Back of the Envelope Estimation Problem for Faraday’s Law

I told my students I would solve this problem for them. It’s a real life problem too.

Is it possible to use a Neodymium magnet and a coil of wire to get an LED to light up?

That’s the real version. But because I was afraid students would be overwhelmed, I added the following:

  • The magnet has a maximum magnetic field of 0.2 Tesla.
  • The LED requires 1.5 Volts and 10 mA to light.

I like this problem because I don’t know the answer. Also, the answer is useful. If you want to do a physics demo showing the voltage induced by a changing magnetic field—what better way than with a hand-held magnet and a small light?

But will it even work? Let’s get started. Here is a diagram.

By changing the magnetic flux through the coil, this will create a curly electric field and an electromotive force (a change in electric potential). The magnetic flux is defined as:

\Phi = N|\vec{B}|A\cos\theta

Where N is the number of turns in the loop, A is the area of the loop and \theta is the angle between the magnetic field and a vector perpendicular to the area. In the diagram above, \theta = 0 since the magnetic field is perpendicular to the coil.

A change in flux produces a voltage according to Faraday’s Law:

\Delta V = \frac{\Delta \Phi}{\Delta t}

Note: yes, I’m different. I think that the number of loops (N) is part of the magnetic flux and that the minus sign in Faraday’s Law doesn’t really mean anything.

Putting the flux into Faraday’s Law, I get (assuming \theta = 0):

\Delta V = \frac{NBA}{\Delta t}

Now for some estimates. I could just estimate everything and then calculate the voltage—but instead I’m going to estimate everything except the number of turns. I can then solve for N and see if it’s reasonable.

Here’s what I have.

  • B = 0.2 T
  • A: radius = 0.01 m
  • Time interval = 1 second
  • Voltage = 1.5 V

Solving for N:

N = \frac{\Delta V \Delta t}{BA}

This is the perfect case to use python for your calculator. You can put your estimates as variables so that you can easily change things up. Here is my code. I get the following output.

Umm….yeah. That’s 23 thousand turns. I’m not going to do that. Even if decreased the time to 0.1 seconds, I would still need 2000 turns. Arg.

Oh, what if I just make a HUGE loop? Nope. That wouldn’t work. In my estimation for the change in flux, I assumed a constant magnetic field—this is obviously not true, but good enough for a small loop. With a big loop, you would have some of the magnetic field creating a negative flux. It would just make things worse.

What if I put the magnet on a spinning stick (run by a motor)?

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