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conantee
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« on: December 31, 2009, 05:19:33 AM »

9. A wire of diameter 0.02 meter contains 1 \times 10^{28} free electrons per cubic meter. For an electric current of 100 amperes, the drift velocity for free electrons in the wire is most nearly
(A) 0.6 \times 10^{-29} \mbox{ m/s}
(B) 1 \times 10^{-19} \mbox{ m/s}
(C)  5 \times 10^{-10} \mbox{ m/s}
(D)  2 \times 10^{-4} \mbox{ m/s}
(E)  8 \times 10^3 \mbox{ m/s}
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« Reply #1 on: December 31, 2009, 05:40:21 AM »

First, look at choices we have. The first two velocities are so slow (it takes more than year to pass the size of atom- 10^{-10} \mbox{ m}). Hence, (A) and (B) can be eliminated. If you are familiar with this subject, you know the correct answer right away since the order of each choice is so different.

However, if you forgot most of things, let's try some sort of logic and dimension tracking.
We know that the current is charge passing through a certain cross-section area per time. So, the dimension of current (ampere) is charge/time
Then, we consider electron current, so we can write current as = electron's charge x number of electrons/ time (dimension : 1/time)
Now, the number of electrons/ time relates to the density (1/length^3), cross-section area (length^2) and drift velocity (length/time).

Doing dimensional analysis, we have number of electrons/time = density x cross-section are x drift velocity

Therefore, we comes down to the relation I = e \rho A v_{d}
That is, v_d = \frac{I}{e \rho A}
Don't forget that the area is circle with diameter 0.02 meter, so A = \pi r^2 \approx 3 \cdot 0.01^2 = 10^{-4} \mbox{ m^2}. We can approximate as long as the order of magnitude does not change much.

Therefore, v_d = \frac{100}{1.6 \cdot 10^{-19} \cdot 10^{28} \cdot 10^{-4}} \approx 10^{2+19-28+4} \approx 10^{-3} \mbox{ m/s}

Hence, the correct answer is (D). 39 of 100 people got the right answer.
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