In this video, our topic is the motion of straight conductors in uniform magnetic fields. We’re going to discover that, in situations like this, where a straight conductor is in constant motion through a uniform magnetic field, an emf, an electromotive force, can be established across the conductor. Let’s consider the parts involved in making this happen.
First, we have a piece of conducting material. This could be a segment of wire or a metal bar, any straight piece of material that easily conducts electricity. Say then that we surround this conductor with a uniform magnetic field. We can call this field 𝐵. And by each one of these small x’s, we can tell that this magnetic field is directed into the screen.
And say that we then start to move our conductor at a constant speed through this field. It may seem like this wouldn’t cause anything to happen. But actually, this movement through this magnetic field creates a force that acts on the charges in the conductor. We can recall that, in general, when we have an electric charge 𝑞 moving with a speed 𝑣 perpendicular to a magnetic field 𝐵, then that charge will experience a force due to this movement.
So thinking about our conductor, if we assume that this is an electrically neutral object, that means it has the same number of positive charges as negative charges. But because it is a conductor, that means that some of those negative charges, some of the electrons, are fairly mobile. That is, they’re easily able to move throughout this conductor. We could say that these minus signs here represent just a few of those electrons. And what we’re saying is, as our conductor moves along through this field, these electrons will experience a magnetic force. But then the question becomes, in what direction does that force act?
To figure this out, we can use what’s called a right-hand rule. The name for this rule comes from the fact that we’ll use our right hand to determine this direction. The way we do it is by considering the directions involved in this equation for force.
The first step is to figure out which way 𝑞 times 𝑣 points. Now, we say it points a certain direction because 𝑣 technically is a velocity vector. And in our diagram of our conductor moving through the magnetic field, we can see that this vector points to the right as we’re looking at it. So that’s the way 𝑣 points. But what we want is the direction of 𝑞 times 𝑣. That’s an important distinction because 𝑞 in our case is negative. Recall that we’re talking about the force on electrons.
So if we take a direction that’s originally pointed to the right and then multiply that by a negative number, the charge of our electron, that shows us that the overall direction of 𝑞 times 𝑣 is not to the right, but instead to the left. Knowing that 𝑞 times 𝑣 points this way, we then arrange our right hand so that our four fingers point in that direction. And then our next step is to consider the direction of 𝐵, the magnetic field.
We’ve said that in this scenario that field points into the screen. So then what we do with our right hand is curl our fingers so they point in that direction. The last thing we do is then point our thumb perpendicular to both these directions our fingers have taken. And when we do that, our thumb is pointing in the direction of the force that would act on these charges.
So we’ve answered our question of which way the force acting on these electrons will point. It will point down toward the lower end of our conductor. And because these charges are mobile, easily able to move all throughout our conductor, they’ll pile up at that end.
What about the force though on the conductor’s positive charges? Well, the nice thing about solving for the force direction on negative charges is that if the sign of our charge changes from negative to positive, then that means the force direction on that charge is simply opposite. So in our case, the magnetic force acting on the positive charges in our conductor would act up.
By itself though, this fact isn’t very important because, unlike these mobile negative charges, the positive charges at our conductor are fairly fixed in place. Despite this, a relative positive charge will develop at the upper end of our conductor. The reason is because the negative charges that used to be here to balance these positive charges out have all been driven downward by the magnetic force. So even though these positive charges have been here all along, now it looks like all these positive charges have accumulated here. Because, in fact, the negative charges have gone away.
So then we have a separation of charges across the ends of our conductor. And if we were to, say, measure the electric potential here in the middle of our positive charges and compare that value to the electric potential here in the middle of our pile of negatives, we would find that there is a difference. And wherever a potential difference is set up, that means that there’s a capacity to move charge.
Now, sometimes a potential difference is also called a voltage. Or it can be called an electromotive force, or emf for short. emf is typically symbolized using the Greek letter 𝜀. So we can say then that an emf is generated across this conductor simply by moving it along at some constant speed through a uniform magnetic field. And in fact, the amount of emf generated depends on these physical parameters.
If we take the speed with which our conductor is moving and we multiply it by the strength of the magnetic field the conductor moves through, and then if we multiply that by the overall length of the conductor, what we can call 𝑙, then that product is equal to the emf generated across the conductor. We see then that there are different ways to increase this value of emf. One way is just to make the conductor longer. Or we could make the magnetic field stronger. Or we could move the conductor more quickly through the field.
Now, by establishing an emf across this conductor, essentially what we’ve created is a battery. We have something with the energy to move charge through a circuit. So imagine that as our bar moves along, we did connect both ends with a conducting wire. And let’s say that all the resistance of this wire, as well as the bar it’s attached to, is summarized by this resistance value we’ll call capital 𝑅.
Well, like we’ve said, our conductor has become a battery. And that means it will tend to make charge flow through this wire. We can see that electrons would move away from the negative end of our conductor. And then they would move toward the positive end. So if the flow of negative charge is clockwise through this loop, then that means the direction of conventional current, that is, the direction that positive charge would move if they did move, is opposite that, counterclockwise.
Now, say that we knew the emf induced across our conductor and we also knew this resistance value 𝑅. If that were so, we could use Ohm’s law with this emf and this resistance to solve for the current 𝐼. So there’s a connection between this emf and voltages we may have seen before. But of course, all this only comes into play if we have a closed loop. If, on the other hand, we just have a straight conductor by itself moving through a magnetic field, then an emf is created, but no current.
Now, at this point, we should say that this equation here is technically a simplified version of a more general form. That’s because, as we’ll see in a moment, the direction this conductor moves can have an effect on the emf generated. To see how that works, let’s consider another sketch.
Let’s say that now our magnetic field lines point like this. And our conductor, this blue bar right here, moves straight down as we’ve drawn it perpendicular to the direction of the magnetic field. We could see that this is the same type of motion as we saw before, where 𝑣 and 𝐵 are perpendicular to one another. And in this case, it’s true that the emf induced in the conductor is equal to 𝑣 times 𝐵 times the length of the conductor. But what if our conductor moved in a different direction? For example, what if it moved like this?
No longer is it moving perpendicular to the field, but now it’s moving at some angle to that field we could call 𝜃. We could say then that there are two components to the motion of our conducting bar. One component, this one here, is perpendicular to our magnetic field. And the other component, here, is parallel to it. It turns out that only this perpendicular component contributes to the emf generated in the conductor.
And so if we were to consider a right triangle, like this, where the hypotenuse of that triangle is the speed 𝑣 with which our bar moves, then it’s only this portion of that triangle that contributes to the emf in the conductor. And we can see that that’s equal to 𝑣 times the sin of 𝜃.
And so to write this equation in a general way, allowing for motion that’s not perpendicular to the magnetic field, we need to add in a factor of sin 𝜃. So this really is what we can use to calculate the emf in the conductor. But notice that under some circumstances, this equation can reduce to the one we saw earlier. That happens when 𝜃 is equal to 90 degrees. An angle of 90 degrees means that the conductor moves perpendicularly to the magnetic field it’s in. And when 𝜃 is 90 degrees, the sin of that angle is one.
In that special case then, the emf developed in the conductor is equal to 𝑣 times 𝐵 times 𝑙. We can see that emf attains a maximum value when 𝜃 is 90 degrees. On the other hand, let’s imagine the opposite extreme, that the motion of our conductor is entirely parallel with the magnetic field. In that case, 𝜃 is zero degrees. And we know the sin of zero degrees is zero. And so with that factor in our equation equaling zero, we know the emf will be zero too. No emf is induced if our conductor moves only parallel to the magnetic field.
Now that we’ve gotten a sense for where this equation comes from, let’s get some practice applying it through an example exercise.
A 7.2-centimeter-long conducting rod moves through a 36-millitesla uniform magnetic field, as shown in the diagram. The rod travels at 4.5 centimeters per second.
Before we get to the first part of our question, let’s consider this diagram. In it, we see a conducting rod with two ends, the top end called 𝐴 and the bottom end called 𝐵, that’s in motion at a steady speed — we can call that speed 𝑣 — through a uniform magnetic field. Now, that speed is 4.5 centimeters per second. And the field that the conducting rod moves through, we can see, is directed into the screen. Let’s call the magnitude of this field 𝐵. And we’re told that it’s equal to 36 millitesla or, in other words, 36 times 10 to the negative third tesla.
Along with all this, we’re told the length of this conducting rod. We’ll call that length 𝑙. And we can see it’s given as 7.2 centimeters. So our conducting rod of this length is moving with this speed through a magnetic field of this strength. Knowing this, let’s now look at the first part of our question.
What is the magnitude of the potential difference across the rod?
Well, it may surprise us that there’s any potential difference, but indeed there is. We know this based on the fact that the emf, or the potential difference, induced in a conducting rod is equal to the rod’s length times the strength of the magnetic field it moves in times its speed, so long as the rod is moving perpendicular to the magnetic field. We see in this case that that is indeed happening, that the rod is moving to the right, while the magnetic field is pointed 90 degrees to that into the screen.
So to calculate the potential difference across the rod, we’ll use the length, magnetic field, and speed values given to us. The rod’s length is 7.2 centimeters. The magnetic field strength is 36 times 10 to the negative third tesla. And it’s moving with a speed of 4.5 centimeters every second.
Before we calculate this potential difference though, let’s convert the distances in these two values from centimeters into meters. If we do that, then we’ll have all of the units in this whole expression in SI base units. Recalling that 100 centimeters is equal to one meter, in both of these cases, we’re gonna be shifting decimal points two spots to the left. Expressed this way, the length of our rod is 0.072 meters. And it’s moving with a speed of 0.045 meters per second.
When we multiply these three quantities together, to two significant figures, we find a result of 1.2 times 10 to the negative fourth volts. That’s the magnitude of the potential difference across the rod. Now, let’s look at the second part of our question.
This part asks, which end of the rod has a higher potential?
Looking at the rod, we see it has these two ends, end 𝐴 and end 𝐵. And as we think about how to answer this question, we can recall that positive electrical charges, by definition, have a higher electrical potential than negative electric charges do. This means that another way to answer this question of which end has higher potential is to figure out which end of the rod has more positive electrical charge.
Now, if we assume that our conducting rod is an electrically neutral object, then we would say it has the same number of positive charges as negative charges. In order for one end of the rod to have a higher potential than the other, there needs to be some sort of charge separation. That is, something has to cause one type of electrical charge to move to one end of the rod and the other type to move to the opposite end.
And indeed, there is such a force involved here. The magnitude of that force is given by this expression here. It says that if we have a charge 𝑞 moving with a speed 𝑣 perpendicularly to a magnetic field of strength 𝐵, then the force magnitude that charge experiences is equal to 𝑞 times 𝑣 times 𝐵.
However, in our case, it’s not the magnitude we’re interested in exactly, but rather the direction. We want to know which way charges are pushed in this conducting rod. Because this rod is a conducting material, that means that all throughout it has mobile electrons. Unlike the positive charges in this conducting rod, these electrons are free to move all throughout the conductor. And they do so in response to forces, specifically this magnetic force.
To figure out the direction in which this force acts on negative charges in our conductor, we can recall a right-hand rule for determining this. The first step involved is to figure out the direction of 𝑞 times 𝑣, where 𝑣 we take to be the velocity vector of our rod and 𝑞 is the value of our charge of interest, in our case an electron.
Now, we’re only concerned about the direction of 𝑞 times 𝑣. So if we look at our diagram, we can see that 𝑣 points to the right. That’s the way the rod overall is moving. But this isn’t the direction of 𝑞 times 𝑣 because recall that 𝑞 is the charge of an electron and, therefore, a negative value. This means that 𝑞 times 𝑣 instead of pointing to the right will actually point the opposite way, to the left.
And now that we know this, we take our right hand and we point our fingers in the direction of 𝑞 times 𝑣. As we said, that’s to the left. Our next step is to figure out the direction of the magnetic field 𝐵. That’s a bit easier. We know from our diagram that 𝐵 points into the screen. So now what we do is we curl the fingers on our right hand in that direction. Once we’ve done that, pointed our fingers into the screen, we then point our thumb perpendicular to both of the directions that our fingers have pointed. And when we do that, our thumb points in the direction of the force that acts on these charges, specifically the negative charges, the electrons, in our conducting rod.
So then as our rod moves along, the mobile negative charges in it will experience a force towards the lower end. And because they are mobile, they’ll actually collect down there at end 𝐵. This means that at the opposite end of the rod, end 𝐴, all the negative charges that left will leave an abundance of positive charge. And now that we have charge separation across the ends of our conductor, we can answer this question of which end of the rod has a higher potential. Because positive charges have a higher electrical potential than negative ones, we can say the end with the most positive charge has the higher electrical potential. And that, as we see, is end 𝐴.
Let’s summarize now what we’ve learned in this lesson. We learned first that a straight conductor moving in a uniform magnetic field may experience an emf. In general, this emf is given by the product of the length of the conductor times the strength of the magnetic field it’s in times its speed multiplied by the sine of the angle between the magnetic field lines and the direction the conductor is moving. We saw that when the conductor moves perpendicularly to the field, that is, 𝜃 is 90 degrees, then the emf induced in the conductor is a maximum value. On the other hand, if the conductor moves parallel to the field, then 𝜃 equals zero degrees and the emf induced is zero.
We also saw that the direction the force on charges in a moving conductor acts is given by what’s called a right-hand rule. And lastly, we saw that when our moving conductor is part of a closed circuit, the emf it generates can drive current in that circuit, where the total circuit resistance is given by the value 𝑅. This is a summary of the motion of straight conductors in uniform magnetic fields.