Lesson Video: Electromagnetic Interactions Between Straight Conductors Physics

Video Transcript

In this video, we will learn how to determine the magnetic forces between parallel lines of current and analyze the net magnetic fields of the currents. Before we do, we must refresh our memory about the magnetic field around a single line of current. In order to understand how to find the net magnetic field around more than one current-carrying wire, we must first understand how to draw in the magnetic field around a single current-carrying wire by doing the right-hand rule.

We need to remember that to use the right-hand rule to determine the direction of magnetic field around a current-carrying wire that our thumb will point in the direction of the current and that our fingers will curl around the wire, pointing in the direction of the magnetic field lines. If we orient a wire such that the current is directed to the right of our screen, then we must position our hand so our thumb is also pointing to the right of our screen. And our fingers will curl around the wire such that magnetic field will be pointing to the bottom of the screen in front of our wire.

This tells us that the magnetic field around our current-carrying wire will create circles all along it, represented by the yellow dotted lines, with the magnetic field being directed into the screen underneath the wire, out of the screen above the wire, and pointing towards the bottom of the screen in front of the wire. If we were to look at our wire in a different direction such that the current was pointing into our screen, applying our right-hand rule with our thumb being directed into the screen and our fingers curling around the wire such that the magnetic fields make concentric circles that are clockwise around our current-carrying wire.

Now that we’ve refreshed our memory on how to find the magnetic field around a current-carrying wire, let’s determine how to find the net magnetic field around more than one wire. To determine the net magnetic field, we need to do vector addition based on the direction of the magnetic field due to each current-carrying wire at the chosen location. We need to do vector addition because our magnetic field is a vector. This means that the two magnetic fields due to the current-carrying wires with magnitudes 𝐵 one and 𝐵 two point in the same direction. We can add the magnitudes together. 𝐵 one plus 𝐵 two would be equal to our net magnetic field.

If, however, our magnetic fields point in opposite directions, then we must subtract their magnitudes. 𝐵 one minus 𝐵 two will be equal to the net magnetic fields. Recall from earlier when I said that a current-carrying wire directed into the screen will have a magnetic field with concentric circles that are directed clockwise around the wire. Now, let’s set up an identical current-carrying wire right next to it.

We have drawn a second current-carrying wire to the right of the first current-carrying wire and represented the magnetic field with pink dotted lines. If we draw a horizontal line that goes through both the current-carrying wires, we can then analyze positions where the magnetic fields overlap to determine whether they are a larger net magnetic field or a smaller net magnetic field. Let’s go through the labeled positions below, one, two, and three, and determine whether we should add or subtract our fields.

At position one, both the yellow magnetic field and the pink magnetic field are pointing directly to the top of our screen. Therefore, they’re in the same direction, which means that we will add the magnitudes of our fields. At position two, the magnetic field in yellow is pointing to the bottom of the screen and the magnetic field in pink is pointing to the top of our screen, which means they’re pointing in opposite directions. So we should subtract the magnitudes. And finally, at position three, both the yellow and pink magnetic fields will be pointing to the bottom of the screen, which means they’re pointing in the same direction. So we will once again add the magnitudes.

We can see that positions one and three will have larger net magnetic fields than position two as we added the individual fields at one and three. To take it one step further, we can even say that the net magnetic field at position three is the largest. This is because position three is closer to the wire on the right side than position one is to the wire on the left side. And position three is also closer to the wire on the left side than position one is to the wire on the right side.

If we remember that the magnetic field around a current-carrying wire 𝐵 is equal to the magnetic permeability 𝜇 times the current in the wire 𝐼 divided by two 𝜋 times the distance from the wire 𝑟, then we would recognize that the position that’s closest to the wire or has the smallest 𝑟 will have the greatest magnetic fields. Of the three positions, position three will have the largest net magnetic field, then position one, and finally position two will have the smallest.

Now, let’s look at the force per unit length between two parallel conducting wires. To determine the force per unit length, we need to recall two equations. First, the magnetic field around a current-carrying wire, 𝐵, is equal to the magnetic permeability, 𝜇 naught, times the current in the wire, 𝐼, divided by two 𝜋𝑟, where 𝑟 is the distance from the wire. The second equation is that the force on a current-carrying wire 𝐹 is equal to the current within the wire, 𝐼, times the length of the wire, 𝐿, times an external magnetic field, 𝐵, that the wire is placed in.

To derive our expression for force per unit length, let’s draw on an example of two wires. Our first wire will have a current 𝐼 one directed to the right of our screen, with a length of 𝐿. Using the right-hand rule, where our thumb will be directed to the right of our screen and our fingers will curl around our wire, we will see the magnetic field goes into the page on the bottom of the wire and comes out of the page on the top of the wire.

We then place a second wire into the field with a current of 𝐼 two directed to the right of the screen and a length of 𝐿. If the distance between the two wires is 𝑑, let’s calculate the force exerted on the second wire by the magnetic field generated by the first wire. We can begin with the equation for force on our wire. To get the equation in the right form, we need to divide both sides by 𝐿. Remember, the 𝐼 is the current in the wire that we’re looking for the force per unit length on. In this case, that would be 𝐼 two. The magnetic field which applies a force to our second wire comes from current-carrying wire one. And we can find the strength of that by using our equation for the magnetic field around a current-carrying wire. The current is current 𝐼 one, and the distance between our wires is 𝑑.

Any time we’re trying to find the force per unit length that two parallel current-carrying wires put on each other, we can use the equation 𝜇 naught 𝐼 one times 𝐼 two divided by two 𝜋𝑑. It doesn’t matter which wire we’re looking at as we’ll come up with the same equation. If we’re looking for the force per unit length on wire one, then we’d use a current of 𝐼 one. And the magnetic field would be due to current 𝐼 two, which once again leaves us with the same relationship.

But what if we have an array of parallel wires, such as in a solenoid? In this case, let’s say that all of the wires on the left-hand side have current going into the screen. And all the wires on the right-hand side have current coming out of the screen. If we plot the individual magnetic fields around each wire, we can see what happens to the net magnetic field. All the wires on the right-hand side are gonna have magnetic fields that are directed counterclockwise. The magnetic field we pointed to the left above the wire will be pointing down to the bottom of the screen at the left of the wire, will be pointed to the right of the screen underneath the wire, and be pointing to the top of the screen on the right of the wire.

Between the wires, the magnetic field’s going to subtract because you have the wire underneath having a magnetic field to the left of the screen but the wire above having a magnetic field to the right of the screen, setting up net magnetic field lines around the wires on the right-hand side that look similar to the ones that we have in yellow going counterclockwise. Applying the right-hand rule to the wires in the left-hand side of the screen gives us a field that is clockwise. Above the wire will be directed to the right of our screen, to the right of our wire will be directed to the bottom of our screen, below the wire will be directed to the left of our screen, and the left of our wire will be directed to the top of the screen.

Just as with the previous wires, the wires on the left-hand side of the screen are gonna have areas where they interact to make a net magnetic field. In between the wires, these magnetic fields are going to be pointing in opposite directions, so they’re going to subtract from each other. This sets up a net magnetic field as shown in the pattern with the yellow dots, which is essentially a solenoid, where the magnetic field in between the two sets of wires adds together to make a stronger net magnetic field. Let’s apply what we’ve learned about electromagnetic interactions between straight conductors to an example problem.

The diagram shows concentric field lines of the magnetic fields of two parallel current-carrying conductors. Both currents are into the plane of the diagram. And the currents have the same magnitude as each other. The increase in radius of the concentric field lines is constant. And the strength of a magnetic field at a point around a current is proportional to the perpendicular distance of the point from the current. Which of the following list of the points shown in the diagram correctly orders the points from greatest to smallest magnitude of net magnetic fields? (A) 𝐵, 𝐶, 𝐸, 𝐴, 𝐷. (B) 𝐵, 𝐸, 𝐶, 𝐴, 𝐷. (C) 𝐴, 𝐵, 𝐶, 𝐷, 𝐸. (D) 𝐸, 𝐵, 𝐶, 𝐴, 𝐷. (E) 𝐷, 𝐴, 𝐸, 𝐶, 𝐵.

On the diagram, we see the two wires set up on the 𝑥-axis along with the five points from 𝐴 to 𝐸. The dotted circles around each wire represent the magnetic fields. To determine which points are greater and smaller in magnitude of net magnetic fields, we need to remember two things. First, when we’re finding the net magnetic field, we need to use vector addition. This means that if the magnetic field lines are pointing in the same direction at the same position, we will add the magnitudes together. And if they’re pointing in the opposite direction of each other, we will subtract the magnitudes.

Let’s apply this to our diagram by using the right-hand rule. We need to remember that we can use the right-hand rule to find the direction of the magnetic field around a current-carrying wire, where our thumb will point in the direction of the current and our fingers will curl around the wire to show the direction of the field lines. When we do the right-hand rule on our diagram, our thumb points into the screen and our fingers curl around. This shows that our magnetic fields will be pointing clockwise around both wires. We can use yellow to show the direction of the field around the right wire and pink to show the direction of the field around the left wire.

Let’s look how both fields are oriented at the five points on our 𝑥-axis. At position 𝐴 and 𝐵, both of our fields are pointing towards the top of the screen. So the magnitude of the field will be added together. At position 𝐶 and 𝐷, the field lines are pointing in opposite directions. Yellow is pointing to the top of the screen, and pink is pointing to the bottom of the screen. So we would subtract the magnitudes. Both fields are pointing to the bottom of the screen at point 𝐸, which means that we will once again add the magnitudes together. Point 𝐷 is in the middle of the wire. And since we know that the magnitude of the magnetic fields is being subtracted at this position, we can say that the net magnetic field at 𝐷 is zero.

With the net magnetic field at point 𝐷 being the smallest, we can eliminate any answer choices that don’t have the last letter as 𝐷. This would be answer choices (C) and (E). To differentiate between any of the other points, we need to remember the equation for the magnetic field around a current-carrying wire. The magnetic field 𝐵 is equal to the magnetic permeability, 𝜇 naught, times the current in the wire, 𝐼, divided by two 𝜋𝑟, where 𝑟 is the distance between the position and the wire. In this example, the current 𝐼 of both wires is the same.

Therefore, no matter where we are in our diagram, we will have the same current, 𝐼, and our magnetic permeability, 𝜇 naught, as a constant as well as two 𝜋. Therefore, the magnetic field will be proportional to one over 𝑟. Looking at our answer choices, we might be able to solve our problem if the net magnetic field is greater at 𝐸 than at 𝐵 by starting to compare those two points.

Let’s label the wires one and two to differentiate them, with one being the left wire and two being the right wire. 𝐸 is one radii, 𝑟, away from the closest wire, in this case wire two, whereas 𝐵 is just shy of two radii away from its closest wire, wire one. Comparing the farther wires, 𝐸 is approximately four radii away from its farthest wire, wire one, where 𝐵 is just shy of five radii away from its farther wire, wire two. Point 𝐸 is closer to the nearest wire as well as the farthest wire when compared to point 𝐵. And at both 𝐸 and 𝐵, we would add the magnitude of the two magnetic fields together because they’re pointing in the same direction.

If we are adding the magnitudes together and 𝐸 is closer, meaning a smaller 𝑟, there will be a bigger net magnetic field at 𝐸 than there would be at 𝐵. With the net magnetic field at point 𝐸 being larger than the net magnetic field at point 𝐵, our final answer will be choice (D), that the magnitude from greatest to smallest of the net magnetic field is 𝐸, 𝐵, 𝐶, 𝐴, 𝐷.

Let’s summarize our lesson.

Key Points

Diagrams can be used to qualitatively determine the net magnetic field around two parallel conducting wires. To find the force per unit length that two parallel conductors put on each other, use 𝐹 over 𝐿 is equal to 𝜇 naught 𝐼 one 𝐼 two divided by two 𝜋𝑑. The net magnetic field of an array of parallel conducting wires can be found by analyzing the individual magnetic fields.