Effect of coil radius on magnetic induction

Question

We've all seen the how dropping a magnet through a coil will produce a "blip" of voltage across the free ends of the coil, e.g.:

https://www.youtube.com/watch?v=jaZfTXUv_lI

How does the radius of the coil affect the magnitude of the blip? Some assumptions:

• number of turns is the same
• neglect coil resistance
• assume the magnet is "small" compared to the coil diameter

For instance, assume the magnet is an inch in diameter and a couple inches in length. How would passing through a 1 foot diameter coil compare to passing through a 3 foot diameter coil? Does the induced EMF fall off as 1/R? or 1/R^2? or something else?

Answer 1

If we consider an idealized magnetic dipole at the origin (infinitesimally small) with dipole moment $m$ and a single-loop coil of radius $R$ that lies in a plane of constant $z$, then it is not too hard to show that the magnetic flux through this loop is $$ \Phi = \frac{\mu_0 m R}{R^2 + z^2}. $$ Assuming that loop's distance $z$ from the magnet is changing at roughly constant velocity, so that $z = vt$, then we have $$ \mathcal{E} = \frac{d \Phi}{dt} = \frac{2 \mu_0 m R v^2 t}{(R^2 + v^2t^2)^2}. $$ From this it is not too hard to find when the peak EMF (which is probably the easiest measure of "how big the EMF" is); it turns out that it is proportional to $1/R^2$. The exact numerical proportionality is left as an exercise to the reader.

This result could be generalized and made more realistic by:

• Using a more reasonable velocity profile for the magnet (including gravitational acceleration, for example—note that in the video, the EMF spike was greater when the magnet was leaving than when it was entering because it was falling faster);
• Considering a coil spanning a certain range of $z$ values (in which case $\Phi$ would have to be integrated over this range);
• Taking into account the self-inductance of the coil.