Four long, parallel power lines each carry 100 A currents. A c...

Question

Four long, parallel power lines each carry 100 A currents. A cross-sectional diagram of these lines is a square, 20.0 cm on each side. For each of the three cases shown in Fig. E28.25, calculate the magnetic field at the center of the square.

Accepted Answer

Welcome back everybody. We are taking a look at four different wires that are passing through the corner of this square right here in their indicated directions, which by the way, just as a reminder X. Means that the flow of current is going into apologies into the page and the dot means it is coming out of the page. We're told that the current flowing through each of these wires is 3.5 amps. And we are tasked with finding what the magnitude of the magnetic field is at the center of our square is. So in order to do this, let me first just kind of draw out some guidelines on our diagram. This is going to help us visualize a couple of things as we are trying to find the magnitude at our center point right here. Now, in order to find the entire magnitude, we're going to have to find the magnetic fields of each of our individual wires here. So our big B is going to be equal to B a, b, b b c plus b D. And we have a formula for the magnetic field. Just for a single wire, this is going to be mu not times the current flowing through it all divided by two pi times the distance away from the center. So the distance for each one of these guys is going to be this diagonal, which I'm just going to label are we need to figure out our and here's how we're going to do this with this diagram. We're given the bit of information that the side length of the square or one of the sides is five centimeters or point oh five m. This means that this distance. And if you think about it, this distance is going to be half of that 50.25 m. So what we can do is we can use pythagorean theorem to find our our. So let's go ahead and do that. Here. We have. That R squared is equal to 0.25 squared plus 0.25 squared. Take the square root of both sides. And you'll see on the left hand side that this gets rid of our radical. You plug this into our calculator. We get that this is 0. m. Right? So now we have this, let's just calculate now what the magnitude is for a single wire. Let's just go ahead and take wire A here. Right. So we are going to have that the magnitude of the magnetic field for wire A is according to this formula up above. So we have not is four pi times 10 to the negative, seventh times the current flowing through of 3.5. All divided by two pi times r distance away from the center 20.353. When we plug this into our calculator, we get that the magnitude of the magnetic field from wire A on the center is going to be 1.98 times to the negative fifth Tesla. But let's think about this conceptually here, all four of these wires are equidistant from the center, they all have the same current. And then the rest of this formula is just a constant meaning the magnitude of the magnetic field is going to be equal for all four of our wires. So now that we know that we're almost ready to calculate our entire magnetic field at the center. But you have to think about this. We're dealing with vectors here. But what direction are these vectors facing? Well, here's what I'll do using the right hand rule, we are able to determine these directions that I'm drawing out before us. Now, I'm gonna write a little key out here. We actually have that our blue arrows are going to represent the direction of the magnetic field for our A wire and our C. Wire. And then we have that our red arrow is the direction for our B. Wire and also our D. Wire. And I'm just gonna put little vector signs because we our dealing with vectors here. Like I said, you find those via the right hand rule a little tough to show you since you can't see my own hand. So since we're dealing with vectors, this means that in order to calculate our entire B. We have to look at both the Y direction of our vectors and the X. Direction of our vectors. Let's think about this conceptually conceptually. And start with our wide direction. If we have two vectors going up and two vectors going down and all the vectors are of equal magnitude R. B. Y are vectors are going to cancel out in the Y direction. So R B Y is just zero. This makes this really easy on us because then that means that our entire magnetic field is just going to be equal to the magnetic field in the positive X direction. So let's go ahead and calculate this here. But how are we going to do that? Well, we need our X component. But what is our X component of these vectors? If you draw kind of this little projection line down to our positive X axis here, you can see that this little bit right here is the X component of all of our vectors. Now, since we're dealing with a square, we know that this angle and therefore this angle is degrees. So here is what we can do. We know that the magnitude for each, for the magnitude of the magnetic field for each of our wires is 1.98 times 10 to the negative fifth. We're gonna multiply that by four. Since we know that that magnitude is equal for all four wires. And then what we can do is we can multiply that by the co sign of our given angle 45. In order to decompose our vectors into the X direction. When we plug this into our calculator, we get a final answer of 5.61 times to the negative fifth Tesla in the positive X. Direction, which corresponds to our answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

Four long, parallel power lines each carry 100 A currents. A cross-sectional diagram of these lines is a square, 20.0 cm on each side. For each of the three cases shown in Fig. E28.25, calculate the magnetic field at the center of the square.

Key Concepts

Biot-Savart Law

Superposition Principle

Right-Hand Rule

Watch next