Magnet Falling Through a Copper Tube
A student investigates electromagnetic braking by dropping a small neodymium magnet through a vertical copper tube. The tube is 1.20 m long, with inner diameter 13 mm and wall thickness 1.5 mm, and is open at both ends. The magnet has mass 7.5 g and its magnetic poles are perpendicular to the tube axis. Timing gates record the magnet taking 1.65 s to travel the length of the copper tube. When the copper tube is replaced by a cardboard tube of the same dimensions, the time is 0.49 s. The magnet is released from rest just inside the top of each tube. The copper tube is then replaced by an identical copper tube that has a narrow vertical slit cut along its full length. Discuss the motion.
1. Show that the magnet reaches a terminal speed in the copper tube and estimate that speed. Explain why the cardboard-tube time is close to free fall.
Read the worked answer
In the cardboard tube the observed time is 0.49 s. Free-fall time from rest over 1.20 m is t = sqrt(2h/g) = sqrt(2(1.20)/9.81) = 0.495 s, so the cardboard tube is essentially free fall and magnetic braking is negligible. In the copper tube the time is 1.65 s, over three times longer. The induced eddy currents produce a magnetic force that opposes the motion and grows with speed. At terminal speed this upward force equals mg, so the acceleration becomes zero. Because the copper-tube time is so much longer than free fall, most of the fall is at a nearly constant speed. Using the average speed as an estimate, v_t ≈ h/t = 1.20/1.65 = 0.727 m/s. The true terminal speed is slightly larger than this average because the magnet starts from rest; the average includes a short acceleration phase.
2. Calculate the average rate at which mechanical energy is dissipated in the copper tube, and compare it with the gravitational power at the estimated terminal speed.
Read the worked answer
The gravitational potential energy lost is mgh = (7.5×10^-3)(9.81)(1.20) = 8.83×10^-2 J. The average dissipation rate over the 1.65 s fall is therefore
P_avg = mgh/t = 8.83×10^-2/1.65 = 5.35×10^-2 W.
At the estimated terminal speed, the gravitational power is P_grav = mgv_t = (7.5×10^-3)(9.81)(0.727) = 5.35×10^-2 W.
These agree because at terminal speed the loss of gravitational energy is almost entirely dissipated as heat in the copper; the kinetic energy is nearly constant. The final kinetic energy is only 1/2 m v_t² = 0.5(7.5×10^-3)(0.727)² = 2.0×10^-3 J, about 2% of mgh, so it does not change the estimate much.
3. Explain why the slit increases the magnet's fall speed, and why replacing copper with an insulating tube of the same dimensions would give a time close to the cardboard-tube time.
Read the worked answer
Eddy currents are driven by the changing magnetic flux through loops of copper. Copper has low resistivity, so for a given induced emf the currents are large and the magnetic braking force is strong. A vertical slit cuts the azimuthal current loops, increasing their resistance and reducing the current for the same flux change. The braking force is therefore smaller, so the magnet falls faster. An insulating tube cannot support significant eddy currents at all, so there is no magnetic braking and the fall time is close to the cardboard-tube free-fall time.