Finding the drift speed of a conduction electrons

[Omar FTM]

Homework Statement

The figure shows wire section 1 of diameter 4R and wire section 2 of diameter 2R, connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire's width. The electric potential change V along the length L = 1.95 m shown in section 2 is 13.5 µV. The number of charge carriers per unit volume is 8.49x10^28 m-3. What is the drift speed of the conduction electrons in section 1?

Homework Equations

v = j/nq
j=E/p = v/pL

The Attempt at a Solution

I got J2 using the provided v2 and p ( of copper = 1.72e-8) and the L2
J2 = (13.5e-6)/(1.72e-8)(1.95) = 402.5 A/m^2
Then I used the relation between the two sections > J1A1 = J2A2 >>> J1(0.25Pi 4^2 ) = J2(0.25Pi 2^2) >>> j1 = 100.625 A/m^2
The drift speed of sec 1 would be >>> V1 = J1 / nq = 100.625/(8.49e28)(1.6e-19) = 7.41e-9 m/s ( which was a wrong answer ) what is my mistake ?

 

[TSny]

I don't see any errors. Be sure that you are using the values for n and ρ from your textbook rather than from the internet. The values could vary a little depending on the source of information.

 

[Omar FTM]

I double checked my p and n , they are correct ( from the book (copper = 1.72e-8 ) , n from the question itself = 8.49x10^28 ) , but it's still telling me that my answer is wrong :(

 

[TSny]

I still can't find any error in your calculation. Maybe we're overlooking something that someone else will catch.

 

[Omar FTM]

Any one ? :)

 

[TomHart]

I am not real familiar with this kind of problem, but the math looked right. I did find an example on Wiki (under "Drift Velocity"), and because of the negative charge of an electron, the result came out to be negative. That's the only possibility I can come up with.

 

[TSny]

I am not real familiar with this kind of problem, but the math looked right. I did find an example on Wiki (under "Drift Velocity"), and because of the negative charge of an electron, the result came out to be negative. That's the only possibility I can come up with.

That's a thought. But speed is the magnitude of velocity, so I would think the answer should be positive. Anyway, thanks for the input, Tom.

 

[TomHart]

But speed is the magnitude of velocity

Yep, good point.

 

[ehild]

Maybe, they meant mV instead of μV. It was a wire almost 2 m long!

 

[TSny]

Maybe, they meant mV instead of μV. It was a wire almost 2 m long!

Yes, could be. I had noticed the current density and drift speed were coming out very small. For j = 400 A/m² and a wire of cross-sectional area 0.5 mm², the current is only 0.20 mA.

 

[Omar FTM]

I get your point , but this is how the question is :D
I think that I won't be able to solve it ...

 

[ehild]

Yes, could be. I had noticed the current density and drift speed were coming out very small. For j = 400 A/m² and a wire of cross-sectional area 0.5 mm², the current is only 0.20 mA.

Yes, the set-up and data for such a practically usual problem should be "real" , with a common voltage source and voltage measurable with a common voltmeter, multimeter. The smallest range of a common multimeter is 100-200 mV.
@Omar FTM Your solution is correct. It is quite possible that it was a misprint in the text of the problem.

 

[Omar FTM]

Tried answering using V as mV instead of micro V , still wrong :(

 

[ehild]

Tried answering using V as mV instead of micro V , still wrong :(

It is also possible that the problem writers made some mistake. It happens quite often.