Problem — Damped magnetic pendulum

Problem

A small cylindrical magnet swings on a thread, at a distance of $2\ \text{mm}$ above a copper plate $1\ \text{cm}$ thick. Explain qualitatively why the amplitude of the oscillations decreases and estimate the characteristic damping time, knowing that the braking force is $\sim b\,v$ with $b\approx (NB)^2/R$, where $R$ is the effective resistance of the copper region involved.

Solution

Why it damps. As the magnet swings, the field impinging on the copper varies in time: eddy currents are induced in the plate. By Lenz’s law these currents oppose the change in flux, i.e. the motion of the magnet: a viscous braking force $F\approx -b\,v$ arises, proportional to the velocity and opposite to it. The mechanical energy is dissipated through Joule heating in the copper, so the amplitude decreases.

Model. Adding the viscous term to the pendulum equation gives a damped oscillator; the amplitude envelope decays as $e^{-t/\tau}$ with $\tau=\frac{m}{b},\qquad b\approx \frac{(NB)^2}{R}$

Order-of-magnitude estimate. For a small magnet of mass $m$ of the order of grams ($m\sim 10^{-3}\ \text{kg}$) and a coefficient $b$ of the order of $10^{-3}\text{–}10^{-2}\ \text{kg/s}$ (highly conductive copper, small $R$, magnet close by), we get $\tau=\frac{m}{b}\sim \frac{10^{-3}}{10^{-3}}\ \text{s}\sim\ \text{a few seconds}$ $\ev{F\sim b\,v,\quad b\approx \tfrac{(NB)^2}{R},\quad \tau=\tfrac{m}{b}\sim\ \text{a few seconds}}$ Bringing the magnet closer to the copper increases $b$ and decreases $\tau$: the damping is faster. This is the principle behind the eddy-current brake.

Links

Topics: Induzione elettromagnetica Concepts: Legge di Lenz · Correnti di Foucault · Oscillazioni smorzate · Effetto Joule