The two wires in FIGURE P27.63 are made of the same material. ...

Question

The two wires in FIGURE P27.63 are made of the same material. What is the electron drift speed in the 2.0-mm-diameter segment of the wire?

Illustration of a wire segment showing electron drift speed and current, with labeled dimensions and values.

Accepted Answer

Welcome back everyone in this problem, we're considering two wires A and B in which the number of electrons flowing per unit of time is the same. The wires are identical in terms of material, but the diameter of wire A is 0.8 centimeters and of wire B is 1.6 centimeters. If in wire, A electron drift speed is 3.2 times 10 to the negative 4 m per second. We want to find the electron drift speed in wire B for our answer choices A is eight times 10 to the negative 5 m per second. B is four times 10 to the negative five C is 12 times 10 to the negative two and D is 16 times 10 to the negative two. And all of those are written in meters per second. Now, we're trying to find the electron drift speed in wire A and recall that the electron drift speed is the current in the wire at a microscopic level. So we're trying to figure out what that drift speed would be, what the velocity there would be. What do we know? Well, we already know that the wires are identical in terms of material and the electrons flowing per unit of time is the same. So since the electrons flowing per unit of time is the same, that means the current through wires A and B is the same. So in other words, we know that the current in wire A would be equal to the current in wire B, we already know the drift speed of wire A, we know that it's 3.2 times 10 to the negative 4 m per second. And we also know that the diameter of wire A is 0.8 centimeters. So that means its radius for wire is going to be 0.4 centimeters. And likewise, we know that the diameter of wire B is 1.6 centimeters. So that means its radius is going to be 0.8 centimeters. OK. Now let's think we have the current, right? And we want to find the drift speed in one of the wires. What links current and drift speed? Well, how about current density? Because remember current density is linked to the current in a wire and it's also linked to the drift speed. Recall that current density J is the current per square meter in a material. OK. But also recall that the current density equals the number of electrons in that material multiplied by the charge of an electron multiplied by the drift speed. So since both of these are here are equal, then that means we can link these together because now that tells us that current divided by cross sectional area equals ne multiplied by e multiplied by the drift speed. OK. Which tells us then that the current would be equal to the number of electrons. Oops, let me get that back. The number of electrons multiplied by the charge of an electron multiplied by the distribute multiplied by the cross sectional area of our, of our wire. Now, if that is true, and we have two wires here, wire A and wire B where as we said before, the currents in both are the same. Then since the current in A equals the current in B, we can create an equation here because then this tells us that the number of electrons multiplied by E multiplied by the drift speed in A multiplied by the area of A would be equal to and E multiplied by E multiplied by the drift speed of B multiplied by the cross sectional area of B. Now the number of electrons would be the same and we know the charge of an electron is constant. So if we simplify, that means the drift speed in A would be equal to, sorry, not the dr speed in A. Yes, the drif speed in A multiplied by the area of wire A equals the drift speed in B multiplied by the area of wire B. But remember we're solving for the drift speed of B. So again, we can transpose to get the juice speed of B being equal to the juice speed of A. OK. Multiplied by the cross sectional area of A divided by the cross sectional area of B. Now we're on to something OK. We're onto something because no, remember we want to solve for the GIF speed and we already know the GIF speed of A. Now we just need to figure out the area, the cross sectional area of wires A and B. But then again, let's think about it. Recall that the area for our wire would be equal to pi R squared. OK. And because we know the radius, OK, we know the radius already remember we know that the radius for A is 0.4 and the radius for B is 0.8 we can substitute that into our equation. Because if we replace A with pi R squared, that means we'll have pi multiplied by the radius of A squared divided by pi multiplied by the radius of B squared P is constant. So we can rewrite this as the juice speed of P being equal to the juice speed of a multiplied by the radius of A squared divided by the radius of B squared. Now that we have that we can go ahead and solve for the drift speed because we know all of these variables. So substituting those values into the equation, we'll have the drift speed of A which was 3.2 times 10 to the negative 4 m per second multiplied by the radius of A which is 0.4 centimeters or 0.4 times 10 to the negative 2 m squared. And all of that is going to be divided by the radius of B squared. So that's going to be 0.8 multiplied by 10 to the negative 2 m squared. No, we're almost there. All we need to do now is to put that into our calculator. Let me, I think I need to make some space. So let me draw a little box here. Now, when you work out what's going on right here, when you put that into your calculator, you'll find that you get the drift speed of wire B to be equal to eight times 10 to the negative 5 m per second. That would be the solution for the drift speed of B when we go back to our answer choices, that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

The two wires in FIGURE P27.63 are made of the same material. What is the electron drift speed in the 2.0-mm-diameter segment of the wire?

Key Concepts

Electron Drift Speed

Current Density

Ohm's Law

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