### Video Transcript

The current supplied to an air conditioner unit is 4.00 amps. The air conditioner is wired using a 10-gauge diameter 2.588 millimeters wire. The charge density in the wire is 𝑛 equals 8.48 times 10 to the 28th electrons per meter cubed. Find the magnitude of current density in the air conditioner’s wiring. Find the magnitude of the drift velocity in the air conditioner’s wiring.

We’re told in this statement the amount of current that runs through the wiring as well as the diameter of the wire and the charge density in the wire of 𝑛 equals 8.48 times 10 to the 28th electrons per meter cubed.

In part one, we want to solve for the magnitude of current density in the wiring. We’ll call that 𝐽. And in part two, we wanna solve for the magnitude of the drift velocity of the electrons in the wiring. We’ll call that 𝑣 sub 𝑑.

Let’s start our solution with a sketch of this wire. We have a current-carrying wire of diameter we can call 𝑑, giving us 2.588 millimeters. We’re told that the wire carries a current we can call 𝐼, equal to 4.00 amps. We’re also told that, based on the material the wires made of, it has a charge carrier density 𝑛 of 8.48 times 10 to the 28th electrons per cubic meter.

Based on all this, we want to solve for the current density 𝐽 and the electron drift velocity 𝑣 sub 𝑑. We can start by recalling a mathematical equation for current density. Current density 𝐽 is equal to the amount of current 𝐼 that flows past an area 𝐴. If we were to draw this in on our diagram, the area 𝐴 would be a cross section of the wire and 𝐼 will be the current flowing through the wire.

Since the area 𝐴 is circular, we’ll recall that the area of a circle is 𝜋 times its radius squared. In our statement, we’re given not radius but diameter. But we can still use that to calculate 𝐴. We replace 𝐴 in our equation with 𝜋 times 𝑑 over two quantity squared.

Since we know both the current 𝐼 and the wire diameter 𝑑, we’re ready to plug in and solve for 𝐽. When we do, we’re careful to write the diameter 𝑑 in units of meters. When we enter all these values on our calculator, we find that, to three significant figures, 𝐽 is 7.60 times 10 to the fifth amps per meter squared. That’s the current density in this wire.

Knowing that, we now want to solve for the drift velocity 𝑣 sub 𝑑. And to do so, we can recall another equation for current density 𝐽. In addition to being equal to current divided by area, 𝐽 equals the number density of charge carriers times the charge of each carrier multiplied by the drift velocity 𝑣 sub 𝑑. In our case, 𝐽 equals 𝑛 times the magnitude of the charge of an electron multiplied by 𝑣 sub 𝑑, the drift velocity of the electrons.

If we rearrange this equation to solve for 𝑣 sub 𝑑, we see it’s equal to 𝐽 over 𝑛 times the magnitude of the charge of an electron. That charge, recall, is equal to negative 1.6 times 10 to the negative 19th coulombs. 𝑛, the charge density, is given in the problem statement. And 𝐽 we solved for in the earlier part.

So we’re now ready to plug in and solve for 𝑣 sub 𝑑. When we do and enter these values on our calculator, we find that 𝑣 sub 𝑑 is 5.60 times 10 to the negative fifth meters per second. That’s the drift velocity of the electrons in this wire.