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Visualize a long conducting rod of length $l$ falling towards the Earth (after being released from a hight of say $5(km)$) perpendicularly to the Earth's magnetic field.

Near the earth's equator, the earth's magnetic field is approximately horizontal.

Reference https://www.physicsforums.com/threads/magnetic-fields-at-earth-equator.16980/

So for symplicity, we release the rod at the equator with the rod being parallel to the equator (I choose $5(km)$ as the height from which the rod is released so the magnetic field stays about the same). It's clear that the electrons in the rod experience a Lorenz force $\vec F=q\vec v \times\vec B$, so there will develop a potential difference. It might be clear that the diameter of the rod plays no role.

Now the strength of the magnetic field of the Earth varies between $0,25$-$0,65$ $(G)$ (Gauss), which corresponds to $25 000$-$65 000$ $(nT)$, (where the $T$ stands for Tesla, the international standard for the strength of a magnetic field). For comparison, the strength of a magnet in a refrigerator is about $10 000 000$ $(nT)$, the international standard for the strength of a magnetic field), or $100$ $G$. Let's take the average value of $0,45$ $G$.

When we attach to both ends of the rod conducting wires and attach those on their turn to a led, a current will flow to it if the rod starts to fall.

My question:

The rod will accelerate upon release and will move faster and faster. Let's ignore friction between the rod (and the things attached to it) and the air, and the resistance in the conducting wires. Will the led at a certain speed light up (so we can see it)? Of course, this depends on the length of the rod, but let's assume $l=1(m)$. Or is the potential difference after falling free for $5(km)$'s too small?

It's my guess that the speed after falling $5(km)$ is big enough ($s=\frac 1 2 \times 9,8 t^2$, $v=9,8t$ and the Lorenz force should bring clarity) the potential difference in the rod is big enough to let the led shine brightly. But of course, I can be wrong.

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  • $\begingroup$ Where is the wire going? If the whole circuit is falling then the voltage on different parts of the wire cancels out (in a uniform B field). If the wire hangs down long enough so that the lower end does not move while the upper end falls, then it will need to be a very long wire! $\endgroup$ – Andrew Steane May 6 at 22:04
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"When we attach to both ends of the rod conducting wires and attach those on their turn to a led, a current will flow to it if the rod starts to fall."

I read this as saying that the rod (AB, say), together with conducting wires and LED, form a complete closed circuit that is falling to Earth. In that case the emf induced in the rod will be in the same direction (AB say) as the emf induced in the wires and the LED. So if we go right round the circuit (A to B via the rod followed by B to A via the wires and LED) the emfs will be in opposite senses and will cancel. [Alternatively, we could say that as the circuit falls, the enclosed area won't change, nor will the flux density (if the distance fallen is much less than the Earth's radius), so there will be no change in flux linkage and no emf.]

To stand any chance of lighting the led we need a set-up such as this imaginable but probably impracticable one ... Let the ends of the rod as it falls brush against stationary vertical rods. The rods are joined at their lower ends by stationary wires with LED. My calculations, based on $\mathscr E =B \ell \sqrt {2gh},$ with $B=30 \mu$T, $\ell=1$m, $h=5$ km then give $\mathscr E =10$ mV, whereas about 1.6 V would be needed to light the LED. In practice the rod will have reached a terminal velocity well before 5 km of fall, so the emf will be even less than 10 mV. A much longer rod might produce the required voltage.

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  • $\begingroup$ That's the answer I needed! In the process of formulating the question, I forgot that the magnetic flux through the circuit isn't changing. You have a nice solution though. What if we let the circuit rotate around the axis parallel to the rod? Will the led, for an $l$ that's big enough, get powered by an AC? $\endgroup$ – descheleschilder May 7 at 14:09
  • $\begingroup$ Thank you. Rotation about an axis parallel to the rod should be fine. Over to you to calculate the rotation frequency needed for a circuit of your chosen area to generate about 2 V. AC should be fine for a LED; it won't light on the 'reverse' half-cycles, but the frequency will probably be high enough for no-one to notice. $\endgroup$ – Philip Wood May 7 at 14:28

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