A Bit on Work: Part II of III

This entry logically follows the preceding entry, so you should read that one first unless you already have a good understanding of the physicist's concept of work.

In that preceding entry we spoke of a person pushing a box up an incline, such that the person's arms were always horizontal or parallel to the ground (and not the incline). That is, the force imparted by the person on the box was entirely horizontal. To find the work done by the person on the box, we had to find the component of the force in the direction of the box's motion. That is, we had to break up (mathematically) the force into a component parallel to the incline and a component perpendicular to the incline, and then we took the component parallel to the incline and multiplied that by the distance up the incline that the box moved. We also reasoned that we could just as easily multiply the total horizontal force applied by the person by the horizontal displacement of the box and arrive at the same answer.

But if we take the second approach for calculating the work done on the box, how do we explain the increase in the vertical position of the box? Was work done on the box in moving it higher above the ground? And by what force? (Really, we face the same question in the first approach to calculating work, but it's easier to conceptualize using the second approach.) Indeed, work was done, and it was done by the person. But it was not the force imparted by the person that moved the box higher. Here, the normal force (if you don't know what the normal force is, you should look it up before continuing), while doing no work itself, redirects the efforts of the person from horizontal to vertical. We say the normal force does no work, because the normal force is always perpendicular to the incline and the box never moves off the inclined plane. But the normal force does have a vertical component that lifts the box by redirecting some of the work done by the person. This is a bit tricky, but understanding it will come in handy later.

The question at the end of the last entry was: What force never does any work? And here is the answer: the magnetic force. There is a law, called the Lorentz force law, which tells us how to compute the magnetic force on a charged particle. It says to take the value of the charge (say, the charge associated with a single electron) and to multiply that value by a vector that is perpendicular to both the magnetic field at the location of the particle and to the velocity vector describing the instantaneous motion of the particle. The magnitude of this mutually-perpendicular vector is equal to the magnitude of the velocity times the magnitude of the magnetic field times sine of the angle between the two vectors. This long description can be written simply in mathematical notation: F = q (v x B), where q represents charge and the "x" signifies the cross product between the velocity v and the magnetic field B. (F, v, and B are all vectors, here.) If you're not familiar with a "cross product," it's more or less explained in the wordy description above, so just digest that and ignore the formula.

In summary, the magnetic force on a charged particle is perpendicular to the particle's direction of motion (as well as the direction of the magnetic field). In other words, no component of the force is ever in the same direction as that in which the particle is moving, and therefore no work can ever be done by this force. (This makes perfect sense if you remember the definition of work, which was stated in the previous blog entry.) As an example, if you had an electron (somehow made visible) that was traveling from the front towards the back of your desk, and then you turned on a magnetic field that was uniform and pointing in the direction of "down", i.e. it is perpendicular to the desk and points down right through the table top, then the electron would immediately get tugged towards the left (while continuing to move forward) and fall into a circular path. Note that as soon as the electron moves a bit forward and to the left along this circular path, the force is not longer exclusively left but is now left and a bit down. I don't know if I've explained this clearly or not, so I'll continue a bit and then move on. As an analogy, picture yourself whirling a ball tied to a string around your head. You hold on to one end of the string and raise your arm and get the ball swinging about in a circular path. The force here, exerted by the string on the ball, like in the case of the magnetic field acting on the electron, is always perpendicular to the direction of motion of the ball. It's a centripetal force, in other words. It does alter the direction of the electron, but it certainly doesn't do work on it. The math here is straightforward, but the idea can seem hard to swallow in certain physical examples.

Let's say I take a magnet from the refrigerator and place it over a paper clip lying on my desk. The paper clip "jumps" to the magnet and sticks to it. Is it really true that the magnet's magnetic force did no work in moving the paper clip the few centimeters from the desk top to the magnet in my hand? Yes, because magnetic forces never do work. So then what did the work? What we're going to find is that, like the normal force mentioned earlier, the magnetic force redirects work done by another agent. But what is this other agent and how does the magnetic force redirect its work?

I'm now going to attempt an explanation of what it is that actually does the work in this example. Now, no good teacher would ever introduce a concept and then quickly jump to a difficult and confusing example involving that concept. They would cover some easy examples first and work their way up to a non-intuitive, challenging problem. But I want to jump right into the difficult explanation as to how the paper clip is pulled upwards towards the magnet. I'll begin this explanation in the next blog entry because this one is long enough.