0
$\begingroup$

Given that the loop of wire is falling down and the magnetic flux is changing, a current is induced in the counter-clockwise direction.

When calculating the force on the wire, $ Ids \times B $, it points in the $-\hat{j} $ for the upper part of the wire. And $\hat{i} $ and $-\hat{i} $ for the left and right side respectively and they cancel each other. The bottom side is out of the magnetic field so there is no force in there.

When it comes to determining the work done by the magnetic force in the falling loop my rationale was that both forces on the sides are perpendicular to the displacement of the loop in the $\hat{j} $ direction so they do no work, and the upper force is opposite to the displacement of the loop so the work done is negative.

However, the correct answer, apparently, is that magnetic forces do not do work as they are always perpendicular to the displacement. I understand that magnetic forces do not work on individual charged particles.

However, this contrasts with my "macroscopic" argument of the wire being "pulled up" while the displacement is down, i.e. negative work.

Where's my understanding wrong? Does the magnetic force do work or not?

enter image description here

$\endgroup$

3 Answers 3

2
$\begingroup$

Your understanding is correct, macroscopic magnetic force (acting on a current-carrying conductor in external magnetic field) in general can do work on the conductor. It does positive work when the conductor moves in direction of the magnetic force, and negative work when the conductor moves in direction opposite to the magnetic force. It's just like work of any other macroscopic force.

The statement "magnetic force does not work" is correct primarily in the special case where the "magnetic force" is actually the magnetic component of the Lorentz force acting on a moving charged point particle in vacuum, which is often expressed as

$$ \mathbf F_{magnetic~part~of~the~Lorentz~force} = q\mathbf v \times \mathbf B , $$

where $q$ is electric charge of the particle, $\mathbf v$ its velocity and $\mathbf B$ is external magnetic field.

This magnetic component of the Lorentz force on a single charged point particle in vacuum is always perpendicular to velocity of the particle, so the work done by it is always zero.

However, in macroscopic electromagnetic theory, "magnetic force on conductor" refers to ponderomotive force (force acting on heavy mass, not on mobile charge carriers), sometimes also called "motor-action" force, or Laplace or Ampere force in French sources. The simplest familiar case of this is magnetic force acting on straight current-carrying wire in external magnetic field. Its magnitude is

$$ \mathbf F_{macroscopic~magnetic} = I\mathbf L \times \mathbf B $$ here $I$ is electric current and $\mathbf L$ is vector whose magnitude is length of the wire and direction is that of the current. This formula depends on electric current, but it does not depend on velocity of the conductor; the latter is assumed to be zero or low enough so that it does not matter (if the conductor moves with very high velocity, this may affect current $I$, but this is usually neglected).

This force acts on the whole body of the conductor and because the conductor moves, in general, with different velocity than mobile charge carriers do, this macroscopic force is not in general perpendicular to velocity of the conductor. So work of this force is not in general zero and the conclusion about zero work from the simple example with single particle above does not apply. We are simply dealing with different kind of "magnetic force" here. There are other similar examples with permanent magnets, electromagnets etc. Magnetic force they exert on other bodies can do work on them. The most useful familiar case is the electric motor - inside, macroscopic magnetic forces do work on the rotor (because it has either moving conductors, or moving magnets).

In your example, magnetic force does negative work on the circuit as it moves down, and thus extracts energy from it. This energy goes to magnetic energy of induced current and some small amount gets radiated out, and then later the magnetic energy dissipates into heat and further small amount of radiation as the current decays to zero.

$\endgroup$
11
  • $\begingroup$ Thanks for your answer. I'm glad you mention the energy case because the provided answer also mentions that is gravity the one that does work and this energy gets dissipated as heat in the wire, with the magnetic field having no influence. How is this compatible with your last paragraph? $\endgroup$
    – Jon
    Commented Oct 18, 2021 at 7:39
  • $\begingroup$ Both gravity force and magnetic force do work here, but their signs are opposite. In the simplest case when the conductor has zero resistance and motion is naturally very slow and its acceleration is negligible (quasistatic motion, happens when self-inductance of the circuit is very high), gravity force does positive work on the conductor, and magnetic force (acting on the top horizontal wire) does equal negative work. As a result, the conductor acquires zero or negligible kinetic energy as it moves downwards, and all the initial potential energy is converted into magnetic energy. $\endgroup$ Commented Oct 18, 2021 at 11:06
  • $\begingroup$ In practice for such simple rectangular circuit, self-inductance is not high enough to generate high enough current to get magnetic force as strong as gravity force, so the circuit will accelerate down quickly and substantial part of potential energy will transform into kinetic energy. In that case, work of magnetic force will still be negative and it will slow down the motion, but its magnitude will be lower than magnitude of work done by gravity force. $\endgroup$ Commented Oct 18, 2021 at 11:13
  • 1
    $\begingroup$ Yes, rate of work microscopic magnetic force (acting on point charged particle) is always zero, but rate of macroscopic magnetic force (acting on the current-carrying conductor as a whole) is not, as in general it does work. The rate of work being done by macroscopic magnetic force in your example is negative, because direction of that force is opposite to velocity of the conductor it acts on. So strictly speaking the provided answer is incorrect about "magnetic force on current-carrying wire", it is only correct about "microscopic magnetic force on single charge particle". $\endgroup$ Commented Oct 19, 2021 at 21:09
  • 1
    $\begingroup$ Magnetic energy is turned into internal energy ("heat") because of negative work of resistance forces of conductor acting against the electric current and against the induced electric field. This "friction" force initially causes the current to start decreasing in time. Current changing in time is associated with induced electric field, which acts back also on the current itself; it slows down decrease of this current. Energy is being extracted from magnetic energy and put into kinetic energy of mobile charges (which as a result decreases slower) and most of this energy ends up as heat. $\endgroup$ Commented Oct 19, 2021 at 21:17
1
$\begingroup$

To start with it might be easier to look at the Earth, loop and magnetic field system and also assume that the mobile charges are positive.

If the loop has a break in it then there is an induced motional emf but no induced current.
This means that the loop accelerates under the influence of the gravitational field and gravitation potential energy is converted into the kinetic energy of the loop.
The $q\,\vec v_{\rm down} \times \vec B$ force to the left on the mobile charges in the loop is exactly balanced by the static electric field force $q \vec E$ to the right because of the accumulation of positive charges on the left and the reduction of positive charges on the right. A motional emf is set up in the loop.

Now assume that the loop is conducting and for ease of explanation, assume that the the downward velocity of the loop $\vec v_{\rm down}$ is constant.
The mobile charge carriers move in an anticlockwise direction and so there is an induced current flowing. In energy terms what is happens?
The rate at which the system loses gravitation potential energy is equal to the rate of heat dissipation (Ohmic heating) in the loop.
Note that the interaction between the moving charges and the magnetic field is the agency which enables energy to be transferred from one form (gravitational potential energy) to another form (heat).

The difference between the two cases that I have described is that in the second case the mobile charge carriers are moving around the loop and work is required for the mobile charge carriers to do that and that work is done as a result of the work done by the gravitational field not the magnetic field.

In the first case and treating the loop alone as the system the only external force acting on the loop is the downward gravitational force which accelerates the loop downwards and the work done by the gravitational force increases the kinetic energy of the loop.
In the second case and again treating the loop alone as the system there are now two external vertical forces acting on the loop, the gravitational force downwards $mg$, and an equal magnitude upward force due to the interaction of the current in the loop and the magnetic field $BIL$.
With a net force of zero on the loop, the loop moves with a constant velocity and thus has constant kinetic energy.

So what is the mechanism for the conversion of gravitational potential energy into heat?
Going back to the loop, magnetic field and Earth system, the downward motion of the wire induces an emf in the loop.
In turn that induced emf exerts a force on the mobile charge carriers which results in the induced current.
However, now note that the direction of the force on the mobile charge carriers due to the motional emf and the direction of travel of the mobile charge carriers is the same so the force does work in driving the charge carriers around the loop.
The mobile charge carriers interact (collide) with the lattice ions which make up the loop making the lattice ions vibrate more which on a macroscopic scale means that the loop is being heated.


System - loop

Apply $\vec F = m\vec a$ with acceleration zero.

Practically this would be very difficult to achieve.

$mg\hat j + BIL (-\hat j) = 0 \Rightarrow mg\hat j\cdot v_{\rm T}\hat j + BIL (-\hat j)\cdot v_{\rm T}\hat j = 0 \Rightarrow mgv_{\rm T} - BILv_{\rm T} = 0$

Which in words is, the rate of work done by the gravitational force minus the rate of work done by the magnetic force is zero.

The induced emf is $\mathcal E = BLv_{\rm T}$ and so the induced current is $I = \dfrac{BLv_{\rm T}}{R} \Rightarrow v_{\rm T} = \dfrac{RI}{BL}$

$mgv_{\rm T} = BILv_{\rm T} = I^2 R$

So the rate of work done by the gravitational force is equal to the rate of heat dissipation in the loop.

Practically one could pull a loop in the horizontal plane through a vertical magnetic field and then the work done by you would become heat in the loop.

$\endgroup$
3
  • $\begingroup$ Thanks! Your answer makes sense in correlation to what the exercise tries to show and also what Griffith writes in "Introduction to Electrodynamics" about the magnetic force being the mechanism to transfer energy, but not doing work on its own. However, Jan's answer seems to indicate that the magnetic force is actually doing work. Could you point to the inaccuracies (if any) in their answer? As yours and theirs contradict each other. $\endgroup$
    – Jon
    Commented Oct 18, 2021 at 17:56
  • $\begingroup$ I wrote, "In the second case and again treating the loop alone as the system there are now two external vertical forces acting on the loop, the gravitational force downwards mg, and an equal magnitude upward force due to the interaction of the current in the loop and the magnetic field BIL." so there is no contradiction between the two answers. I will add a bit more to my answer which assumes that the velocity of the falling loop is constant. $\endgroup$
    – Farcher
    Commented Oct 19, 2021 at 10:43
  • 1
    $\begingroup$ BTW Could you point to the inaccuracies (if any) in their answer? From my experience this is very unlikely to be so as @JánLalinský has an extremely far reaching knowledge and understanding of Physics. $\endgroup$
    – Farcher
    Commented Oct 19, 2021 at 11:06
0
$\begingroup$

The idea that the $B$ field does no work comes from Poynting’s theorem where the work done on matter is given by $E\cdot J$. This term includes all the work done on matter, including both electrical work and mechanical work. But the $B$ field and the $E$ field are interrelated, so it is possible to expand this term and separate out the mechanical work.

If a piece of material is moving at velocity $ v\ll c $ then we can make the following transformations:

$ E' = E + v \times B $

$ J' = J - \rho v $

where the primes indicate values in the reference frame where the material is at rest.

Substituting those in we have:

$ E \cdot J = (E' - v \times B) \cdot (J' + \rho v) $

$ =E' \cdot J' + E' \cdot \rho v - (v \times B) \cdot J' - (v \times B) \cdot \rho v $

$ =E' \cdot J' + \rho v \cdot (E+ v \times B) - (J-\rho v) \cdot (v \times B) $

$ =E' \cdot J' + \rho v \cdot E - J \cdot (v \times B) $

So finally we end with

$ E \cdot J=E' \cdot J' + v \cdot (\rho E + J \times B) $

Since $E'$ and $ J'$ are in the reference frame of the material, I would interpret the term $E' \cdot J’ $ as being Ohmic losses or other non-mechanical forms of work. Then $E \cdot J $ contains not only the Ohmic losses, but also the material velocity times the Lorentz force density, which I would call the Lorentz power density.

I don't have a reference for this, so I don't know if someone else has already done this derivation, but I thought you might like to see it. It indicates that even with Poynting’s theorem you can conclude that magnetic fields can do work. If you dig a bit deeper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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