 Now let's take a look at some example problems finding the current density. I'm going to start with this form of the equation for the current density, where your current density is given by J, and it's the current per unit area. If they actually give you the values for the current in the area, it's pretty straightforward to plug them in. Let's say they gave you 10 amps, and they gave you 0.2 meters squared for your area. Well, then your current density would be 50 and is amps per meter squared. And there's no special reduction of that unit. It's just amps per meter squared. Now you might be able to write it depending on your professor as amps per meter squared. Just make sure it's really clear that the amps is divided by the meter squared. Of course, we could also rearrange this equation. And for simplicity, I'm going to actually use the same values here. If I was going to solve this equation for the current, then I would take the current density, J, and multiply it by the area. And you'll see in this example that, again, our 50 times our 0.2 gives us our 10 amps. So that works out the way it was before. And again, that's amps per meter squared times meters. So the meter squared is canceled out and leaves you with just amps for your unit. Checking units is a good way to make sure you've done the algebra correctly. The other way we could rearrange this equation would be to solve it for the area. And if we're solving this for the area, you would take your current divided by your current density. And that's going to give you your area in meter squared. Now, in my introduction video for the current density, I cautioned students to be very careful because we've got an A which stands for area, which is measured in meter squared, and a current which is measured in amps. Don't confuse the amps unit for the current with the A variable symbol for area. It'd be nice if we had more letters in the alphabet when we start doing algebra and physics, but it's definitely easier for the kindergartners to only have a few. Now, this is a good example problem of working with the current density, but we often have problems in a little bit of a different form. So rather than giving you the current and the area, a lot of times what you're given is something like a word problem saying that you have something like a 6 milliamp current in a wire with a diameter of 4 millimeters. So this involves prefixes as well as measurements that are given in word problems. So for your current, you would want to make sure that you turn that milliamps into amps by making it 10 to the minus 3. And here you actually have two steps. They give you the diameter as 4 millimeters, but what you're really gonna need to use in the calculation is the radius, which is then gonna be 2 millimeters, which is 2 times 10 to the minus 3rd meters. Now, it might be easy and tempting to go ahead and plug your current in on the top and try to plug in your radius on the bottom, but that's not the area. Your diameter is not the area either. So how are you gonna actually go through and find the area? And the clue here is the word diameter, because if you've got a diameter, it's pretty clear that they're gonna be talking about a circle. And so your area is pi r squared, and that's why I said you really wanna be using the radius. So for this particular case, if you wanna double check, make sure you understand how to do the math, the area would be pi times, again, the radius, which is half of the diameter, squared. And in this particular case, it comes out with this calculation. Again, remember to use your parentheses so that it squares the entire thing 2 times 10 to the minus 3. Once we've got that value for the area, then we can go ahead and plug it back into our current density formula, plugging in our current, our six milliamps, and now the area of that four millimeter diameter wire. And that's gonna give us our actual current density, which in this case comes out to be 476 amps per meter squared.