Question Video: Finding the Density of Free Electrons in a Material Physics

Video Transcript

A current of 77 milliamperes in a conducting wire of an unknown material is carried by free electrons. The cross-sectional area of the conductor is 1.5 times 10 to the negative six meters squared. Find the density of free electrons in the material if the average speed of the free electrons along the wire is 0.18 millimeters per second. Use a value of 1.6 times 10 to the negative 19 coulombs for the electron charge. Give your answer in scientific notation to one decimal place.

We are being asked in this question about a conducting wire, and let’s suppose that this here is a section of that wire. We are told that this wire has a cross-sectional area, which we’ve labeled as 𝐴, equal to 1.5 times 10 to the negative six meters squared. Now in this wire, there’s a large number of free electrons, which we’ve represented here as blue dots. These electrons will move along the wire. And as we’ve drawn it here, they’re moving to the right. We’re told that their average speed is 0.18 millimeters per second, which we’ve labeled as 𝑣. As these electrons move along to the right, they pass through this cross-sectional area 𝐴.

Since electrons are charged particles and in fact we’re given a value of 1.6 times 10 to the negative 19 coulombs for the electron charge 𝑒, then the motion of these electrons will create a current in the wire. We are told that in this case that current is equal to 77 milliamperes, and we’ve labeled it as 𝐼. Let’s recall that electric current is defined as the rate of flow of electric charge over time. Mathematically, if an amount of charge 𝑄 passes through a cross section of a wire over a time of 𝑡, then the current 𝐼 is equal to 𝑄 divided by 𝑡.

Now, the total amount of charge through a cross section in a given amount of time, that’s 𝑄, is equal to the charge of each electron 𝑒 multiplied by the number of electrons passing through the cross section during that time. And we’ve labeled that number here as capital 𝑁. If we divide both sides of this equation by the electron charge 𝑒, then we can cancel the 𝑒 in the numerator and the 𝑒 in the denominator on the right-hand side. We have then that 𝑁 is equal to 𝑄 divided by 𝑒. If we now take this equation, which relates the current 𝐼 and the charge 𝑄, and we multiply both sides of it by the time 𝑡 so that on the right-hand side the 𝑡’s cancel each other out, then we have that the charge 𝑄 passing through a cross-sectional area in a time of 𝑡 is equal to the current 𝐼 multiplied by that time 𝑡.

We can then use this equation to replace 𝑄 by 𝐼 multiplied by 𝑡 on the right-hand side of this equation here. If we do this, we have that 𝑁, the number of electrons passing through the cross-sectional area in a time of 𝑡, is equal to 𝐼 multiplied by 𝑡 divided by the electron charge 𝑒. If we then divide both sides of the equation by 𝑡 so that the 𝑡’s on the right-hand side cancel out, we have that 𝑁 over 𝑡 is equal to 𝐼 over 𝑒. We can notice that on the left-hand side, 𝑁 over 𝑡 is just the number of electrons passing through the cross section with area 𝐴 per second of time.

Let’s remember that the reason these electrons are passing through this cross section is that they’re moving to the right with an average speed that we’ve labeled as 𝑣. The number of free electrons per second that are passing through the cross section of the wire must be equal to the number of these free electrons that there are per unit length of the wire multiplied by the average speed with which these free electrons are moving in the direction that takes them through this cross section. In our case, we know that this average speed is the value of 0.18 millimeters per second, which we labeled as 𝑣.

We can also notice that the number of free electrons per unit length of the wire must be equal to the number of free electrons per unit volume multiplied by the wire’s cross-sectional area 𝐴. Then, the number of free electrons per unit volume is just equal to the density of free electrons in the material, which is what we are asked to find in this question.

Let’s label this free electron density, that is, the number of free electrons per unit volume, with a lowercase 𝑛. So then, this electrons per volume here is just equal to this lowercase 𝑛, the free electron density. This means then that the number of free electrons per unit length is equal to 𝑛, the free electron density, multiplied by 𝐴, the cross-sectional area. We have then that the number of free electrons per second passing through the cross-sectional area 𝐴 is equal to 𝑛 multiplied by 𝐴 multiplied by 𝑣, the average speed.

Then, let’s recall that this term on the left-hand side, the number of electrons passing through the cross section per second, is equal to this quantity capital 𝑁 divided by 𝑡. We have then that capital 𝑁 divided by 𝑡 is equal to lowercase 𝑛, the free electron density, multiplied by 𝐴, the wire’s cross-sectional area, multiplied by 𝑣, the average speed of the free electrons. We can then use this equation in order to replace the capital 𝑁 over 𝑡 with lowercase 𝑛 times 𝐴 times 𝑣 on the left-hand side of this equation here. When we do this, we find that lowercase 𝑛 times 𝐴 times 𝑣 is equal to the current 𝐼 divided by the electron charge 𝑒.

Notice that by using this equation here to replace the capital 𝑁 divided by 𝑡 in this equation, we’ve managed to get ourselves an equation here in which we know the values of all of the quantities except for one. On the right-hand side of the equation, we know the current 𝐼 in the wire and we know the value of the electron charge 𝑒. Then, on the left, we know the wire’s cross-sectional area 𝐴 and we know the average speed 𝑣 of the free electrons.

The only unknown quantity is this lowercase 𝑛, the free electron density. And that’s what we’re trying to find out. That means that we want to make 𝑛 the subject of the equation. And to do that, we divide both sides by 𝐴 and 𝑣. On the left-hand side, we see then that the 𝐴s and the 𝑣’s in the numerator and denominator cancel each other out. This gives us an equation that says the free electron density lowercase 𝑛 is equal to the current 𝐼 divided by the electron charge 𝑒, the cross-sectional area 𝐴 of the wire, and the average speed 𝑣 of the free electrons. This equation we’ve derived here is going to be the key to answering this question because it tells us how to work out the thing we want to find in terms of four quantities that we know the values of.

Let’s now clear some space on the board so that we can make use of this equation.

Before we substitute our values into the right-hand side of this equation, we want to check that all the quantities have consistent units. On the left-hand side, this lowercase 𝑛 is the density of free electrons. That is, it’s the number of free electrons per unit volume. The SI unit for 𝑛 will therefore be units of per cubic meter, which we can write as meters to the negative three. To calculate a value of 𝑛 in this SI unit of meters to the negative three, we’ll need all of the quantities on the right-hand side to be expressed in their own respective SI units.

Our value for the electron charge 𝑒 is expressed in units of coulombs, which is indeed the SI unit for electric charge. The cross-sectional area 𝐴 is given in meters squared, which is the SI unit for area. However, the speed 𝑣 is expressed in millimeters per second, and the current 𝐼 is expressed in milliamperes. That means we need to convert these values into units of meters per second and amperes, respectively. To do this, we can notice that in both cases, we’ve got this unit prefix of lowercase m or milli-, which means a factor of one over 1000. That means that to convert from millimeters per second to meters per second, we’ll need to divide by a factor of 1000. And similarly, to convert from milliamperes to amperes, we’ll again need to divide by 1000.

We have then that the speed 𝑣 is equal to 0.18 divided by 1000 meters per second, and the current 𝐼 is equal to 77 divided by 1000 amperes. The speed works out as 0.00018 meters per second, while the current works out as 0.077 amperes.

We are now ready to take our values for the current 𝐼, the electron charge 𝑒, the cross-sectional area 𝐴, and the speed 𝑣 and substitute them into this equation. When we do that, it gives us this expression here for the free electron density 𝑛. That’s the current of 0.077 amperes divided by the electron charge of 1.6 times 10 to the negative 19 coulombs, the wire’s cross sectional area of 1.5 times 10 to the negative six meters squared, and the average speed of the free electrons of 0.00018 meters per second.

Since all these quantities on the right are in their own SI units and we know that the value of 𝑛 we’ll calculate will have units of meters to the negative three, that means that on the right-hand side, we can replace all the individual units with these overall resulting units of meters to the negative three. Then, if we evaluate this expression by typing it into our calculator, we get a result of 1.7824 et cetera times 10 to the 27 meters to the negative three.

We are told in the question to give our answer in scientific notation to one decimal place. This value that we’ve calculated is already expressed in scientific notation, so we just need to round it to one decimal place. To one decimal place, the result rounds up to 1.8 times 10 to the 27 meters to the negative three. Our answer then is that the density of free electrons in the material is equal to 1.8 times 10 to the 27 meters to the negative three, or 1.8 times 10 to the 27 per cubic meter.