# Question Video: Solving for the Resistance of a Straight Conductor Moving through a Uniform Magnetic Field Physics • 9th Grade

A conducting rod moves on conducting rails that form a circuit that contains a resistor, as shown in the diagram. The rod travels the full distance across the rails in a time of 36 s, moving at a constant speed. The magnetic field around the circuit has a strength of 275 mT. The current in the circuit is 32 𝜇A. Find the resistance of the rod.

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### Video Transcript

A conducting rod moves on conducting rails that form a circuit that contains a resistor, as shown in the diagram. The rod travels the full distance across the rails in a time of 36 seconds, moving at a constant speed. The magnetic field around the circuit has a strength of 275 millitesla. The current in the circuit is 32 microamperes. Find the resistance of the rod.

In our diagram, we see this conducting rod stretched across two parallel conducting rails. The rod begins at the left side of the rails but moves at a constant speed all the way across their length. As it does so, it moves through this uniform magnetic field. Since the rod is a conductor moving through such a field, a potential difference is generated across its ends. The equation describing the potential difference generated is given here. The symbol 𝐸 stands for EMF, or equivalently potential difference. 𝑙 is the length of the conducting rod, 𝑣 is its velocity, 𝐵 is the strength of the magnetic field the rod moves through, and 𝜃 is the angle between the velocity 𝑣 and the magnetic field 𝐵.

In our situation, it’s not exactly the potential difference we want to solve for, but rather the resistance of the rod itself. That resistance added to the resistance here of 24 ohms is equal to the entire circuit’s resistance. If we recall Ohm’s law that the potential difference across a circuit is equal to the current in the circuit times that circuit’s resistance, we can see that to solve for the resistance 𝑅, we will need to know the potential difference 𝑉 and the current 𝐼. Note that in our situation, we’re given the current 32 microamperes. But we still need to solve for the potential difference 𝑉, and that’s where this equation comes in.

The potential difference generated across our moving rod, we’ll call it 𝑉, equals the length of the rod times its velocity times 𝐵. Note that we haven’t written the sin of 𝜃 here because, as we noted, 𝜃 for this case is 90 degrees. The sin of 90 degrees equals one, so we can leave out that factor without changing our result. What we now want to do then is to find 𝑙 and 𝑣 and 𝐵 so that we can solve for the potential difference 𝑉 across the conducting rod. As we go forward, note that this uppercase 𝑉 stands for potential difference, while this lowercase 𝑣 stands for speed.

In general, the speed of some object equals the distance that object travels divided by the time taken to travel that distance. In our case, the distance traveled by the conducting rod equals the length of one of these rails. Instead of writing that length in centimeters though, let’s write it in meters. 125 centimeters is 1.25 meters. So, that’s the distance traveled by a rod. And now, we’ll divide that by the time it takes to travel that distance. 1.25 meters divided by 36 seconds is the speed of our conducting rod. The length 𝑙 of that rod is given to us in the diagram; it’s 9.5 centimeters. Once again, we’ll convert this result into base units before we plug it in to our equation. 9.5 centimeters is 0.095 meters.

Lastly, there’s the magnetic field with a strength of 275 millitesla. To convert the units from millitesla to tesla, we’ll shift the decimal place three spots to the left. At this point, we could multiply these values together to solve for the potential difference 𝑉 induced across our conducting rod. But if we take an extra step and divide both sides by the current 𝐼, then notice we have this fraction 𝑉 over 𝐼, which is equal to the resistance 𝑅 in the circuit. The current 𝐼 in the circuit is 32 microamperes. A microampere is one one millionth of an ampere. So, when we write this in units of amperes, we can write it as 32 times 10 to the negative six amperes. This entire fraction is equal to 𝑉 divided by 𝐼. And therefore, it’s equal to the total resistance of the circuit 𝑅.

In our next step, we need to be a little careful. Our question statement doesn’t actually ask us for the resistance of the circuit, but rather for the resistance of the rod. This value here, 𝑅, is the total resistance in this entire circuit. But earlier, we saw that some of this is due to this 24-ohm resistor here. Clearing a bit of space to work, we can write that the total resistance in our circuit, 𝑅, is equal to the resistance of the rod — we’ll call that 𝑅 sub 𝑟 — plus 24 ohms. We add these resistance values like this because the resistors are in series. If we subtract 24 ohms from both sides of this equation, we find that the resistance of the rod equals the total resistance in the circuit minus 24 ohms.

We now recall that this big fraction down here is equal to 𝑅, the total circuit resistance. Therefore, if we subtract 24 ohms from this fraction, just like our equation up here tells us, all that will be equal to 𝑅 sub 𝑟, the rod’s resistance. That’s the value we really want to solve for. When we enter this entire expression on our calculator, we find a result of 4.347 and so on ohms. Notice that for the values we were given, some of them have as few as two significant figures. Therefore, we’ll limit our final answer to two significant figures of precision. That means we’ll report our final result to one decimal place. And when we round to one decimal place, we get 4.3 ohms. This is the resistance of the conducting rod to two significant figures.