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I want to devise a simple procedure that can be followed by students to discover that a current in a wire produces a magnetic field near that wire.

I am of course restricted to very simple materials, but I do have a strong neodymium magnet with which the students magnetize needles, which we then use to discover and map the magnetic field.

I have four $1.5V$ AA batteries connected in series, which yield $5.5V$ (one battery is kinda weak). I'll use 1mm thick copper wire, as well as some aluminum foil and tape to secure the connections. I assume $1\Omega$ resistance (Wikipedia says $0.1\Omega$ internal resistance for a typical AA battery, compared to which the wires should be negligible, so I'd rather overestimate the resistance)

According to my calculations, at a distance of 3cm from the wire (about the size of the needle) this should net me $33\mu T$ of magnetic field, which should be somewhat detectable by the needle, considering it visibly picks up the Earth's magnetic field which is $65\mu T$ again according to Wikipedia.

However, the needle does not seem to care one whit about the current in the wire.

Did I mess up my calculations, or did I make a mistake in my setup?

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Why not use a plotting compass instead of a home-made magnetised needle? Plotting compasses are available from educational suppliers. They used to be very cheap. They are easy to use.

With a current of 2 A through a long wire you should get a field of about 40 $\mu$T at 1 cm from the wire. This is of the same order as the Earth's magnetic field, so the compass should respond clearly.

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  • $\begingroup$ With the 5.5V I have, I should have way more than 2A through the wire. However the needle does not respond. Is it possible that the batteries have more internal resistance than expected? $\endgroup$ – Riccardo Orlando May 13 at 10:21
  • $\begingroup$ I am measuring $3mA$ (that's milli-Amperes) across the battery. How did I lose three orders of magnitude in current? That's weird (but explains the absence of a magnetic field) $\endgroup$ – Riccardo Orlando May 13 at 10:27
  • $\begingroup$ Why not use a new battery (say 2 or 3 cells)? This should give you a current of a few amperes on dead short circuit (which is pretty much what you'll have if you connect a piece of wire across it). Leave it connected just for long enough to see what the compass does, so that you don't run the battery down in a single go! $\endgroup$ – Philip Wood May 13 at 10:46
  • $\begingroup$ yeah, you're right. This quarantine is making me hoard my supplies. $\endgroup$ – Riccardo Orlando May 13 at 10:53
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    $\begingroup$ It's not usually good to connect cells in parallel, as small differences in their emfs can drive a current through them, even with no external circuit. I wouldn't do so, at least not before you've tried a single cell. Nor, at first, would I use cells in series. As you say, most of the circuit resistance comes from the cell or cells, so you'll get pretty much the same current with more than one cell in series as with a single cell! My recommendation: TRY IT ! $\endgroup$ – Philip Wood May 13 at 11:48
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You may want to go with D batteries. An old computer or stereo-amp that someone is throwing out may have a good multi-amp DC power supply. Some of the little black boxes (often available at garage sales) may be rated for 2 amps DC. Harbor Freight carries a low cost digital multi-meter. I use to string a wire from ceiling to floor next to a table. Boards on each side extending from the table permitted the use of a small compass or iron fillings on a sheet of paper. Connect your ammeter in series with the long wire.

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As a simple, but engaging alternative, you could try having them make their own homopolar motors with wire and a battery. As a starting point, Wikihow has some instructions on the basic set-up.

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You may want to consider using a suspended magnet attached to a mirror. A laser reflecting off of the mirror will detect the magnetic field produced by a AAA battery 12 feet away. Here’s a link to my video where I conducted this experiment. The portion related to your question starts a 9:33 into the video.Here it is.

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