A 15-centimeter-long conducting rod has a potential difference across it, as shown in the diagram. The rod moves through a uniform magnetic field at 0.32 meters per second. The magnitude of the induced potential difference is 9.6 millivolts. What is the strength of the magnetic field? Which side of the field-containing region is the rod moving toward?
In our diagram, we see a conducting rod that is moving through a uniform magnetic field. We’re told that a potential difference is induced across the rod. And we can see this because of the positive and negative signs at either end. The first part of our question asks us to solve for the strength of the magnetic field that the rod moves through.
This field, we see, is pointing out of the screen towards us, and let’s label this field as B. Clearing some space on screen to work, we’re told that the length of the rod — we’ll call it 𝑙 — is 15 centimeters. The speed of the rod, which we’ll call 𝑣, is 0.32 meters per second. And the magnitude of the induced potential difference across the rod — we’ll refer to it using the Greek letter 𝜀 — is 9.6 millivolts.
Clearing a bit more space on screen, we can remember this general equation that for a conductor of length 𝑙 moving with a speed 𝑣 through a uniform magnetic field of magnitude 𝐵, the emf, or electromotive force, equivalent to potential difference induced across the rod is equal to 𝑙 times 𝐵 times 𝑣 times the sin of this angle 𝜃. Now, 𝜃 is the angle between the direction of the rod’s motion and the direction of the magnetic field.
Something interesting about our particular situation is we don’t know the direction of the rod’s motion. But we can safely assume that the rod is moving in the plane of the screen, in other words either up, down, left, or right, as we see it here. In all those cases, the angle between the direction of the rod’s motion and the direction of the magnetic field B is 90 degrees. In our situation then, 𝜃 is 90 degrees, and therefore the sin of 𝜃 is one, since the sin of 90 degrees is one.
We can therefore apply our general equation in a simplified form. The potential difference induced across the rod equals 𝑙 times 𝐵 times 𝑣 times one, or just 𝑙 times 𝐵 times 𝑣. At this point, let’s recall that it’s not the potential difference but rather the magnetic field strength 𝐵 that we want to solve for. Dividing both sides of this equation by 𝑙 times 𝑣, we see those factors cancel out on the right. The magnetic field strength 𝐵 equals the potential difference induced across the rod divided by the length of the rod times its speed. Note that we’ve been given all three of these values on the right-hand side of our expression.
Before we calculate our final result, we’ll want to make sure the units in this expression are all on the same footing. Right now, in our numerator, we have units of millivolts and in our denominator units of centimeters. We’d like to convert these two units of volts and meters, respectively. We recall the conversions that 1000 millivolts equals one volt and 100 centimeters equals one meter. Therefore, if we divide 9.6 millivolts by 1000, we’ll get this number in volts. 9.6 divided by 1000 is 0.0096. That’s the number of volts of potential difference is established across our conducting rod. Then, to convert the length of our rod from centimeters to meters, we’ll divide by 100. 15 divided by 100 is equal to 0.15. In units of meters then, our rod’s length is 0.15 meters.
We’re now ready to calculate this fraction and solve for 𝐵. When we do, we get a result of exactly 0.2 teslas. This is the strength of the magnetic field in our scenario. We can now store this result off to the side and clear some space for the second part of our question.
That part reads “Which side of the field-containing region is the rod moving toward?”
We now want to answer that question we raised earlier of in which direction is the rod moving. A clue that will help us figure this out is that positive charge accumulates at the top of the rod and negative charge accumulates at the bottom of it. A key fact about this rod is that it is a conducting rod. This means the rod contains electric charges that we can think of as mobile. They’re capable of moving all throughout the rod as the rod moves through this magnetic field.
In general, when a charge 𝑞 moves with a velocity 𝑣 through a magnetic field of strength 𝐵, then that charge experiences a magnetic force equal in magnitude to 𝑞 times 𝑣 times 𝐵. Here though, we’re less interested in force magnitude and more in the force direction. To figure that out, we’ll use something called a right-hand rule. If we take the fingers on our right hand and point them in the direction of 𝑞 times 𝑣, that is, in the direction of the velocity of the moving charge times the charge 𝑞 itself, and then as a next step curl the fingers on our right hand so that they point in the direction of the magnetic field B, in this case we see that that field points out of the screen towards us. Then, in that case, our thumb points in the direction of the magnetic force 𝐹 sub B on this charge 𝑞.
Often, this right-hand rule is used in order to determine the direction of 𝐹 sub B. In our scenario though, we already know something about this force. For charges in our moving conducting rod, the magnetic force on negative charges is towards the bottom of the rod and the magnetic force on positive charges is towards the top. Just to pick one of these two charge types, let’s consider positive charges in a rod. We see that the direction of the magnetic force on positive charges is upward. This is why those positive charges accumulate at the top of the rod. We also see that the magnetic field B points out of the screen towards us. This is actually consistent with the direction of the magnetic field we used in our example.
So, thinking of our right-hand rule, if after curling our fingers our right hand looked like this, that would mean we find the direction of 𝑞 times 𝑣 by uncurling our fingers. So, for positive charges 𝑞 in our rod, 𝑞 times 𝑣 points to the left. Since 𝑞 is positive, it doesn’t change the overall sign of 𝑞 times 𝑣. Therefore, since 𝑞 times 𝑣 points to the left, so does 𝑣 itself, the velocity of the rod. This means that the direction of the rod’s motion is to the left. It’s by moving this way that positive charge accumulates at the rod’s top and negative charge at the bottom of the rod.
In answer to our question then, we say that the side of the field-containing region the rod is moving toward is the left side.