Question Video: Ranking Magnetic Field Strengths from Current-Carrying Wires Physics

The diagram shows concentric field lines of the magnetic fields of two parallel current-carrying conductors. The right-hand current goes into the plane of the diagram and the left-hand current goes out of the plane of the diagram. The currents have the same magnitude as each other. The increase in the radius of the concentric field lines is constant, and the strength of a magnetic field at a point around a current is inversely proportional to the perpendicular distance of the point from the current. Which of the following lists of the points shown in the diagram correctly orders the points from the greatest to the smallest magnitude of net magnetic field? [A] 𝐴, 𝐡, 𝐸, 𝐢, 𝐷 [B] 𝐴, 𝐸, 𝐢, 𝐡, 𝐷 [C] 𝐢, 𝐷, 𝐸, 𝐡, 𝐴 [D] 𝐷, 𝐸, 𝐢, 𝐡, 𝐴 [E] 𝐸, 𝐡, 𝐴, 𝐢, 𝐷

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Video Transcript

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

After all that, what we’re considering here is this scenario. In it, we have two conductors carrying current. There is one carrying current out of the screen towards us here and one carrying current into the screen away from us here. These currents have the same magnitude, and they create magnetic fields around themselves. Concentric circles indicating these magnetic fields are shown in these gray, dashed lines. We see one set of lines centers on the one current-carrying conductor and the other set of circles centers on the other.

Earlier in our problem statement, we were told that the strength of the magnetic field, we’ll call it 𝐡, is inversely proportional to the straight-line distance between a current-carrying conductor and the point where we want to calculate the field due to that conductor. Generally then, this tells us that the farther we are away from a given current-carrying conductor, the weaker the field due to that conductor will be. But this relationship also gives us some specific information. For example, let’s say we’re considering the magnetic field due to this particular current-carrying conductor. We are told that these concentric rings are the same distance apart from one another.

Therefore, if we were to move straight away from the current-carrying conductor, from say the first circle out to the second circle, we will be doubling our distance from that conductor. This inverse relationship tells us that for a transition like that, from the first circle out to the second circle away from a given current-carrying conductor, because our distance is doubling, our magnetic field strength is being divided in half. That is, the field at this location due to this conductor is twice as strong as the magnetic field at this location due to that same conductor.

As we’ve seen, in this example, we have not just one current-carrying conductor, but two. And each one creates a magnetic field. These fields overlap. And in the midst of the fields, we see we have these five label points: 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸. What we ultimately want to do is rank or order these five points according to the magnitude of magnetic field. In other words, we’re comparing the overall magnetic field magnitude at point 𝐴 with that at point 𝐡 with that at point 𝐢 with that at point 𝐷 and with that at point 𝐸.

To be able to compare these magnitudes, first we’ll need to calculate them. As we start doing that, let’s recall that the magnetic field strength at perpendicular distance π‘Ÿ away from a current-carrying wire carrying current 𝐼 is given by this expression. Here, this symbol πœ‡ naught is the permeability of free space, also called the permeability of vacuum. That’s a constant. And as we’ve seen, 𝐼 is the current carried in a conductor and π‘Ÿ is the perpendicular distance between the axis of that conductor and a point where we’re calculating the magnetic field strength 𝐡. Note that this general equation in green is what leads to our relationship of 𝐡 being inversely proportional to π‘Ÿ. At each of our five points 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸, we want to calculate the relative magnitude of the magnetic field strength.

As we begin that process, let’s start with our last point, point 𝐸. For point 𝐸, as will be the case for all of our points, the net magnetic field at that location is due to the two current-carrying conductors. The conductor on the right carries a current that points into the screen. Therefore, using a right-hand rule, where we point the thumb of our right hand in the direction of the current, that’s into the screen, the direction the fingers on that hand are able to curl, in this case clockwise, indicates the direction of the magnetic field created by that current. Our conductor carrying current into the screen creates a clockwise magnetic field. At point 𝐸 then, we can see that field will point downward.

At this point, let’s establish a convention that magnetic fields that point upward are positive, while those that point downward point in the negative direction. By our convention, the magnetic field created at point 𝐸 due to the right current-carrying conductor is negative. The total magnetic field at point 𝐸 though will include the effects of the conductor on the left. This conductor, we know, carries current pointing out of the screen toward us. And so once again using our right-hand rule to find the direction of a magnetic field, we see that this magnetic field points counterclockwise. Therefore at point 𝐸, the magnetic field due to this current-carrying conductor is positive.

Notice that we’ve drawn the pink arrow, the one due to the conductor on the left, as shorter than the orange one due to the conductor on the right. This difference is because, as we see, point 𝐸 is closer to the conductor on the right than that on the left. Our relationship of 𝐡 being inversely proportional to π‘Ÿ comes into effect. Since the perpendicular distance between any two adjacent concentric rings is the same, let’s give that distance a name. Let’s call it lowercase π‘Ÿ.

Now we can write the total magnetic field at point 𝐸 as the sum of the magnetic fields due to the conductor on the left and the magnetic field due to the conductor on the right. We can use this expression to solve for these two magnetic fields based on the distance each conductor is from the point 𝐸 and also use our sign convention to eventually be able to solve for the total magnetic field at this point.

The first thing we can do is figure out just how far the conductor on the left is from point 𝐸. Starting at this conductor, the one carrying current out of the screen towards us, we can count one, two, three, four concentric rings away from this conductor to reach point 𝐸. And we can write the magnetic field at point 𝐸 due to the conductor on the left as πœ‡ naught times 𝐼, this will be the magnitude of the current carried in either conductor, divided by two πœ‹ times four π‘Ÿ, the perpendicular distance between point 𝐸 and our left conductor.

Since the magnetic field at point 𝐸 due to the leftmost conductor points upward, it has a positive value. On the other hand, the magnetic field at this point due to the rightmost conductor will have a negative value. That is, it will point downward. So we can actually replace this positive with a negative. And then seeing from our diagram that the rightmost conductor is one concentric circle away from the point 𝐸, we write the magnetic field at point 𝐸 due to this conductor as negative πœ‡ naught 𝐼 divided by two πœ‹ times π‘Ÿ. Notice that in both of these terms, we have πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ. That gives us πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ all multiplied by the quantity one-eighth minus one-half.

If we multiply one-half by four divided by four, which is equal to one, so it doesn’t change the overall number, we get four divided by eight. One-eighth minus four-eighths is negative three-eighths. Now let’s recall that we want to compare the magnitude of the net magnetic field at each of our five points. For each of those points, what we’re really going to be comparing is the magnitude of the number in these parentheses. All of our points 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸 will have this prefactor of πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ. When it comes then to the meaningful value of the net magnetic field at point 𝐸, we’ll write that this way. We’ll say the magnitude of the magnetic field at point 𝐸 is proportional to three divided by eight.

We can say this because this magnitude is equal to πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ multiplied by three-eighths. And this entire prefactor in front of the negative three-eighths in this expression can be considered a constant. We could write then that the magnitude of the net magnetic field at point 𝐸 is equal to a constant, we’ll call it π‘˜, multiplied by three-eighths. And mathematically, this means the same thing as our earlier expression. All that said, this is how we’ll record the important information about the total magnetic field at point 𝐸.

Now that we’ve done this for point 𝐸 on our diagram, let’s move on and do the same thing for point 𝐷. Once again, the net magnetic field at this point is due to the magnetic field created by the left conductor and the right conductor. The point 𝐷 lies at the origin of our axes and is therefore the same distance away from each of these conductors, and that distance, we see, is 1.5π‘Ÿ. The conductor on the left, because it has a counterclockwise magnetic field, will create a positive magnetic field at point 𝐷, while the conductor on the right, which has a clockwise magnetic field, will also create an upward-pointing or positive magnetic field at this point.

Note that now the magnetic fields created by both of these conductors are adding to one another rather than subtracting from one another. Here then is how we can write that total magnetic field. Once again, we use the base formula of πœ‡ naught 𝐼 divided by two times πœ‹ times the perpendicular distance between the conductor and the point of interest.

Now, that perpendicular distance in each case for both conductors is one and a half or three-halves π‘Ÿ. This is because point 𝐷 is right between the first and the second concentric ring for each of these conductors. Before we combine these two terms, notice that the one-half cancels out in the denominator with the factor of two. So when we add these terms together, we get two times πœ‡ naught 𝐼 divided by πœ‹ times three π‘Ÿ. In this expression, we’re treating πœ‡ naught 𝐼 divided by πœ‹ times π‘Ÿ as a constant. What we have then is that constant value multiplied by two-thirds. And two-thirds is the critical information we want to retain about the magnetic field at point 𝐷. To do this, we’ll write that the magnitude of the net magnetic field at point 𝐷 is proportional to two-thirds.

We now have a way to compare, for example, between the net magnetic field magnitudes at point 𝐷 and point 𝐸. We want to be able to do this with all five of our points, so now let’s move on to point 𝐢 in our diagram. Once again, the magnetic fields from each of our two current-carrying conductors point upward or in the positive direction at this point. And if we count the concentric rings between our conductor on the left and point 𝐢, we count one. So the overall distance between them is lowercase π‘Ÿ, while if we do the same thing for our conductor on the right, we count two rings. So the overall distance between point 𝐢 and the conductor on the right is two times π‘Ÿ.

If we factor out from these two terms what we’re calling our constant πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ, we have in our parentheses one-half plus one-fourth. One-half multiplied by two divided by two is equal to two divided by four. And two-fourths plus one-fourth is equal to three-fourths. So we’ll write that the net magnetic field magnitude at point 𝐢 is proportional to three-fourths.

Now let’s move on to consider point 𝐡. And we see that this point doesn’t lie exactly along the concentric rings on our diagram. However, it is very close to this concentric ring. For that reason, we’ll say that the magnetic field at point 𝐡 is approximately equal to the field at a point that does lie along that ring. The magnetic field at point 𝐡 due to the conductor on the left points downward, while the field at this point due to the conductor on the right points upward. Therefore, 𝐡 sub L, the magnetic field at point 𝐡 in this case due to the conductor on the left, will be negative, and the distance we’ll use will be two times π‘Ÿ. Again, this is an approximate distance very close to where point 𝐡 is actually located.

To see the distance to use for the magnetic field due to the conductor on the right, we can count rings from that conductor. We count one ring, two, three, four, which is as far as we can get from the conductor on the right. But then we see one ring distance out beyond this is where point 𝐡 is approximately located. Therefore, for 𝐡 sub R, we’ll write πœ‡ naught 𝐼 divided by two πœ‹ times five times π‘Ÿ.

Factoring out as usual πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ, we have negative one-quarter plus one-tenth. If we multiply negative one-quarter by five divided by five and one-tenth by two divided by two, we get negative five twentieths plus two twentieths. This is equal to negative three twentieths. This value in parentheses is the relevant value for the magnetic field at point 𝐡. We’ll write then that the magnitude of the net magnetic field at point 𝐡 is proportional to three twentieths.

Our last step is to follow a similar process for the magnetic field at point 𝐴. Notice that, just like point 𝐡, point 𝐴 doesn’t lie exactly along one of the concentric rings. However, it’s very close to the one, two, three, third ring out from the conductor on the left. We’ll approximate point 𝐴 as being located at that point. The conductor on the left will create a downward or negative magnetic field at this point, while the conductor on the right will create an upward or positive magnetic field. At point 𝐴, the conductor on the left then will have an approximate contribution of negative πœ‡ naught times 𝐼 divided by two πœ‹ times three π‘Ÿ.

On the other hand, the conductor on the right has a distance of one, two, three, four concentric rings plus one, two more such rings. That gives us a total of approximate distance of six π‘Ÿ between point 𝐴 and the conductor on the right. Factoring out πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ, in the parentheses we have negative one-sixth plus one twelfth. Negative one-sixth times two divided by two is equal to negative two twelfths. And negative two twelfths plus one twelfth is just negative one twelfth. Here then is how we can write the magnitude of the net magnetic field at point 𝐴. It’s proportional to one twelfth. Just as before, here we’ve treated πœ‡ naught times 𝐼 divided by πœ‹ times π‘Ÿ as a constant.

We now have all the information needed to compare the net magnetic field magnitudes at each of our five points. We’ll start by finding the largest of the five values boxed in orange. Looking throughout these choices, we see that the largest value is three-quarters. This means that in our ranking of net magnetic field magnitudes, point 𝐢 must come first. As it turns out, only one answer option shows this. Just to be sure that option (C) is correct though, let’s compare the remaining points. After three-quarters, the next largest value we see boxed in orange is two-thirds. That corresponds to point 𝐷, which agrees with the order of answer option (C).

After that, the next largest value is three-eighths. That’s the value corresponding to point 𝐸, which we see is also next in this answer list. After that, the next largest value is three twentieths followed by one twelfth. These are the corresponding values for points 𝐡 and 𝐴, respectively. We see then that answer option (C) is correct. In terms of the net magnitude of the magnetic field at each of these five points, point 𝐢 has the largest such value then point 𝐷 followed by point 𝐸 followed by point 𝐡 and lastly followed by point 𝐴.

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