# Deflection of suspended coil (fluxmeter)

1. Jul 14, 2013

### Order

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
$$\frac{d \Phi}{dt}=-V$$
$$I=V/R$$
$$T=I\frac{\partial \Phi}{\partial \theta}$$

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
$$\frac{d \Phi}{dt}=k \frac{d \theta}{dt}=-V$$
which leads to the current in the curcuit
$$I=V/R=-\frac{k d \theta / dt}{R}$$
Now there is a torque on the coil given by
$$T=I\frac{\partial \Phi}{\partial \theta}=Ik=-\frac{k^2 d \theta / dt}{R}$$
And from the laws of mechanics it is given by
$$T= I_m \frac{d^2\theta}{dt^2}$$
Where $I_m$ is the moment of inertia for the Circuit. This leads to the differential equation
$$\frac{d^2 \theta}{dt^2}+a\frac{d \theta}{dt}=0$$
where $a=\frac{k^2}{RI_m}$. The solution is
$$\frac{d \theta}{dt}=Ce^{-at}=\omega (t)$$
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. Jul 15, 2013

### rude man

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.

3. Jul 15, 2013

### Order

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
$$\frac{d^2 \theta}{dt^2}+\frac{k^2}{R I_m} \cos^2 \theta \frac{d \theta}{dt}=0$$
This is obviously too difficult to solve (for a textbook problem).

But lets say there is a restoring torque per unit twist $\alpha$. Then
$$T=I\frac{\partial \phi}{\partial \theta}=\cos^2 \theta \frac{d \theta}{dt} k^2/R=\alpha \Delta \theta$$
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.