Worked exercise — magnet in a tube, terminal velocity

Problem

A cylindrical magnet of mass $m=25\ \text{g}$ falls inside a vertical copper tube $L=1{,}2\ \text{m}$ long. The transit time is $\tau=3{,}2\ \text{s}$. (a) Estimate the average velocity and the terminal velocity (the steady state is established after a few centimetres). (b) Estimate the power dissipated in the copper. (c) Compare with the time of free fall from the same height.

Solution

(a) Average velocity and terminal velocity. $v_{\text{avg}}=\frac{L}{\tau}=\frac{1{,}2}{3{,}2}\approx 0{,}375\ \text{m/s}$ Since the steady state is established almost immediately, for nearly the whole fall the magnet travels at terminal velocity, so $\ev{v_{\text{lim}}\approx v_{\text{avg}}\approx 0{,}38\ \text{m/s}\approx 38\ \text{cm/s}}$

(b) Power dissipated. At steady state the velocity is constant: all the work of gravity is converted into heat (Foucault currents in the copper). $P=m\,g\,v_{\text{lim}}=0{,}025\cdot 9{,}8\cdot 0{,}38\approx 9{,}3\cdot 10^{-2}\ \text{W}\approx 93\ \text{mW}$ $\ev{P\approx 93\ \text{mW}}$

(c) Comparison with free fall. $L=\tfrac{1}{2}g\,t^2\ \Rightarrow\ t=\sqrt{\frac{2L}{g}}=\sqrt{\frac{2\cdot 1{,}2}{9{,}8}}\approx 0{,}49\ \text{s}$ $\ev{t_{\text{free}}\approx 0{,}49\ \text{s}}$ In free fall the magnet would arrive about six times faster. The induced currents (Lenz’s law) brake the magnet: almost all the kinetic energy it would have in free fall (~99%) is instead dissipated as heat in the tube.

Links

Topics: Electromagnetic induction Concepts: Lenz’s law · Foucault currents · Joule effect