# Energy from pickup coil

If I have an inductive coil generating AC voltage due to an oscillating magnetic field, is there a way to determine the current (or energy) being generated in the coil? Further, if a capacitor were placed in parallel, would the energy get transferred to the capacitor and create a decaying oscillation?

I know that the coil will generate a voltage to resist the changing magnetic field, but does that mean a current flows through the coil equal to the voltage divided by coil resistance? If so, would that current be in phase with the voltage, or 90 degrees out of phase?

I see a lot of high-level discussion of this topic related to tank circuits, LC oscillators, and inductive charging, but I am not sure where to find a more physics/electron-level explanation of this phenomenon.

• I recommend that you read "Matter & Interactions," by Chabay and Sherwood, 3rd edition (or newer, if it happens by the time you get it.) It's a pure physics treatment but very, very easy to read and understand while also being thoroughly quantitative and qualitative. So you'll get a high level view as well as a very detailed view with probably the easiest book I know of that achieves this. It's as simple as possible, but no simpler.
– jonk
Commented Jun 15, 2017 at 2:35
• youtube.com/watch?v=nGQbA2jwkWI&t=12m30s states the current is only affected by coil resistance (doesn't mention inductance), so I guess V=IR applies
– abc
Commented Jun 15, 2017 at 23:02
• I think I was looking for something like this: rspa.royalsocietypublishing.org/content/472/2191/20160338 ('A physical model for low-frequency electromagnetic induction in the near field based on direct interaction between transmitter and receiver electrons')
– abc
Commented Jun 17, 2017 at 22:35

If I have an inductive coil generating AC voltage due to an oscillating magnetic field, is there a way to determine the current (or energy) being generated in the coil?

Yes this is quite simple, connect a load to the terminals using a fixed value resistor of a known value and measure the voltage across that resistor. Use ohms law etc..

Further, if a capacitor were placed in parallel, would the energy get transferred to the capacitor and create a decaying oscillation?

Yes, an initial amount energy would be taken from the field to produce a constant oscillation (if the field is maintained). Energy cyclically interchanges between the coil's inductance and the capacitor and this is improved by choosing the capacitor value to match the coil inductance and operating frequency of the (presumably) AC magnetic field. There is a simple formula to calculate this.

I know that the coil will generate a voltage to resist the changing magnetic field

No, the induced voltage does not resist the applied changing magnetic field.

but does that mean a current flows through the coil equal to the voltage divided by coil resistance?

Without a load connected to the coil terminals, no current flows and if a load is connected then that current is determined by the self inductance of the coil, the incident magnetic field amplitude, the number of turns, the coil shape/dimensions, the operating frequency and the load impedance.

• Concerning voltage being generated in response to the changing magnetic field, perhaps I worded it incorrectly, but I do see a voltage appear on the oscilloscope proportional to the magnetic field intensity (affected by proximity to source).
– abc
Commented Jun 15, 2017 at 15:50
• And concerning the energy generated from the coil, is there a way to calculate that without empirically measuring?
– abc
Commented Jun 15, 2017 at 15:53
• Re: my first comment, maybe I'm closing the circuit and allowing a very small current to flow via the high Z osc probe?
– abc
Commented Jun 15, 2017 at 16:47
• There is energy in the coil only when you take current and, if you know the inductance of the coil, that energy is $\dfrac{LI^2}{2}$ but remember it is cyclic because of the field alternating. Maybe try and explain what it is you are trying to achieve by estimating the energy? Commented Jun 15, 2017 at 17:07
• A capacitor does muddy the water if you are trying to infer field values. Consider Faraday's law of induction as a starting point - V = N d(phi)/dt. The peak voltage seen directly equates to the maximum rate of change of flux through the coil and, if it's a sinewave you should be able to infer total flux through the coil aperture. From that you can infer flux density. Adding any load will take current and somewhat disturb/alter the original field and rubbish potential measurements. Commented Jun 15, 2017 at 17:18