Hold a permanent magnet above a paper clip and the paper clip "jumps" up to the magnet. How does this happen given that the magnetic field is not doing any work on the paper clip?
First off, the paper clip is made of steel, which contains a large amount of iron. Iron is a ferromagnetic material, meaning it can become magnetized when placed in a magnetic field and remain magnetized when removed from that field. Non ferromagnetic materials, on the other hand, would lose their magnetization upon removal of the external magnetic field. (In this case, we're not removing the magnetic field, but it's still nice that we're working with a ferromagnetic material. You'll see why later.) When we place a permanent magnet above a paper clip, the magnetic field produced by the magnet induces magnetism in the paper clip by applying a torque to the magnetic dipoles in the iron, lining them up.
What's a magnetic dipole? A small current loop (say, electrons flowing around a tiny loop of wire) is, more or less, a magnetic dipole. We call the small current loop a magnetic dipole because it produces a magnetic field, at some distance, that is strictly "dipolar" in nature. Not all systems produce a magnetic field that is dipolar in nature. Some systems produce a field that is not dipolar at all, but perhaps "quadrupolar" or "octopolar." Other systems might produce fields that are largely dipolar but a little quadrupolar, too. What does it mean for the field to be strictly "dipolar" in nature? It means that, as you move away from the system, the strength of the magnetic field drops off as one over the distance cubed. There's no component of the field that drops off as one over the distance or one over the distance to the fourth power, etc. Not too many systems can actually produce a field that is strictly dipolar. It's easy to produce one that is largely dipolar, but quite difficult to produce one that is strictly dipolar. A tiny current loop does the trick, however. But it has to be really tiny, as in infinitesimally small. How do you make such a thing in the lab? You don't. Luckily, I suppose, they already exist in nature as electrons whizzing about nuclei inside of atoms. (No wires necessary.) Previously, I said that the magnetic field of a permanent magnet applies a torque to the magnetic dipoles in a sample of iron, lining them up. Now it should be fairly clear that it is atomic electrons (acting in their capacity as magnetic dipoles) that are doing the lining up. Note: not every electron in a sample of iron experiences this torque. Only the unpaired electrons do. (Each iron atom has 4 unpaired electrons.)
OK, so how can an electron point in a particular direction? Really, it can't. What I mean is that this whirrling electron, this magnetic dipole, points its magnetic dipole moment in a particular direction. This so-called dipole moment is a property of the dipole and can act to represent the physical dipole. It's a vector (with, of course, a magnitude and a direction) that quantifies the contribution of a system's internal magnetism to the external dipolar magnetic field produced by the system. That is, a measure of how much what's going on inside the system is effecting the magnetic field observed outside the system. The moment may be non-physical, but it often proves useful to picture an electron as a little vector when doing calculations or thinking through problems like the one we're addressing here. Therefore, to say that dipoles are lined up is to say that their dipole moments are lined up (or parallel to one another).
Magnets come in different strengths, which we quantify through the concept of magnetization. Something with a large magnetization is both strongly affected by external magnetic fields and the source of its own strong magnetic field. We define magnetization as the amount of magnetic dipole moment per unit volume. Therefore, given a unit volume, we perform a vector sum of all the little moments (vectors) in that volume, and we see how strong our magnet is. Two vectors of equal magnitude pointing in opposite directions sum to zero. Likewise, a large number of arbitrarily directed vectors also sums to zero. This explains why the paper clip, before being magnetized by the permanent magnet, isn't magnetic. It has plenty of little moments (or vectors) inside, but they are arbitrarily directed (well, sorta) and so the net sum of these moments, per unit volume, is pretty close to zero. Once the permanent magnet acts on the dipole moments in the paper clip and lines them up, the vector sum no longer equals zero. Rather, it is now rather large, and the paper clip now has a large magnetization and acts outwardly like a magnet.
Electrons, bound to atoms, move in two ways. This leads to two magnetic dipoles or, better put, two contributions to a single magnetic dipole. (It's a simple vector sum.) Firstly, an electron "orbits" the nucleus. Even though it's not accurate, people often picture this motion as being like a planet orbiting the sun. That's a good way to think of it at the present. Secondly, an electron "spins." You might think of this as an electron spinning about its own axis, just as the Earth spins about its axis once every 24 hours. But this is a rather horrible and misleading analogy, because this "spin" is not really a physical rotation about an axis. The electron is a point particle with no physical size, so there's really no way it could have a component off-axis that could move around some central point. For this reason, physicists say the electron has an intrinsic magnetic dipole moment that originates with its spin. It just exists.
Let's step back now and look at what we have. A permanent magnet (which itself is comprised of magnetic dipoles, with the moments all pointing in the same direction, hence its large magnetization) produces a magnetic field that exists in the space around the magnet. This space includes the paper clip, sitting on a table. The magnetic field interacts with the electrons in the iron/steel paper clip, changing the magnetic dipole moments of these electrons, and inducing magnetism in the paper clip. How exactly?
The magnetic field produced by the permanent magnet has the property, determined through experimentation, that it can exert a force on a moving charged particle, like our atomic electrons. (This force, called the Lorentz force, was defined in the previous blog entry.) Acting on each magnetic dipole, this force (actually, torque) acts to twist the dipole moments such that they line up parallel to the field. The result: countless magnetic dipole moments in our paper clip are now pointing in the direction of the field. (It is at this point we appreciate the paper clip being made of iron, a ferromagnetic material. If it was not, the magnetic force would have a more difficult time turning all of the dipole moments and would ultimately manage to turn only some of them, diminishing the strength of the magnetization induced in the paper clip.) The paper clip is now a magnet.
How else does the magnetic field interact with the atomic electrons? Surprisingly, it can change the speed with which the electrons orbit their nuclei! Before the magnetic field enters the picture, the electrons are held in their orbits by electrical forces alone. (Unlike charges, i.e. protons and electrons, attract.) When the magnetic field shows up, it produces a force that acts in the opposite direction as the electrical force, at the location of each orbiting electron, and serves to weaken the pull on the electron towards the center of the orbit. The electron no longer needs to travel so quickly to maintain its orbital radius and it slows down. Now, think back to the previous blog entry and the example of you holding on to a string, at the other end of which is attached a ball. You're whirling this ball about your head. In this case, there is a centripetal force pulling the ball towards the center of its circular path, thereby acting, at all times, perpendicular to the direction of the ball's motion. (This is like the electric force holding the electron in its orbit about the atom's nucleus.) Likewise, the magnetic force, once introduced, acts perpendicular to the circling electron. The magnetic force, however, is acting not towards the center of the circle but radially outward, away from the center. As stated before, this diminished centripetal force causes the electron to slow down. We'll soon see that this slowing of the electrons in the atoms of the paper clip is key to the lifting of the paper clip by the permanent magnet.
Now is the time to point out that the magnetic field produced by the permanent magnet is non-uniform. It generally is pointing downwards, assuming the north pole of the magnet is nearer the paper clip than the south pole, but it also flares out. It's the vertical component of the field which acts to slow down the electrons but the horizontal component, existing in the plane of the orbiting electrons, that provides an upward force. Adding together these two components we end up with a force (represented by a vector) that is pointing up and out (away from the center of the electron's circular orbit). We know this force must be perpendicular to the motion of the electron, and indeed it is, as the electron begins to move upwards along a helical path.
The motion of an electron in its orbit constitutes stored kinetic energy. It's this energy that is tapped to lift the paper clip off the table. As the paper clip (acting as a magnet) rises, the unpaired electrons inside slow down and the stored kinetic energy decreases. The magnetic force redirects this energy into lifting the paper clip/magnet against the force of gravity. The net magnetic force, like the normal force mentioned in the previous blog entry, is responsible for the vertical motion of the object (previously a box and now an electron) despite the fact that it doesn't do work on that object. Both of these forces (the magnetic and the normal) are redirecting work done by another agent. In the case of the box being pushed up the incline, the other agent is a person. And in this case, it's ... Well, who or what is this agent?
Uhhhhhh. Well, the agent is whatever got all those electrons circling around all those nuclei to begin with. Whatever it was, it did work and imparted kinetic energy to each little electron. Trying to trace the formation of these iron atoms back to the original source of energy would lead us back to the Big Bang. So it was God, I guess. "God" lifted the paper clip.
Monday, September 21, 2009
Wednesday, September 9, 2009
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.
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.
Tuesday, September 1, 2009
A Bit on Work: Part I of III
The concept of work has a specific meaning in the sciences. It is best described by stating how it is calculated, which is the force applied to an object in the direction of its motion times the distance it moves. A force is, more or less, a push or pull. Newton stated that force, mass, and acceleration are linked through the equation F = m a. (He stated it in different terms, but this is how it is normally written today.) Some prefer to write it as a = F/m, to emphasize that a force causes acceleration and not the other way around.
If I push on something (say, a piece of paper taped to a wall) and it does not move, am I applying a force to the paper? Yes, but in this case I am not the only thing applying a force. The wall is pushing on the paper just as hard as I am but in the opposite direction. The net force F on the paper is zero and both the left and right sides of the equation a = F/m are zero and all is well.
Am I doing work on the paper? I'm applying a force but there is no displacement of the paper, so the answer is no. If I manage to push the paper through the wall then indeed I have done work (and I will feel very stupid). New example. Let's say a heavy box is at rest on an incline. I push on it with my arms parallel to the ground (not the incline), and it moves a bit up the incline. Let's say I push with a force of 20 newtons. (Newtons are the SI units of force. One newton is equal to the amount of force required to give a mass of one kilogram an acceleration of one meter per second squared.) Let's also say that the box moves up the incline a meter. Is the work done equal to 20 newtons x 1 meter? No, it's not, because we are concerned not with the overall force but with the force in the direction of the object's motion. We therefore need to mathematically subdivide this overall force (exerted by myself) into a component in the direction of the object's motion (which is of interest to us) and one in the direction perpendicular to this motion (which is not of interest to us). We then multiply the force component in the direction of the object's motion (which is going to be the overall force times cosine of the angle of the incline, with respect to the ground) with 1 meter to get the amount of work done. Yes, you can also calculate work by taking the total force I exert and multiplying this by the horizontal displacement of the object, which will be less than 1 meter. By the way, the SI unit of work is the joule, which is equal to, of course, a newton times a meter.
Let's see if I can confuse you. You push a bag of flour across the kitchen counter. According to Newton's third law, the bag of flour pushes back on you with an equal (in magnitude) but oppositely directed force. Why does the bag of flour move if the forces cancel out?
If you need a hint, take another look at the equation a = F/m. The forces exerted by you and by the bag may be equal in magnitude but your masses surely are not. Therefore you should have different accelerations, which indeed you do. This problem is complicated by the force of friction, which is robbing you of your acceleration and reducing the acceleration of the bag of flour. If you could get rid of the friction, you too would accelerate backwards (but not very quickly because of your relatively large mass). Do you see the difference between this example and the paper against the wall?
I want to talk more about work and a special kind of force that, oddly enough, never does any work. There aren't many forces in nature, so perhaps you can figure out which one I'm talking about. I'll discuss it in my next blog entry. Thanks for reading.
If I push on something (say, a piece of paper taped to a wall) and it does not move, am I applying a force to the paper? Yes, but in this case I am not the only thing applying a force. The wall is pushing on the paper just as hard as I am but in the opposite direction. The net force F on the paper is zero and both the left and right sides of the equation a = F/m are zero and all is well.
Am I doing work on the paper? I'm applying a force but there is no displacement of the paper, so the answer is no. If I manage to push the paper through the wall then indeed I have done work (and I will feel very stupid). New example. Let's say a heavy box is at rest on an incline. I push on it with my arms parallel to the ground (not the incline), and it moves a bit up the incline. Let's say I push with a force of 20 newtons. (Newtons are the SI units of force. One newton is equal to the amount of force required to give a mass of one kilogram an acceleration of one meter per second squared.) Let's also say that the box moves up the incline a meter. Is the work done equal to 20 newtons x 1 meter? No, it's not, because we are concerned not with the overall force but with the force in the direction of the object's motion. We therefore need to mathematically subdivide this overall force (exerted by myself) into a component in the direction of the object's motion (which is of interest to us) and one in the direction perpendicular to this motion (which is not of interest to us). We then multiply the force component in the direction of the object's motion (which is going to be the overall force times cosine of the angle of the incline, with respect to the ground) with 1 meter to get the amount of work done. Yes, you can also calculate work by taking the total force I exert and multiplying this by the horizontal displacement of the object, which will be less than 1 meter. By the way, the SI unit of work is the joule, which is equal to, of course, a newton times a meter.
Let's see if I can confuse you. You push a bag of flour across the kitchen counter. According to Newton's third law, the bag of flour pushes back on you with an equal (in magnitude) but oppositely directed force. Why does the bag of flour move if the forces cancel out?
If you need a hint, take another look at the equation a = F/m. The forces exerted by you and by the bag may be equal in magnitude but your masses surely are not. Therefore you should have different accelerations, which indeed you do. This problem is complicated by the force of friction, which is robbing you of your acceleration and reducing the acceleration of the bag of flour. If you could get rid of the friction, you too would accelerate backwards (but not very quickly because of your relatively large mass). Do you see the difference between this example and the paper against the wall?
I want to talk more about work and a special kind of force that, oddly enough, never does any work. There aren't many forces in nature, so perhaps you can figure out which one I'm talking about. I'll discuss it in my next blog entry. Thanks for reading.
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