# Lesson Video: The Magnetic Field due to a Current in a Solenoid Physics

In this video, we will learn how to calculate the magnetic field produced by a current in a solenoid.

12:45

### Video Transcript

In this video, our topic is the magnetic field due to a current in a solenoid. We’ll see how this field is created, what it looks like, and we’ll also learn how to calculate the field strength within the solenoid.

To begin, we can recall that any current-carrying wire like this one here will produce a magnetic field around itself. And further, if we take the ends of this wire and we connect them together to form a loop, then that current-carrying circular loop also creates a magnetic field within as well as outside of itself. Now, if we take this loop of wire and we angle it a bit, so we’re looking at it almost side on, what happens — even though it’s a little hard to see with all the magnetic field lines — is that inside this loop of current-carrying wire, a net magnetic field to the left is created.

This is the result of one circular loop. But what if we had a number of loops like this, all with current pointed in the same direction? If that were so, then each individual loop would also contribute to this magnetic field pointing to the left inside the loops. The net result is a fairly strong and fairly uniform field pointed that way. Now, what if instead of being separate, these loops were all joined together into one continuous wire? If this were so, we would find that the magnetic field that’s created is essentially the same as that created by our several loops. The name we give to a wire that’s arranged this way is solenoid.

If we looked at a solenoid end on, it might just look like this, like a circle. But seen from the side, we see all the different turns or loops that are a part of this solenoid coil. When a solenoid carries current, as this one does, it indeed creates a strong magnetic field within the turns of the solenoid. We can represent that with these multiple field lines. And it also creates a magnetic field, albeit a weak one, outside the solenoid.

Now, considering this end-on view of the solenoid here, let’s say that we got that view by putting our eye over at this end. This means that if we could see the magnetic field within the turns of the solenoid, it would look something like this. Thanks to the total effect of all the different turns in the solenoid, the strength of the magnetic field in this cross section that we’re looking at right here is effectively constant. Whether we’re talking about the magnetic field strength here or here or here in our cross section or at any other point, that value is essentially the same.

There’s an equation to describe the strength of the magnetic field inside the loops of the solenoid. If we call the strength of that field 𝐵, then 𝐵 is equal to this constant, 𝜇 naught, which represents the permeability of free space — essentially, how easily magnetizable free space or a vacuum is — multiplied by 𝐼, the magnitude of the current in the solenoid, times capital 𝑁, where capital 𝑁 represents the number of loops or turns the solenoid has. And then all this is divided by the length of the solenoid along its axis referred to as capital 𝐿.

In the solenoid we’ve sketched on screen then, capital 𝑁 will be equal to one, two, three, four, five, six, seven, since we have seven turns in our solenoid. And 𝐿, the solenoid’s length, would be equal to this distance right here. Those are the values we would use in this equation to calculate the strength of the magnetic field inside the turns of the solenoid.

It’s worth mentioning that just as the magnetic field magnitude is the same in any given cross section of the solenoid’s interior, we’re also saying it’s the same in magnitude along the length of the solenoid. So the magnetic field strength here inside the turns of the solenoid is equal to the magnetic field strength here, say, also within the solenoid core. Basically, there’s this volume, we’ve shown it here in pink, and we’re saying that the magnetic field strength at any point within that volume is given by this equation here.

Now, if we rearrange this equation a bit so it reads 𝜇 naught times 𝑁 times 𝐼 over 𝐿, sometimes we’ll see this equation written in an alternative form where capital 𝑁 over 𝐿 is replaced by a lowercase 𝑛. When the equation is written this way, we say that it’s written in terms of the turn density of the loops in the solenoid. In other words, how many loops or how many turns there are per unit length. So, the equation for magnetic field can be written either way, and each form means the same thing. But this form, as we’ve seen, is a bit more condensed and that it uses this factor called the turn density. When we use lowercase 𝑛, it means that we don’t need to know specifically how many total turns a solenoid has or exactly how long it is. We’re just interested in the ratio capital 𝑁 to 𝐿.

So that’s how we calculate the magnitude of the magnetic field within the loops of a solenoid. And as we saw, within the turns of a solenoid, the direction of the field is always the same. The direction of the field will depend on the way the current points in the solenoid, but it will always point along the solenoid’s long axis, either to the left or to the right when we view a solenoid side on. Notice though what happens when we move outside these turns. In this case, if we follow a field line, say this one, we see that it starts to loop back in on itself and eventually forms a closed loop. By the way, this is true of magnetic field lines in general.

So, on a diagram showing us the magnetic field created by a solenoid, we can expect that whichever direction the field within the loops of the solenoid points, in this case that direction is to the left, the field lines outside the solenoid will point the opposite direction as they turn back on themselves. But then, once these field lines enter the solenoid loops, they do point in the same direction as all the other field lines. Knowing all this, let’s get a bit of practice with these ideas through an example.

A solenoid has a length of 3.2 centimeters and consists of 90 turns of wire. The wire carries a current of 1.2 amperes. Calculate the strength of the magnetic field at the center of the solenoid. 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 from 𝜇 naught.

Okay, so in this exercise, we’re working with a solenoid, which is a series of turns or loops of wire. Now, we’re told that our solenoid, has 90 turns, and we can see there definitely aren’t 90 here, but we can just pretend there are. So this solenoid has 90 turns, and we’ll say that’s represented by the letter capital 𝑁. Along with this, we know that the length of our solenoid, we’ll call it capital 𝐿, is 3.2 centimeters and also that a current, we’ll call it 𝐼, of 1.2 amperes is carried by this wire.

Because of this current, the turns in the solenoid will create a magnetic field around themselves. We want to calculate the strength of that cumulative or overall field at the center of the solenoid. So, that would be about at this point along its length. And if we were looking at the solenoid end on through its circular cross section, we would find that point right here. So it’s there that we want to solve for the strength of the magnetic field, and we’ll call that field strength capital 𝐵.

At this point, we can recall a mathematical relationship describing this magnetic field. The strength of the field at the center of the solenoid is equal to 𝜇 naught, a constant called the permeability of free space, multiplied by the total number of turns the solenoid has times the current that exists in this wire all divided by the overall length of the solenoid from end to end. In our problem statement, we’re told what 𝑁, 𝐼, and 𝐿 are, and we’re also given the value to use for 𝜇 naught. So, we can substitute those given values into our equation for 𝐵. So then, the value we use for 𝜇 naught is four 𝜋 times 10 to the negative seventh tesla meters per ampere. Capital 𝑁 is 90. The current 𝐼 is 1.2 amperes. And the length of the solenoid 𝐿 is 3.2 centimeters.

Before we calculate this magnetic field strength, notice that the units of amperes, A, here cancel out in our numerator and that also, in the numerator, we have these units of meters while in the denominator we have a distance unit in centimeters. We’re going to give our answer in teslas. We see that’s the unit right here represented by capital T. And to do this, we’ll need our distance unit to cancel out. To help that process along, let’s convert the length of our solenoid from centimeters into meters. 100 centimeters is equal to one meter. And therefore, 3.2 centimeters is 0.032 meters. And now, the distance units in our numerator and denominator, written in meters, will indeed cancel out and we’re left with units of teslas.

And when we calculate 𝐵, we get a result in scientific notation to one decimal place of 4.2 times 10 to the negative third teslas. This is the strength of the magnetic field at the center of solenoid.

Let’s look now at another example exercise.

A wire that carries a constant current of 0.9 amperes is formed into a solenoid of length 310 millimeters. The strength of the magnetic field at the center of the solenoid is measured to be 7.7 times 10 to the negative fourth teslas. Calculate the number of turns used to form the solenoid, giving your answer to the nearest whole number of turns. Use a value of four 𝜋 times 10 to the negative seventh tesla meters per ampere for 𝜇 naught.

In this example then, we have a solenoid that carries a constant current, we can call it capital 𝐼, of 0.9 amperes and whose length is 310 millimeters. And we’ll call this length capital 𝐿. The solenoid is made of an unknown number of turns. We’ll call that number capital 𝑁, and that’s what we want to solve for. To help us do this along with the current 𝐼 and the length 𝐿, we’re told the strength of the magnetic field at the solenoid center. We can call that field 𝐵, and its strength is given as 7.7 times 10 to the negative fourth teslas. In order to calculate capital 𝑁, the total number of turns in the solenoid, we’ll want to recall how this variable is related to the variables of magnetic field strength, current, and length.

The strength of the magnetic field at the center of a solenoid is given by 𝜇 naught, this constant called the permeability of free space, multiplied by the total number of turns in the solenoid times the current that exists in it all divided by the length of the solenoid along its axis. In our case, it’s not 𝐵 that we want to solve for but the number of turns, 𝑁. So to do that, let’s multiply both sides of the equation by 𝐿 over 𝜇 naught times 𝐼. Over on the right-hand side, this means that 𝐿, 𝜇 naught, and 𝐼 all cancel out. And we find that the number of turns in a solenoid is equal to its length times the magnetic field strength at its center divided by 𝜇 naught times the current in the solenoid 𝐼.

When it comes to the factors on the left-hand side of this expression, we’re given all four of them. We know 𝐿, 𝐵, and 𝐼. And we’re told to use a value of four 𝜋 times 10 to the negative seventh tesla meters per ampere for 𝜇 naught. Substituting all these values in, we find this expression for the number of turns 𝑁. Before we calculate this value though, let’s convert this length of our solenoid, which is currently in units of millimeters, to units of meters. To help us do that, we can recall that 1000 millimeters equals one meter, which means that to convert 310 millimeters to meters, we’ll shift our decimal place one, two, three spots to the left, giving us a result of 0.310 meters.

And now, let’s look at the units in the numerator and denominator of this fraction. We see, first of all, that these units of meters in numerator cancel with meters here. And then also, the unit of teslas cancels out from top and bottom, as does the unit of amperes because the ampere is in the numerator and denominator, we could say, of our overall denominator. So just like we would hope for, this result is going to be unitless because we’re calculating a pure number.

Now, when we enter this fraction on our calculator, we find that we don’t actually get a whole number result. This may happen practically due to the construction of a solenoid where, say, at the ends of the solenoid, one complete turn may not be finished. So, there’s nothing necessarily wrong that 𝑁 is not a whole number. But our question statement does tell us to round our result to the nearest whole number. When we do this, we find a result of 211. This is the number of turns this solenoid has, to the nearest whole number.

Let’s summarize now what we’ve learned about the magnetic field due to a current in a solenoid. In this lesson, we learned that a solenoid is a wire that’s arranged in a series of loops or turns. We saw further that when a solenoid carries an electric current, it creates a magnetic field. That field is essentially constant within the loops of the solenoid while outside those loops, the field is much weaker and of varying direction.

Within the loops of the solenoid, the strength of the magnetic field 𝐵 is given by 𝜇 naught, the permeability of free space, times the number of turns in the solenoid times the current magnitude that it carries all divided by its total length. We saw further that this is equal to 𝜇 naught times lowercase 𝑛 times 𝐼, where lowercase 𝑛 is the turn density. This is a summary of the magnetic field due to a current in a solenoid.