# Induced EMF in wireless LED

I have already asked this question here but this may be a better place.

I have built the following circuit to power a wireless LED but 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 Faraday's 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}

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.

• pardon me if this is obvious since I'm not experienced in magnetic calculations - but why is the magnetic field divided by 2R ? Commented Oct 21, 2022 at 19:55
• @user253751 You can calculate this using the Biot Savart Law: hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html. You multiply by N if there are N loops. Commented Oct 21, 2022 at 19:56
• That's just calculating the field at the center (where it's weakest!), not the total field Commented Oct 21, 2022 at 19:57
• @user253751 I know! So the voltage should be smaller if anything, not larger... Commented Oct 21, 2022 at 19:57
• I think you should calculate a voltage that's too low since you're calculating a field that's too weak, and weaker magnetic field produces less voltage. Commented Oct 21, 2022 at 19:58