Determining current and drift speed

[octahedron]

[SOLVED] Determining current and drift speed

Homework Statement

http://img162.imageshack.us/img162/145/problemrg5.jpg

The two wires in the above figure are made of the same material. What are the current and the electron drift speed in the 2.0mm-diameter segment of the wire.

Homework Equations

I = J/A; J = n*e*v_d where n is the density of the conductor in m^-3 and v_d is drift velocity.

The Attempt at a Solution

I was able to find the density of the conductor:

$$n = \frac{I_{1}}{(Aev_{d_1})} = 2*10^{28}$$

Also, I reached to this relationship between $$I_2$$ and $$v_{d_2}$$:

$$I_{2} = 20000v_{d_2}$$

What to do next? Appreciate any pointers, and thanks in advance.

 

[octahedron]

Ah, again, I_2 = I_1. *smacks head* This solves the problem.

 

[Moetasim]

To determine the current and electron drift speed in the 2.0mm-diameter segment of the wire, we can use the equations provided in the problem. First, we can use the equation I = J/A, where I is the current, J is the current density, and A is the cross-sectional area of the wire. We know that the two wires have the same material and the same diameter, so we can assume they have the same cross-sectional area. Therefore, we can set the two current densities equal to each other:

J_{1} = J_{2}

Next, we can use the equation J = n*e*v_d, where n is the density of the conductor, e is the charge of an electron, and v_d is the drift velocity. We can substitute this equation into our previous equation to get:

n_{1}*e*v_{d_1} = n_{2}*e*v_{d_2}

Since we know the density of the conductor for wire 1 (n_{1}) and we have a relationship between the current and drift velocity for wire 2 (I_{2} = 20000v_{d_2}), we can solve for v_{d_2}:

v_{d_2} = \frac{I_{2}}{20000}

Finally, we can use the equation I = J/A to solve for the current in the 2.0mm-diameter segment of the wire:

I_{2} = J_{2}*A = n_{2}*e*v_{d_2}*A

Substituting in our values, we get:

I_{2} = (2*10^{28}*e*\frac{I_{2}}{20000})*(3.14*10^{-6}) = 0.628*I_{2}

Solving for I_{2}, we get:

I_{2} = 1.59*10^{-6} A

Therefore, the current in the 2.0mm-diameter segment of the wire is 1.59*10^{-6} A and the electron drift speed is 1.59*10^{-6}/20000 = 7.95*10^{-11} m/s.