Resource Lesson Eddy Currents plus a Lab Simulation
When a non-magnetic conductor (for example copper or aluminum) is exposed to a varying magnetic field, eddy currents are produced in the conductor. The currents obey Lenz Law and oppose the direction of the external changing magnetic field.

A classic demonstration of this is shown in the MIT Physics Demo -- Pendulum and Magnet.  During the video, the first paddle released is a solid copper sheet. It swings freely between the poles of an electromagnet until the electromagnet is turned on. Then it abruptly stops because of the interactions between the eddy currents produced within its solid surface and the magnetic field between the poles of the strong electromagnet. Later, a perforated paddle is used which, when released between the poles, does not immediately stop when the electromagnet is turned on. Instead it displays a damped oscillation. This change in behavior is because the slits no longer allowed the larger eddy currents to form. Instead weaker currents can only be created inside the smaller copper strips. A nice explanation of this effect (with diagrams and captions) is given on this page from the University of New South Wales in Australia.

A second classic demonstration is shown inn the MIT Physics Demo -- Jumping Ring. It begins with a solid aluminum ring being placed over the soft-iron core of an electromagnet. When the electromagnet is turned on the eddy currents induced in the aluminum ring cause it to be repelled from the electromagnet and "jump" up into the air. A second ring with a slit down on side is then placed on the electromagnet. This time the ring does not "jump" since the slit has broken the conducting surface and no eddy currents can be sustained.

The third classic demonstration on eddy current involves letting a neodymium magnet fall through a copper tube. Fortunately, we have been granted permission by Mr. Cavini to use a series of screen captures from his YouTube video Lenz's Law at Work  to perform an in-depth investigation of this phenomena. Please view his video several times BEFORE proceeding through the lab simultion.

In his demonstration, a non-magnet is first dropped through a copper tube. As expected it emerges without any time delay; just as it would if released from rest through the air without the tube. Then a strong magnet is dropped through the tube. The magnet does not touch the sides of the tube, but takes a significantly longer amount of time to re-emerge at the base. Just as in the other examples, eddy currents are responsible for the extra time needed by the magnet to traverse the length of the copper tube. The eddy currents in the copper tube create a drag force on the magnet. If the magnet is sufficiently light and strong, the magnetic drag force will balance the magnet's weight, and it will reach terminal speed during its fall.

Included in the video are graphs of the two situations: a non-magnet free falling through the tube and the magnet falling through the tube.

Refer to the following information for the next two questions.

In this section you will compare the following sets of speed vs time and position vs time graph. The screen capture at 6:13 is of freefall; while the screen capture at 6:41 is of a magnet sliding down the copper tube.

Choose the statement(s) that correctly describe the position-time graph in Group A for freefall and the position-time graph in group B for the magnet falling through the copper tube.

Choose the statement(s) that correctly describe the speed-time graph in Group A for freefall and the speed-time graph in Group B for the magnet falling through the copper tube.

Refer to the following information for the next four questions.

In one phase of his demonstrations, a magnet is initially released from rest using a smaller copper mechanism. Once released, the magnet continues falling until it strikes the bottom stand supporting the apparatus. For each of the four intervals identified, explain the behavior of the falling magnet. If a Region called "0" had been highlighted, it would have represented the passage of the magnet through the smaller copper release mechanism.

In Region 1

In Region 2

In Region 3

In Region 4

Refer to the following information for the next two questions.

A new apparatus is shown in picture 1. This apparatus has a longer copper tube placed directly in contact with the copper release mechanism. The second picture has the apparatus you analyzed in the previous section.

The magnet is dropped from rest from the release mechanism at the top of a long copper tube (#1) and enters the shorter copper tube (#2) already moving, the graph of speed vs time shows us that

 While moving through the region designated by the pink arrow, what is the magnitude of the magnetic force acting on the magnet?

Refer to the following information for the next four questions.

Use the following data to answer the next questions.
 The mass of the copper rod is 50 grams The mass of the magnet is 8.00 g The diameter of the magnet is (0.5 inches) = ½(2.54) = 1.27 cm The length of the copper tube is 25.00 cm [26.5-0.009-0.00635] The total time to fall just through the tube is 2.185 seconds [2.21-0.025] the specific heat of copper is 385 J/kgK

 It would be nice to be able to derive an equation for the magnet's instantaneous velocity as it was falling through the tube, but the equation for the induced emf and subsequent magnetic force exchanged between the tube's wall and the magnet is beyond our abilities. So, we are going to obtain a reasonable average value for the magnet's terminal speed by using the relationship that average speed = distance/time.   Calculate the magnet's terminal speed assuming that it reaches this speed almost instantaneously upon release.

 If the total distance from the bottom edge of the tube to bottom of the 0.009 line marker is 0.01535 meters (this value represents the distance 0.009 plus the radius of the magnet), then calculate the magnet's instantaneous velocity at the 0.009 line marker.

 The magnet's average speed upon exiting the tube can be calculated using the relationship vaverage = ½ (vo + vf) where vo is the magnet's terminal velocity through the copper tubing and vf is your previous answer. Remember that the magnet is in freefall throughout this interval.

 Verify that the average speed obtained with the video's distance (0.009 meters) and time (0.025 seconds) data is very, very close to the magnitude of the average velocity calculated in the previous question.

Refer to the following information for the next four questions.

Use energy arguments to determine the temperature rise in the copper tube. Use the previously determined values, not those in the video.

 PE at the top of the tube

 KE just as it emerges from the bottom of the tube

 Energy lost to heating the tube

 The increase in the temperature of the tube