# 2 solenoids: emf and frequency

I'm studying a circuit in AC. I use a function generator and set a waveform. I have a solenoid and I put a small solenoid inner it. Could you tell me if there is a relation between frequency and emf inducted on the second solenoid?

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Aside from the mutual coupling between the solenoids, the primary (the one you are driving) will be highly inductive to the driving circuit. If nothing else, beyond some point, as you increase the frequency of the function generator at a fixed voltage, you'll drive less and less current through the primary due to the inductive impedance increasing with frequency. Now thinking of the coupling between the two coils, less current through the primary means less magnetic fields to couple into the secondary coil, meaning less induced voltage on the secondary.

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Assume the inner solenoid is open-circuited, so no current flows in its wire. Then its induced emf $v_s$ is equal to the rate of change of the total magnetic flux linkages $\Lambda_s$ through it (Faraday's law): $$v_s = \frac{d \Lambda_s}{dt}$$

Here $\Lambda_s$ is the sum of the magnetic fluxes passing through each of the turns of the solenoid.

In the outer solenoid, the same law governs time rate of change of its flux linkages $\Lambda_p$ induced by the function generator $v_p$, which drives current through the outer solenoid and so produces a magnetic field: $$\frac{d \Lambda_p}{dt} = v_p$$ There's no contribution from the inner solenoid because no current flows there.

Finally, the ratio between the secondary and primary flux linkages is a constant, fixed by the solenoids' geometry and independent of the details of the flux waveforms: $$\frac{\Lambda_s}{\Lambda_p} = k$$

Putting all that together, one finds that the ratio of secondary emf to primary voltage is a constant, independent of frequency: $$\frac{v_s}{v_p} = k$$ This configuration is an example of a transformer.

Limitations:

Note that this relationship breaks down at DC, since the rate of change of flux linkages is 0 there and $v_s=0$. In this case, the current $i$ in the outer solenoid is a maximum, limited only by the resistance $R$ of the wire. In fact, even in the AC case, as the frequency is reduced and the current increases, more and more of the voltage $v_p$ goes to support the coil $iR$, less is due to flux generation, and $v_s$ will be lower than expected. The analysis above neglected the coil resistance.

At high frequencies, stray circuit capacitances upset the strict proportionality.

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You have not specified the nature of the signal generated - sinusoidal or non sinusoidal, periodic or non periodic. You have the relations: $$e_1 = L\frac{d i_1}{dt} + M\frac{d i_2}{dt}$$ $$e_2 = M\frac{d i_1}{dt} - L\frac{d i_2}{dt}$$ If they are wound on the same core, you also have $e_1 = e_2$ since the same flux passes through both of them. Taking Fourier transforms, you get: $$\tilde{e_1} = -i \omega( L\tilde{i_1} + M\tilde {i_2})$$ $$\tilde{e_2} = -i \omega( L\tilde{i_1} + M\tilde{i_2})$$ So you can determine the frequency dependence from the above.

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