# Why does a wire with both ends connected to the same earth wire has no current?

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.

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.

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 $$ASGEDA$$.

• 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 $G$).
– Andy
Jan 29, 2022 at 13:55
• 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 $SB$. Jan 29, 2022 at 14:17
• 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
– Andy
Jan 29, 2022 at 14:58
• You might note that the 25 turn coil must be inductively linked to the 90 turn coil. Jan 29, 2022 at 17:06