In this problem, we are going to look at the whole effect and the whole effect occurs in usually thin films. Um and they are actually a way to measure material properties inside of the film. But here we have a thin film of silver and it's carrying a fairly large current to the right of 100 and 20 amps with a magnetic field 200.95 Tesla, which is pointing perpendicular to the film. And that's kind of what we need is that magnetic field pointing perpendicular, so the direction of motion of the charges. So what happens, we could use the right hand rule with uh I. L. Crosby and we would figure out that positive charges would have a tendency to gather on the front part of the film. And the opposite charge sign would have a tendency to gather on the other side. And what this does is it sets up an electric field inside the film that points in this case from the back to the front. And those charges will continue to get built up until equilibrium occurs. And the equilibrium condition is that the electric force on moving charge would be equal to the magnetic force on that same moving charge and the drift velocity goes along with the current. So are velocity of those charges is perpendicular to be. Um And the electric field is perpendicular to both the V and the B. Um And we can write then the condition for the balance as E equals. Let's get the right color here. E magnitude wise is equal to V drift times B. Um Now a reminder that the magnitude of E is also proportional to the um voltage difference between plates which were kind of thinking of the two sides of the film as being plates divided by their spacing. And we see that the spacing is Z. One. Yeah. Um And that delta V. Is known as the hall voltage. And it's usually the parameter along with current that you measure while you're doing your experiment on the material. So we know the magnetic field. But in order to find that hall voltage, what will need to do is first of all find our drift speed salt for the electric field. And then we can solve for the hall voltage. So let's kind of take that in steps are. First step is to find the drift speed which is something we can relate to the properties of the material. And of course then the electric field will pop out and finally the hall voltage will be determined um from the electric field. So there is a relationship drift speed is related to the current in a fairly simple way, solve rate that equation down that the drift speed of the carriers in the material. And since this is a metal, we know that they are the electrons. But that drift speed is determined by the current divided by the number density of the carriers, number per cubic meter times the charge per carrier. And then cross sectional area. And in our slab, what's presenting the cross section is um fairly easy to determine kind of color. That in with a sharp truce has the area. So that's just a product of the Y one, C one. And um we know the charge on an electron of course, and the density. So we can quickly write down the numbers that we need in this equation. And since I'll put everything in S. I. Units, I know that the drift speed will come out in meters per second. Yeah. Okay, let me just get all these numbers in here. And then the area is um C one times Y one. So I'll get those numbers in there as well. But they multiply out to be 2.71 times 10 to the minus six in terms of square meters. And when I put that together, not surprisingly, the drift velocity inside of things like metals and wires is surprisingly small. Um it's usually of the order of a fraction of a meter per second. So you don't want to rely on electrons whizzing from one place to another inside of a material to make your circuit work. It's usually um the electric field which travels at the speed of light inside the material that causes your circuit to turn on and respond to changes anyway. So that was number one and then number two is find the electric field. So that is um 4.73 times 10 to the minus three times the 0.95 Tesla all in Si units. And so our electric field turns out to be quite small. Let's see 4.49 times 10 to the minus three. And I'll put this in volts per meter. And then finally our hall voltage is the electric field times see once in terms of magnitude. Anyway, um, and of course the side with positive charges gathering on them is going to be the higher potential side and get everything, get everything in S. I. Units and you're voltage will come out vaults. So it's a really typically a very small voltage. So in order to pick up the hall effect and detect it, you really need um, some careful measuring devices that can measure of the order of micro volts and block out other signals that could come in and ruin the experiment.