of total resistance ##R##. If the instrumental constant for the fluxmeter is ##k## (That is for small ##\theta## you can approximate the flux as ##\Phi = k \theta + \Phi_0##), determine the deflection ##\Delta \theta ## of the fluxmeter for a change of flux ##\Delta \Phi## through the search coil.

    2. Relevant equations
    [tex]\frac{d \Phi}{dt}=-V[/tex]
    [tex]I=V/R[/tex]
    [tex]T=I\frac{\partial \Phi}{\partial \theta}[/tex]


    3. The attempt at a solution
    I don't know what the relation is between the search coil and fluxmeter coil, but i am first examinining the search coil. I use the relation ##\Phi = k \theta + \Phi_0## to give
    [tex]\frac{d \Phi}{dt}=k \frac{d \theta}{dt}=-V[/tex]
    which leads to the current in the curcuit
    [tex]I=V/R=-\frac{k d \theta / dt}{R}[/tex]
    Now there is a torque on the coil given by
    [tex]T=I\frac{\partial \Phi}{\partial \theta}=Ik=-\frac{k^2 d \theta / dt}{R}[/tex]
    And from the laws of mechanics it is given by
    [tex]T= I_m \frac{d^2\theta}{dt^2}[/tex]
    Where ##I_m## is the moment of inertia for the Circuit. This leads to the differential equation
    [tex]\frac{d^2 \theta}{dt^2}+a\frac{d \theta}{dt}=0[/tex]
    where ##a=\frac{k^2}{RI_m}##. The solution is
    [tex]\frac{d \theta}{dt}=Ce^{-at}=\omega (t)[/tex]
    But since ## \omega (0) = 0## then ##C=0## so ##\theta = constant## and there is no deflection at all in the search coil. So (if this is right) why should there be a deflection in the fluxmeter?
     
  2. jcsd
  3. Jul 15, 2013 #2

    rude man

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    Homework Helper
    Gold Member

    Since no one else is responding:

    This description of a fluxmeter doesn't match anything I've looked at. There seem to be two coils, why I can't imagine.

    Anyway, if the expression is flux = k(theta) + flux0 for small angles theta, then maybe it's proper to assume that for larger angles theta the expression is flux = k sin(theta) + flux0.

    So then, solve for theta in the above expression.
     
  4. Jul 15, 2013 #3
    Yes i guess it is hard even for trained physicists. Or maybe I screwed it up so much noone wants to help.

    No I don't understand it either. Thats why I ask for help

    According to my calculations above this would lead to the very unlinear differential equation
    [tex]\frac{d^2 \theta}{dt^2}+\frac{k^2}{R I_m} \cos^2 \theta \frac{d \theta}{dt}=0[/tex]
    This is obviously too difficult to solve (for a textbook problem).

    But lets say there is a restoring torque per unit twist ##\alpha##. Then
    [tex]T=I\frac{\partial \phi}{\partial \theta}=\cos^2 \theta \frac{d \theta}{dt} k^2/R=\alpha \Delta \theta[/tex]
    and for ##\alpha=0##, as stated in the problem, I get an infinite ##\Delta \theta##, that is it never stops spinning.

    So I never seem to get a finite value (either zero or infinite) for ##\Delta \theta##. That's a big problem. But if noone wants or can help me there is nothing I can do about it.
     
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