Video Transcript
A current of 1.4 amperes in a copper wire is carried by free electrons. The cross-sectional area of the wire is 2.5 times 10 to the negative six meters squared. Find the average speed at which free electrons pass through the wire. Use a value of 1.6 times 10 to the negative 19 coulombs for electron charge and a value of 8.46 times 10 to the 28 per cubic meter for the density of free electrons in copper. Give your answer in scientific notation to one decimal place.
Let’s say that this is a section of our copper wire and that here is a cross section of that wire whose area we’re given. In our copper wire, there are free electrons. In fact, there are an almost unbelievable number of free electrons. And these electrons moving along, say to the right, pass through this area and create a current in the wire. In this example, we want to solve for the average speed at which these free electrons pass along the wire. Considering that average speed, we can realize that if the average speed of these electrons is very small, then the current in the wire will be small as well. This is so because the current in a wire 𝐼 is equal to the amount of charge 𝑄 that passes through a cross section of that wire in some amount of time 𝑡.
So, if the speed of free electrons in the wire is relatively small, then the amount of charge 𝑄 that passes a cross section of the wire over some amount of time will also be small. And therefore, the current 𝐼 will be small too. On the other hand, the faster the average speed of these free electrons, the more current there will be in the wire.
The fact that we’re told how much current is in the wire, 1.4 amperes, in a sense sets the average speed for these free electrons. One ampere of current is equal to one coulomb of charge passing through a wire’s cross section in one second of time. The fact then that we have a current of 1.4 amperes in our copper wire means that we have 1.4 coulombs of charge passing through a cross section every second of time. One important fact to recall here is that each free electron in our wire contributes a certain amount of charge. Each of these electrons we’re told contributes 1.6 times 10 to the negative 19 coulombs of charge.
This means that if we want to know the number of free electrons that pass through a cross section of our wire in every second of time, and let’s say we call this number of free electrons capital 𝑁, then 𝑁 is equal to 1.4 coulombs, the total amount of charge passing through a cross section in one second, divided by the charge of a single free electron, 1.6 times 10 to the negative 19 coulombs. Notice that in this fraction the units of coulombs cancel out and our result is unitless.
Getting back to our sketch of the copper wire, let’s say, just for argument’s sake, that this volume of the wire is equal to one cubic meter. This would mean that the wire is very thick. But let’s just think this way for now. In this one cubic meter of the wire, there are 8.46 times 10 to the 28 free electrons. That’s over a billion, billion, billion free electrons. And all these free electrons we know are spread out over our volume. Let’s say that we want to figure out how many free electrons there are in a one-meter-long length of our wire. In other words, if this say is one meter of our wire, how many free electrons are there in this stretch of wire? We can figure this out by taking our free electron density, which is the number of free electrons that occupy a cubic meter of our wire, and multiplying this by the wire’s cross-sectional area.
Notice what happens to the units when we do this. Two factors of meters cancel from denominator and numerator. And we end up with a number of free electrons that are in one meter of length of the wire. At this point, let’s clear some space on screen so we can expand on our sketch. Let’s say that this is an expanded version of our copper wire with one meter marked out along its length. In that one meter, there are this many free electrons. To create the given current though, 1.4 coulombs of charge passing our cross section in one second of time, the number of free electrons we need to pass through our cross section is this number.
The number 𝑁 determines how fast all of these free electrons are moving on average so that the right number passes through our cross section every second. What we’ll do is we’ll take 𝑁 electrons per one second of time, and we’ll divide that by the number of electrons there are in one meter of wire. Notice that for this value, we have units of inverse meters. In our earlier designation, we specified electrons per meter, but we can note that electrons aren’t a unit per se and that we included this symbol just to clarify what particle we were considering in our wire. The actual units of this product then are simply inverse meters, as we’ve written up here.
The reason we’ve taken this number of electrons per second of time and divided it by this number of electrons per meter of wire is that if we multiply both the numerator and denominator by units of meters canceling that unit in the denominator, we get a result with units of meters per second. This is promising because recall that we want to solve for a speed of our electrons.
Our claim then is that if we take 𝑁 electrons, where 𝑁 equals 1.4 divided by 1.6 times 10 to the negative 19, and we divide that by the number of electrons in one meter of length of our wire, then that will give us the average speed in units of meters per second of the free electrons in the wire. Calculating this value to one decimal place in scientific notation, we get a result of 4.1 times 10 to the negative five meters per second. This is the average speed the free electrons in the wire must have in order to create a current of 1.4 amperes. Note also how small this average speed is; it’s less than one millimeter every second.
