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I am watching a video on how a crystal radio works. It's very nice and I get the gist, there's a bit on tuning that claims that if you have (see picture below) a wire connected to earth from both sides, and you move a magnet next (in blue) to it (so Farady induction law is happening) there will be no current through the wire (in black)

The video is https://www.youtube.com/watch?v=0-PParSmwtE&ab_channel=RimstarOrg (see 1:10)

Why would this be true? The induction law is $$ \nabla \times E = dB/dt$$ so if the magnetic field is changing we have EMF around the coils which should be moving electrons.

enter image description here

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I agree that the video is a little misleading.

I have taken the circuit diagram for the crystal radio from How to Make/Build a Crystal Radio and added a few labels.

enter image description here

The important part of the circuit is the loop $ASGEDA$ which is a tuned circuit with rectification done by the diode $D$ so that an audio frequency voltage is applied across the earpiece $E$.

In the section of the coil $SB$ there is an induced voltage and that does produced an induced current in the loop $SBGB$.
However that induced voltage/current will be very,very small as there is virtually no capacitance associated with that circuit and so no enhancement of the signal due to a resonant $LC$ circuit as there is for circuit $ASGEDA$.

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  • $\begingroup$ Your picture is much better. There's something I still don't get: How do you analyze the effect of $S$ on the LC (i.e why the current doesn't ignore $S$ and keeps going to $B$). In fact is there any reason that $B$ should be connected to the capacitor and not straight to $G$? If that was the case, I'd find your answer intuitive (I don't like all the mixing of cables happening at $G$). $\endgroup$
    – Andy
    Commented Jan 29, 2022 at 13:55
  • $\begingroup$ As drawn the slider $S$ changes the inductance in the tuned circuit, ie changes the frequency of the station being listened to. Since the circuit $SBGS$ has a very small resistance as compared with any other loop it is connected to that is where the current will flow as a result of the voltage induced in $SB$. $\endgroup$
    – Farcher
    Commented Jan 29, 2022 at 14:17
  • $\begingroup$ I see, so even if some of that current chooses to go to $E$ it's minor compared to the resonance of the one with the capacitor $\endgroup$
    – Andy
    Commented Jan 29, 2022 at 14:58
  • $\begingroup$ You might note that the 25 turn coil must be inductively linked to the 90 turn coil. $\endgroup$
    – R.W. Bird
    Commented Jan 29, 2022 at 17:06

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