Video Transcript
The ends of a conducting rod are connected to a loop of conducting wire, as shown in the diagram. The rod enters a region that contains two oppositely directed uniform magnetic fields of equal magnitudes, where exactly half of the length of the rod is in each of the fields. The rod moves perpendicularly to both of the fields. The conducting wire does not enter either of the fields. The rod is two centimeters long and moves at one centimeter per second, and the magnetic fields are both 20 milliteslas in strength. The total resistance of the rod and the wire is 0.5 ohms. What is the potential difference between the ends of the rod as it moves through the fields?
Here in our diagram, we see this conducting rod moving perpendicularly through two uniform magnetic fields. The top half of the rod moves through a field that’s pointing out of the screen towards us. We’ll call this 𝐵 sub out. The bottom half of the rod though moves through a field pointed in the opposite direction. We’ll call this 𝐵 sub in. The magnitudes of these two magnetic fields are the same, and we’re told that that magnitude is 20 milliteslas. Since our moving rod is a conducting rod, that tells us that it contains mobile electric charges. As these charges move through the two magnetic fields, 𝐵 sub out and 𝐵 sub in, they experience a force. That force distributes the charges in a particular way across the conducting rod.
Depending on that distribution, a potential difference is then induced across the ends of the conducting rod. To understand how mobile charges in the conducting rod are pushed due to that rod’s motion through the magnetic fields, we want to understand something called a right-hand rule. In this application of the right-hand rule, we’ll learn the direction of the force that acts on a charge moving at some velocity through a magnetic field.
Let’s say that the object we’re considering has a charge 𝑞 and that the object is moving with a velocity 𝑣. If we point the fingers on our right hand in the direction of 𝑞 times 𝑣 and we do this in such a way that we can curl our fingers closed so that they point in the direction of the magnetic field, in this case, out of the screen towards us, then our right thumb pointing perpendicularly both to 𝑞 times 𝑣 and to 𝐵 will point in the direction of the force that acts on this charge 𝑞.
As we apply this right-hand rule to our scenario, let’s note that the mobile charges that are in this conducting rod are negative; they’re electrons. Those negative charges, along with the rest of the conducting rod, are moving to the right. This means that negative 𝑞 times 𝑣 points to the right, and that tells us that positive 𝑞 times 𝑣 points in the opposite direction to the left. According to our right-hand rule then, we point the fingers of our right hand in that direction. Then, we curl our fingers in the direction of the magnetic field. We see though that that field direction depends on whether we’re considering 𝐵 sub out or 𝐵 sub in. In this case, let’s consider just the top half of our conducting rod, the one that moves through the magnetic field pointing out of the screen.
Curling the fingers of our right hand so they point out of the screen, we note that our thumb points straight up. This shows us the direction of the magnetic force on our mobile charges, our negative charges. Recall that here we’re just talking about the top half of our conducting rod. In this top half, the force on negative charges is upward, so negative charges will accumulate up towards the top of the rod. And this means there will be a majority of positive charge at the bottom of this top half of our conducting rod. This separation of charge will create an electric potential difference between the ends of this half of the rod.
Note that in our question though, we’re considering the potential difference established between the ends of the entire conducting rod. This means we also need to account for the effects on the bottom half of that rod. Let’s use the right-hand rule once again to understand the charge distribution on this part. Just like before, the negative mobile charges are moving to the right. And this means that positive 𝑞 times 𝑣 — that is, a positive charge times the velocity of the rod — is to the left. Now, our next step will be different than it was for the top half of the conducting rod because in this case our magnetic field points in the opposite direction.
Basically, we’ll want to arrange our right hand so that the fingers can point to the left but that those fingers can curl so that they point into rather than out of the screen. We can do this by orienting our right hand this way. Now, our palm faces into the screen so that while our fingers do point in the direction of 𝑞 times 𝑣, when we curl them in the direction of 𝐵, our thumb now points downward. This is the force direction on negative charges in the bottom half of our conducting rod. Those negative charges then accumulate down here. And that means a majority of positive charge collects at the top of this bottom half of our rod.
Once again, we have charge separation that leads to a potential difference across this half of the rod. Notice that in the two halves of our rod, the polarities of these potential differences are opposite. They have the same magnitude, but they point in opposite directions. If we were to relate this to a circuit diagram, this is like having a circuit with a cell that’s oriented one way and putting immediately after it another cell that’s oriented oppositely. If these cells both produced a potential difference of the same magnitude, then the fact that they point in opposite directions means that their overall effect would be cancelled out. The total potential difference across both cells together would be zero. This is exactly what we see here in our conducting rod.
Even though there is some nonzero potential difference across each half of the rod, those potential differences cancel one another out. When we consider the rod from one end to the other then, the total potential difference across it is simply zero volts. This is due to the fact that half of the rod moves through a magnetic field pointing in one direction. And the other half moves through a magnetic field pointing the opposite way. Overall, zero volts of potential difference are established between the ends of the rod.
Let’s look now at part two of our question.
What is the current through the wire as the rod moves through the fields?
We’ve seen that because of this rod’s motion, it functions essentially as a pair of cells. Since each end of the rod is connected to a common wire, a closed electrical circuit is formed. This means current can exist in the circuit, and it will do so according to Ohm’s law. This law says that the potential difference across a circuit equals the current in the circuit multiplied by the circuit’s overall resistance. By dividing both sides by the resistance 𝑅, we can rearrange this equation so that 𝐼 is the subject.
In our problem statement, we’re told that the total resistance of our circuit, we’ll call it 𝑅, is 0.5 ohms. And this means the total current in our circuit equals the potential difference across the circuit divided by 0.5 ohms. From our answer to part one though, we know what that potential difference is. Zero volts of potential difference are established across the moving conducting rod. Since zero is in the numerator of this fraction, the overall result is zero amperes. There is no current in the wire as the rod moves through the fields.
Let’s now clear some space and look at the last part of our question.
What is the magnitude of the potential difference between either end of the rod and the center of the rod as the rod moves through the fields? Give your answer in scientific notation to one decimal place.
Earlier, we saw that if we consider just one-half of the conducting rod, either the top or the bottom half, a potential difference is established across it. In this part of our question, we want to solve for the magnitude of that potential difference. And indeed, whether we consider the top half or the bottom half, the magnitude of that potential difference is the same. To figure out what that is, let’s recall the mathematical relationship for the EMF, also called the potential difference, of a conducting rod of length 𝑙 moving with a velocity 𝑣 through a magnetic field of strength 𝐵.
In this equation, the angle 𝜃 is equal to the angle between the velocity vector of the rod and the magnetic field the rod moves through. In the case of our conducting rod, both the top and the bottom half move to the right, that’s the direction of their velocity 𝑣. But then, the top half moves through a magnetic field pointing out of the screen, while the bottom half moves through a magnetic field pointing into it. The important point though is that in both cases, the angle between the velocity of our rod and the magnetic field that it moves through is 90 degrees. Therefore, whether we consider the top or the bottom of the rod, 𝜃 equals 90 degrees. And because of that, the sin of 𝜃 in our case is equal to one.
This means that when we write the equation for the EMF, or potential difference, established across one-half of our rod, we can use this simplified form, where that potential difference equals the length of the half of the rod multiplied by its speed or the magnitude of the velocity times the strength of the magnetic field the rod moves through. Let’s recall from our problem statement that the total length of our rod is two centimeters. Therefore, half the length of the rod, the length we’re using as 𝑙 in this equation, is one centimeter. We can then further recall that the speed 𝑣 with which a rod moves is given as one centimeter per second. Lastly, the magnitude of both magnetic fields in this example is 20 milliteslas.
Before we multiply all these values together, let’s convert the distances of centimeters into units of meters and the units of milliteslas into teslas. This will make all the units in this expression reducible to SI base units. We recall that 100 centimeters equals one meter and 1000 milliteslas equals one tesla. Therefore, to switch centimeters into meters, we’ll move the decimal point two spots to the left in both cases. That gives us 0.01 meters and 0.01 meters per second, respectively.
Then, because of the way milliteslas convert into teslas, we’ll move the decimal point in this instance three spots to the left. The potential difference across one-half of our conducting rod, either the top or the bottom, equals 0.01 meters times 0.01 meters per second times 0.020 teslas. To one decimal place in scientific notation, this equals 2.0 times 10 to the negative six volts. This is the magnitude of the potential difference established across either the top or the bottom half of our moving conducting rod.
