Worked exercise — Waltenhofen's disc

Problem

A copper disc of radius $r=5\ \text{cm}$ and thickness $s=2\ \text{mm}$ oscillates in a uniform field $B=0{,}5\ \text{T}$, with typical speed $v=0{,}5\ \text{m/s}$. Roughly estimate the power dissipated by eddy currents. Resistivity of copper $\rho\approx 1{,}7\cdot 10^{-8}\ \Omega\cdot\text{m}$.

Solution

Typical EMF in a ring of radius $r$ moving in the field: $\text{EMF}\sim B\,r\,v=0{,}5\cdot 0{,}05\cdot 0{,}5=1{,}25\cdot 10^{-2}\ \text{V}$

Resistance of the ring (cross-section $\sim s\cdot r$, length $\sim 2\pi r$): $R\sim \rho\,\frac{2\pi r}{s\,r}=\frac{2\pi\rho}{s}=\frac{2\pi\cdot 1{,}7\cdot 10^{-8}}{2\cdot 10^{-3}}\approx 5{,}3\cdot 10^{-5}\ \Omega$

Power dissipated: $P\sim \frac{\text{EMF}^2}{R}=\frac{(1{,}25\cdot 10^{-2})^2}{5{,}3\cdot 10^{-5}}\approx 2{,}9\ \text{W}$ $\ev{P\sim 2{,}9\ \text{W}}$

A power of this order is enough to brake the disc within a few cycles (eddy currents, by Lenz’s law, oppose the motion). If radial slits are cut into the disc, the current loops can no longer close: the resistance $R$ rises by several orders of magnitude and the dissipation collapses, so the disc oscillates almost freely.

Collegamenti

Argomenti: Electromagnetic induction Concetti: Eddy currents · Lenz’s law · Joule effect