Lesson Video: The Magnetic Field due to a Current in a Circular Loop of Wire | Nagwa Lesson Video: The Magnetic Field due to a Current in a Circular Loop of Wire | Nagwa

Lesson Video: The Magnetic Field due to a Current in a Circular Loop of Wire Physics • Third Year of Secondary School

In this video, we will learn how to calculate the magnetic field produced by a current in a circular loop of wire.

12:21

Video Transcript

In this video, our topic is the magnetic field due to a current in a circular loop of wire. We’re going to learn how to calculate the magnitude of this field strength at the center of these circular loops. And we’ll also learn how to determine field direction.

A good place for us to start is to recall that any current-carrying wire will produce a magnetic field around itself. And this holds true if we take the ends of that wire and we shape it into a circular loop. If that loop looked like this, then it would create a magnetic field whose effects could be experienced both inside and outside this circle. Our focus is going to be at a particular point, the very center of this loop. So if we were looking at this loop from the side on like this, we’re interested in the magnetic field strength at this point here.

That field magnitude, and we can refer to it as capital 𝐵, depends on a few different factors. First, like most equations for a magnetic field strength, it depends on 𝜇 nought, known as the permeability of free space. This is a universal constant that measures the resistance of free space, of vacuum, to the formation of a magnetic field in it. The value of 𝜇 nought is four times 𝜋 times 10 to the negative seventh tesla meters per ampere. So the magnetic field strength of the center of our current-carrying loop depends on 𝜇 nought, and it also varies with two different physical parameters. The first is the current magnitude 𝐼 that’s in this circular loop. And the second is the radius of this loop. We can call that lowercase 𝑟.

When we combine 𝜇 nought, 𝐼, and 𝑟 to solve for the field strength 𝐵, here’s how they all come together. The magnetic field strength 𝐵 is equal to 𝜇 nought times 𝐼 divided by two times 𝑟. Now, this relationship is specifically for the magnetic field created at a particular point right at the center of our circular loop in the plane of the circle. This equation doesn’t apply for the magnetic field strength anywhere else.

Considering this equation, we can see how changes in the current 𝐼 and in the loop radius 𝑟 would affect the field strength. An increase in the current magnitude would lead to a proportional increase in the field strength 𝐵, while, on the other hand, the loop radius and the field strength are inversely related. This means the bigger our circular loop — all other factors held constant — the smaller the magnetic field strength will be at its center.

This makes a bit of sense when we consider that it’s the current 𝐼 which creates this magnetic field in the first place. So for a larger radius loop, as our circle gets bigger and bigger, the distance between the current and the loop’s center increases. From this perspective then, we would expect 𝐵 to get smaller as 𝑟 grows.

Now, so far, we’ve been considering a circular current of just one loop or just one turn. In general, though, it’s possible to have more than this. We could have two or 10 or hundreds of loops. In many cases. All of these different turns are part of the same circuit. We could think of it as a coil of wire, all carrying the same current 𝐼. In the case of multiple loops, as we’ve seen here, that has an effect on the strength of the magnetic field created. Because the current in these loops all points in the same direction, it reinforces the magnetic field so that if we have a situation where there are 𝑁 loops of current, where 𝑁 is an integer value, then the strength of the magnetic field at the center of all these loops reflects that change.

We take the strength of the field created by a single loop and multiply it by the number of loops there. So this is how we calculate the magnitude of the magnetic field at the center of a circular loop of current. But what about that field direction? We know, after all, that magnetic field is a vector, having magnitude and direction.

To figure this out, we use a rule known as the right-hand screw rule. If we take a right-handed screw and imagine the direction that we would need to turn this screw in order to drive it into some surface — say, a piece of wood — then by considering these two directions — the way we’re turning the screw and the direction of the screw as it moves into a surface — we can figure out the way that a magnetic field in the center of a current-carrying circle points. What we do is we orient the screw such that the direction we would have to turn it to drive it into something matches the direction of current in our circular loop.

So, for example, in the case of this loop here, we can see that from this perspective, current is moving counterclockwise, which means to figure out the direction of the magnetic field at its center, we would position a right-handed screw — like this one here — so that if we turn the screw in the direction of the current, counterclockwise, the screw would be driven into some surface. So right now, we’re looking at the pointy end of the screw. It’s pointing out of the screen at us. And that tells us the direction of the magnetic field at the center of this current-carrying loop.

So we could go here and say that 𝐵 points out of the screen at our particular point of interest at the very center of the circular loop of current. So applying the right-hand screw rule means taking such a screw and orienting it so that we can rotate it in the direction of the current 𝐼 and then, in so doing, the direction the screw would travel, if we were screwing it into some surface by doing this, indicates the way the magnetic field 𝐵 points. Now that we know about solving for field magnitude as well as direction, let’s get a bit of practice with these ideas through an example.

A circular loop of wire is carrying a constant current 𝐼 in a clockwise direction as viewed from above. The current creates a magnetic field. Based on the diagram, state the direction of the magnetic field at the center of the coil.

Okay, in this diagram, we see our circular loop of wire and also that it carries a current 𝐼 in a clockwise direction around this wire as we’re looking at it. We’re told that this current creates a magnetic field, and we want to solve for the direction of that field at the very center of the coil, at this point P. To figure this out, we can recall what’s known as the right-hand screw rule. This rule recognizes that if we have a right-handed screw — say, this one here — then if we were to turn that screw in the direction that would make it go into some surface — say, a bit of wood or metal — then if that turning direction can be made to match up with the direction of a current in a circular loop of wire, then the direction the screw would sink into or be driven into that surface gives the direction of the resulting magnetic field at the center of the circular loop.

So for a right-handed screw, the direction the screw turns can be made to match current direction. And in that case, the tip of the screw points in the direction of the resulting magnetic field, specifically the magnetic field at the center of a current-carrying circular loop. So then, over here on our diagram, if we took a right-handed screw and arranged it so that we would turn that screw in the direction of the current in this loop, then we can see that the tip of the screw, as well as the direction the screw would travel, would be into the screen. So then that’s our answer for the direction of the magnetic field created at point P.

We can symbolize that field direction this way or simply write that it points into the screen. And we knew this thanks to the right-hand screw rule.

Let’s look now at a second example exercise.

A circular loop of wire carries a constant current of 0.9 amperes. The radius of the loop is 13 millimeters. Calculate the strength of the magnetic field at the center of the loop. Give your answer in teslas expressed in scientific notation to one decimal place. Use a value of four 𝜋 times 10 to the negative seventh tesla meters per ampere for 𝜇 nought.

In this exercise, we have a circular loop of wire. And we’re told that it carries a constant current, we’ll call it 𝐼, of 0.9 amperes. Along with this, the radius of this circular loop, we can call it 𝑟, is given as 13 millimeters. What we want to do is calculate the strength or the magnitude of the magnetic field at the center of the loop. On our sketch, that’s right here at this point. We’ll call that magnetic field magnitude at that point capital 𝐵. And to solve for it, we can recall this relationship that that magnetic field strength at the center of a current-carrying circular loop equals a constant called the permeability of free space, that’s 𝜇 nought, times the current in the loop, 𝐼, all divided by two times the loop’s radius.

Since we’re told the specific value to use for this constant 𝜇 nought, and we also know the current 𝐼 as well as the radius 𝑟, we’re ready to substitute in to solve for 𝐵. Before we calculate this fraction, though, we’ll want to make one change to it. So long as the radius of our circular loop is in units of millimeters, it doesn’t match the other SI base units in this expression, for example, the unit of meters in 𝜇 nought. So let’s convert the radius of our circle from millimeters into meters.

To do that, we can recall that 1000 millimeters equals one meter, which means that 13 millimeters equals 0.013 meters. See that we’ve moved our decimal point three spots to the left. Now, we are ready to calculate the field strength 𝐵. And when we do, to one decimal place, we get a result of 4.3 times 10 to the negative fifth teslas. This is the magnitude or the strength of the magnetic field at the very center of our circular loop.

Let’s look now at one last example exercise.

A circular loop of wire with a radius of 9.5 centimeters carries a constant current of 𝐼 amperes. The strength of the magnetic field produced by the current is 5.2 times 10 to the negative fifth teslas at the center of the loop. Calculate 𝐼, rounding your answer to one decimal place. Use the value of four 𝜋 times 10 to the negative seventh tesla meters per ampere for 𝜇 nought.

So here, we have a circular loop of wire with a radius we’ve called 𝑟 given as 9.5 centimeters. And we’re told that the loop carries a constant current of 𝐼 amperes. So 𝐼 is some pure number. And this is a current value expressed in amperes. Due to this current, a magnetic field is produced at the center of this circular loop. If we call the magnitude of that field 𝐵, we’re told that it’s equal to 5.2 times 10 to the negative fifth teslas.

Knowing all this, we want to calculate the magnitude of the current 𝐼. To do this, we can recall that the magnetic field magnitude at the center of a current-carrying circular loop is equal to this constant 𝜇 nought, the permeability of free space, multiplied by the current magnitude in the loop divided by two times the loop’s radius. Now, in our statement, we’re given 𝜇 nought, we’re given 𝑟, and we’re also given 𝐵, wanting to solve for the current 𝐼.

Now, if we multiply both sides of this expression by two 𝑟 divided by 𝜇 nought, then over on the right-hand side, the factors of two cancel and the factors of 𝑟 and 𝜇 nought, leaving us just with the current 𝐼. So two times 𝑟 times 𝐵 over 𝜇 nought equals 𝐼. And if we substitute in the values we’re given for 𝐵, 𝑟, and 𝜇 nought, then we come up with this expression here.

We’re just about ready to calculate 𝐼. But before we do, let’s change the units of our radius, which are in centimeters, so that they match the SI base units in the rest of our expression. In other words, let’s convert our radius from centimeters into meters. 9.5 centimeters is 0.095 meters. So now, when we go ahead and calculate 𝐼, to one decimal place, we find a result of 7.9 amperes.

For our final answer, though, we’ll just box the number portion of this quantity because recall that our problem statement tells us we have a constant current of 𝐼 amperes. So to solve for 𝐼, we just want a number. And that result is 7.9.

Let’s now summarize what we’ve learned about the magnetic field due to a current in a circular loop of wire. In this lesson, we saw that the magnitude of the magnetic field produced at the center of a circular current-carrying loop is equal to this universal constant 𝜇 nought called the permeability of free space multiplied by the current magnitude 𝐼 in the loop divided by two times the radius of this circle.

We saw further that when there are 𝑁 identical current-carrying loops in a given coil, that in that case the magnetic field magnitude at the center of that coil is equal to the field magnitude created by a single one of the loops multiplied by their number.

And lastly, we learned that the direction of the magnetic field at the center of a circular current-carrying loop is given by what’s called the right-hand screw rule. Using this rule. If we turn a right-handed screw in the direction of the current in a circular loop, then the tip of the screw and the direction the screw moves tells us the direction of the magnetic field at the center of such a loop.

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