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We know that when we release a magnet in conducting pipe (aluminum, copper, etc..) it will be subjected to a magnetic force induced by the induced current. This magnetic field will be created in a manner opposite to the rate change of flux caused by the falling magnet. This in turn will slow down the magnet as you can see here.

I have seen on this forum message #6, the equation of the terminal velocity reached by the magnet during the fall where $v$ is the terminal velocity, $m$ is the mass of the magnet, $g$ is gravitational acceleration, $\rho$ is the resistivity of the tube, $R$ is the inner diameter (I am not sure why it is symbolized by R), $B$ is the magnetic flux density, $b$ is the length of the tube and $T$ is its thickness:

$$v = \frac{mg\rho}{4 \pi RB^2bT}$$

My questions are:

  1. I have only seen the above equation of Physics Forums, I am not sure how it is derived? If this is difficult to answer, I would like to know if it has a name or if you have seen this equation before?

  2. I want to perform this experiment and I would be capable of getting all the above factors except for the magnetic flux $B$. Do you have any idea how I can determine B experimentally?

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  • $\begingroup$ What would "the" magnetic flux density be? The field will be wildly inhomogenous around the magnet. The easies way to measure the flux would be to use a hall sensor. $\endgroup$
    – Jasper
    Commented Aug 13, 2019 at 12:01
  • $\begingroup$ Magnet falling inside a conductive tube and Electromagnetic braking. $\endgroup$
    – Farcher
    Commented Aug 13, 2019 at 12:25
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    $\begingroup$ In the equation from the forum post, it's not clear which value of $B$ to use; the magnetic field of a dipole varies from point to point in space. Another derivation (which is somewhat long & involved) is an exercise in Zangwill's Modern Electrodynamics; the result (up to a proportionality factor) is provided in my answer here. But that derivation depends on several assumptions which may or may not be valid. $\endgroup$
    – Michael Seifert
    Commented Aug 13, 2019 at 13:19
  • $\begingroup$ Has anyone checked the velocity equation for dimensional consistency? I have tried at least 3 times, and can't seem to verify that equation. $\endgroup$
    – David White
    Commented Aug 13, 2019 at 15:58
  • $\begingroup$ @DavidWhite: The units do work out. I'm not sure how accurate it is, though; the equation I posted in the other answer has different dependencies. $\endgroup$
    – Michael Seifert
    Commented Aug 14, 2019 at 12:02

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  1. How can we experimentally determine the velocity of the falling magnet and compare it to the equation above? I know that in this case it reaches terminal velocity in fractions of a second, so can we use v=d/t where we measure time experimentally? Can this work as a rough comparison? :)

Get two copper pipes of the same inner diameter and wall thickness, but with somewhat different lengths. Drop the magnet in each tube, and time how long it takes to travel the length of each tube.

In the first few inches, the magnet will be decelerating as it induces a current in the copper tube. This deceleration will not be constant, as it will be velocity dependent. However, it is safe to assume that the length and time that it takes for the magnet to reach a constant velocity will be the same for both tubes. That length and time will be unknown, but an equation can be set up for the time that it takes for the magnet to fall through each tube. When these two equations are subtracted from each other, everything that remains constant between the tubes, including the distance and time that it takes for the magnet to reach a constant velocity, drops out of the answer. The resulting equation will be: $t_2 - t_1 = (L_2 - L_1)/v_{constant}$, where $v_{constant}$ is the terminal velocity of the magnet as it falls down the tube.

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  • $\begingroup$ @Physics110, for the purposes of timing the drop time of the magnet, the tube should be much longer than the uncertainty in the timing. If you use a stop watch, there is probably 0.2 s of timing error on both ends of the tube, which would require the shorter tube to be approximately 50 cm long to account for this. The longer tube should be 40-50% longer than the short tube. Note that if you are using some sort of electronic timing, the tube lengths can be somewhat shorter because the errors in timing are much less. $\endgroup$
    – David White
    Commented Aug 14, 2019 at 2:15
  • $\begingroup$ The assumption that it is safe to assume that the magnet will reach a constant velocity for both tubes in the same amount of time and in the same distance down the tube must be made if you believe that your experiment is reproducible. The magnet doesn't know which tube it is being dropped down, and you should get the same results for both tubes regarding when the magnet reaches terminal velocity. $\endgroup$
    – David White
    Commented Aug 14, 2019 at 2:18