# Calculating Induced EMF in Wireless LED Circuit

I have built the following circuit to power a wireless LED and my calculations and measurements do not give the same values. I am measuring a voltage nearly 6 times larger than I would expect in my receiver circuit.

I measured the AC frequency of the circuit to be $$\frac{\omega}{2\pi} = f = 350 \ kHz$$. Assuming the magnetic field is given by

\begin{align} B &= \frac{N \mu_0 I}{2R}, \quad I = I_0\cos(\omega t) \approx 0.052A \cos(\omega t), \end{align}

and I use faradays law of induction, the induced EMF in the receiver circuit should be given by

\begin{align} \mathcal{E} &= - \frac{d\Phi}{dt} = - A \frac{dB}{dt} \\ &= A \frac{N \mu_0 I_0}{2R} \omega sin(\omega t), \quad A = \pi R^2 \\ & = \frac{N \pi R I_0 \mu_0}{2} \omega \sin(\omega t) \\ & = \frac{(30)\pi (0.05m)(0.052 A) (4\pi \times 10^{-7}H/m)}{2} (2\pi \times 350 \times 10^3 Hz) \sin(\omega t) \\ & \approx 0.339 \sin(\omega t) \text{ Volts } \end{align}

However when I read the voltage in the receiver, I am actually getting a much larger value of over $$1.9 V$$.

From searching, I believe this has something to do with resonant frequencies, but I do not understand this. If someone can show me why my calculation is wrong and provide the right one, I would appreciate it!

• The fact that it is not a sine wave may indeed be the culprit in my faulty calculation. A voltage spike would make $|\frac{dB}{dt}|$ larger. If I was able to measure the inductance, 'L' of my coils, is there a way that I could obtain an analytical solution of the waveform without an oscilloscope? Oct 21, 2022 at 21:00