I came across a YT video where allegedly the Hall effect is measured.

No matter how I tried, I couldn't reproduce it. I used a lab power supply, applied up to 10 amperes of current. The voltmeter does register some low voltage, but it does not change when I apply magnetic field.

Is the video a hoax? Can the Hall effect actually be measured that way?

  • 1
    \$\begingroup\$ I looked at several videos, none are convincing IMO. It is hard to get a voltage gradient across a conductor, hall effect devices normally use a semiconductor, probably doped silicon. Also, you need to get the magnetic field to flow in the same direction, the magnetic fields are flowing through the aluminum foil and back. You want something more like a horseshoe magnet with N on one side and S on the other side. \$\endgroup\$
    – Mattman944
    Commented Jul 31, 2022 at 14:30
  • \$\begingroup\$ Thanks for your reply. I found this laboratoriofisica.uc3m.es/guiones_ing/sse/… It's a rig to measure Hall effect. And it looks like magnetic field should be applied from opposite sides so that it flows in the same direction, like you said. And it looks like they are using a thin copper plate. (a regular PCB?) It also mentions electrons exhibit drift, so measuring poles should not be opposite of each other, they should be offset. \$\endgroup\$ Commented Jul 31, 2022 at 14:37
  • \$\begingroup\$ Behind every “simple” experiment is a bunch of laboratory technique. Lots of resources either assume you have the background knowledge, or just copy information from elsewhere, without the author actually successfully reproducing the experiment. Good on you to actually try stuff out and report a negative result. Most such setups don’t work on the first try. It is my assumption that 9 out of 10 “experimental” YT videos are not showing the effect they intend to demonstrate. Mistakes in metrology can make the most intricate of setups useless. \$\endgroup\$ Commented Jul 31, 2022 at 16:56

3 Answers 3


It is very difficult to measure Hall effect with available tools at home.

The Hall coefficient of Aluminum, and of metals in general is low (because there are so many conducting carriers; that's partially why semiconductors are used.

The voltage created by the Hall effect measured with a cuboid (i.e. rectangular) conductor depends on the product of current and magnetic field (so it's good to get both high), and inversely on the thickness. It does not depend on the width (or length) of the conductor. The thinnest commonly (home) available metal is household aluminum foil (about 50 um thick), or candy wrapping foil (about 20-25 um thick).

Use a piece as wide as you think you can create a magnetic flux over. Wider than that will reduce the measured signal because the inactive portions of the conductor will 'short out' the Hall voltage. Wide conductors reduce current density and therefore heating/fusing -- this does not affect the Hall measurement.

The length doesn't matter -- use as long as is convenient. It is important to stabilize the system -- use thumbtacks (with plastic heads if possible) to pin the foil to a wooden board or similar.

Clip your voltage measurement probes to the middle of the sheet. An offset doesn't matter.

Likely the Hall voltage you measure will be in the uV range -- this will require a voltmeter or amplifier capable of sensing such low values.

Apply a DC current (with no magnetic field) and measure the voltage. It will be non-zero because there will likely be an offset in the pickup positions. Ensure the voltage stabilizes (if the system heats up, the offset may change). Then apply the magnetic field, being careful to not disturb the physical arrangement. You will likely first notice quite large voltage shifts as the changing magnetic field induces voltages in the measurement loop. This is not the Hall effect. The voltage measured after the system stabilizes is the Hall voltage. Try again with the opposite field direction -- you should measure the opposite polarity change.

If you don't see the opposite polarity change, then likely what you are observing is slight shifts in the position of the (probably magnetic) leads or pickup points. This would be the same polarity, independent of the applied static field.

In practice, the voltage you might observe is likely less than 1 uV and will require a special (lab-grade) meter to observe.


There's no reason to call it out as a hoax necessarily, but it would be difficult to reproduce as shown.

The thin metal conductor has magnets stuck top and bottom. As they are stuck to the conductor, this necessarily means they are the correct polarity to set up a strong field in it.

The main problems are the low current from the PP3 battery giving low sensitivity, and the positional stability of the croc-clip connectors, you would surely move them as you attach/detach magnets.

When I performed this experiment in UK 6th form (age 17, a long time ago before neo magnets were available), we used aluminum kitchen foil. As the main problem is the stability of the pickoff connections, the foil was stuck down to card, and then cut in a cross shape, like this ASCII art

-------  ----------
-----  --  --------
     ||  ||      
     B    C

A low value potentiometer was connected between B and C, and adjusted to get zero potential difference between its wiper and A, with current flowing, and no magnetic field. This eliminated the potential gradient along the foil as a contributor to the potential across the foil.


I don't know what material is used in the video but, it looks like Aluminium foil (maybe 0.01 mm thick) so, here's my best estimate of what hall voltage would be produced.

With an aluminium sheet of maybe 0.00001 metres (0.01 mm) thickness, a magnetic field strength of (say) 1 tesla and (say) a current of 1 amp (unlikely with a PP3 battery), the hall voltage would be about 10 μV.

This is based on this formula: -

$$V_H = \dfrac{I\cdot B}{n\cdot t\cdot e}$$

  • \$I\$ is current (maybe 1 amp as unlikely as it sounds from this type of battery)
  • \$B\$ is magnetic flux density
  • \$n\$ is the electron mobility of aluminium (\$6\times 10^{28}\$)
  • \$e\$ is charge of an electron (\$1.6\times 10^{-19}\$)

The formula result (about 10.4 μV) is supported by this hall voltage calculator: -

enter image description here

Is the video a hoax? Can the Hall effect actually be measured that way?

Can you tell what the material is? Can you tell what the thickness is? Can you tell what the current is and what the flux density is?

To me, you don't know any of those numbers hence, the video has zero worth except for the easily impressed non cynical types (not me).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.