8
$\begingroup$

A ball falls from a magnet but the magnet still exerts an upwards force against gravity yet the ball falls anyway. However, the ball slows down and thus the sound when it hits the floor is less signalling that some energy has been lost during its descent. What I'm wondering is where that energy has gone? Has the magnet gained magnet energy? Or has earth gained energy. Or has the ball not lost energy but its remaining energy just wasn't turned into sound upon contact with the ground? Or is it something else?

$\endgroup$
  • 1
    $\begingroup$ Is the metal ball made of iron or steel? $\endgroup$ – cms Jan 6 at 15:02
5
$\begingroup$

If the ball is made of iron

You put “magnetic” potential energy into the system when you brought the metal ball up to the magnet. The sign of that potential energy is negative indicating an attractive force.

When you drop the ball, the ball loses gravitational potential energy, but adds magnetic potential energy and kinetic energy. Therefore less energy was available for kinetic energy and the corresponding impact speed was less.

edit:

We can write this as an energy relationship: $$ \begin{align} \Delta E &= \Delta K + \Delta U \\ 0 &= (K_f-K_i) + \Delta U_G + \Delta U_{M} \\ K_f &= -\Delta U_G - \Delta U_M \end{align} $$ Where $K_i = \Delta E = 0$. In the situation where the ball falls the gravitational potential energy decreases ($\Delta U_G < 0$) but the magnetic potential energy increases ($\Delta U_M > 0$), since the metal ball has moved further from the magnet. The type of forces two magnets experience is a conservative force (since it's path independent) and so it makes sense to talk about a magnetic potential energy.

The ball would need to be iron (or one of its alloys like steel) because in terms of everday materials, iron is the only one that can be temporarily (induced) magnetized. For most other materials, $\Delta U_M \approx 0$.

$\endgroup$
  • $\begingroup$ Why only iron and not any metal? $\endgroup$ – yolo Jan 6 at 18:11
  • 1
    $\begingroup$ @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys. $\endgroup$ – Peter Shor Jan 6 at 18:40
  • $\begingroup$ So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference? $\endgroup$ – yolo Jan 6 at 21:56
  • $\begingroup$ Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy $\endgroup$ – yolo Jan 6 at 22:01
5
$\begingroup$

My favorite version of this demonstration is dropping a magnet through a conducting tube. In that case the gravitational potential energy in the system is converted, by the electromagnetic field, into thermal energy in the tube. The mechanism is Joule heating by the induced currents as the magnetic field at different parts of the tube changes. The terminal velocity decreases if the conductivity of the tube is improved; a superconducting tube would prevent the magnet from falling altogether, such is another well-known demonstration.

In your case, with a fixed magnet and a falling metal ball, the mechanism is the same: eddy currents in the metal cause resistive heating, for which the energy comes from the electromagnetic field. But the geometry there makes computations trickier.

$\endgroup$
  • $\begingroup$ What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball. $\endgroup$ – cms Jan 6 at 15:04
  • 1
    $\begingroup$ Oh, true. I'll see if the question gets clarified. $\endgroup$ – rob Jan 6 at 15:05
4
$\begingroup$

The magnetic force acting on the ball as a whole (ponderomotive force due to magnet) slows down the fall, thus does negative work on the ball. This means energy is taken away from the mechanical energy of the system (ball+Earth). Since the force is due to electromagnetic field, the only obvious destination where it goes is the electromagnetic energy of the system (ball+magnet). The electromagnetic field is everywhere, and due to descent of the ball, it changes everywhere. In some places, the EM energy density increases, in some other places, it decreases. If there are no losses of energy to heat, net EM energy has to increase by the same amount that mechanical energy decreased.

$\endgroup$
4
$\begingroup$

The magnet does negative work on the ball. The energy becomes internal energy in the magnet/ball.

You might have heard that magnetic forces do no work. However, this is only true for magnetic fields acting on classical particles. The permanent magnetic dipoles in the atoms of the magnet and ball, which are responsible for the force between them, can only be explained in quantum mechanics, as shown by the Bohr--van Leeuwen theorem, and in the quantum case work can be performed. The end result is that the quantum energy levels of the ball and magnet will be higher than they were before. This internal energy may then be manifested as heat.

$\endgroup$
  • $\begingroup$ Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory. $\endgroup$ – Ján Lalinský Jan 9 at 15:48
3
$\begingroup$

Magnetic force makes work, for example, in electromotors, in electrogenertors, etc. There are many devices that use magnetic force to do some useful work. The energy conservation is applicable to the sum of all different forms of energy of the system.

$\endgroup$
  • $\begingroup$ I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = \vec F\cdot dx = dt\vec F\cdot\vec v$ is exactly zero for the magnetic force $q\vec v\times\vec B$ due to a magnetic field. A little more explanation would be valuable. $\endgroup$ – rob Jan 6 at 15:02
  • 2
    $\begingroup$ @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them). $\endgroup$ – Vladimir Kalitvianski Jan 6 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.