Question 5.3: A rectangular loop of wire, supporting a mass m, hangs verti...

First of all, the current must circulate clockwise, in order for (I × B) in the horizontal segment to point upward. The force is

where a is the width of the loop. (The magnetic forces on the two vertical segments cancel.) For F_{mag} to balance the weight (mg), we must therefore have

I=\frac{mg}{Ba}.                   (5.18)

The weight just hangs there, suspended in mid-air!
What happens if we now increase the current? Then the upward magnetic force exceeds the downward force of gravity, and the loop rises, lifting the weight. Somebody’s doing work, and it sure looks as though the magnetic force is responsible. Indeed, one is tempted to write

W_{mag} = F_{mag}h = I Bah,                  (5.19)

where h is the distance the loop rises. But we know that magnetic forces never do work. What’s going on here?
Well, when the loop starts to rise, the charges in the wire are no longer moving horizontally—their velocity now acquires an upward component u, the speed of the loop (Fig. 5.11), in addition to the horizontal component ω associated with the current (I = λω). The magnetic force, which is always perpendicular to the velocity, no longer points straight up, but tilts back. It is perpendicular to the net displacement of the charge (which is in the direction of v), and therefore it does no work on q. It does have a vertical component (qωB); indeed, the net vertical force on all the charge (λa) in the upper segment of the loop is

F_{vert} = λa\omega B = I Ba             (5.20)

(as before); but now it also has a horizontal component (quB), which opposes the flow of current. Whoever is in charge of maintaining that current, therefore, must now push those charges along, against the backward component of the magnetic force.

The total horizontal force on the top segment is

F_{horiz} = λauB.                            (5.21)

In a time dt, the charges move a (horizontal) distance ω dt, so the work done by this agency (presumably a battery or a generator) is

which is precisely what we naïvely attributed to the magnetic force in Eq. 5.19. Was work done in this process? Absolutely! Who did it? The battery! What, then, was the role of the magnetic force? Well, it redirected the horizontal force of the battery into the vertical motion of the loop and the weight.^{7}

It may help to consider a mechanical analogy. Imagine you’re sliding a trunk up a frictionless ramp, by pushing on it horizontally with a mop (Fig. 5.12). The normal force (N) does no work, because it is perpendicular to the displacement. But it does have a vertical component (which in fact is what lifts the trunk), and a (backward) horizontal component (which you have to overcome by pushing on the mop). Who is doing the work here? You are, obviously—and yet your force (which is purely horizontal) is not (at least, not directly) what lifts the box. The normal force plays the same passive (but crucial) role as the magnetic force in Ex. 5.3: while doing no work itself, it redirects the efforts of the active agent (you, or the battery, as the case may be), from horizontal to vertical.

^7If you like, the vertical component of F_{mag} does work lifting the car, but the horizontal component does equal negative work opposing the current. However you look at it, the net work done by the
magnetic force is zero.