Painless Surge

Electricity and Magnetism Level 3

Consider a wire of length 4 m and cross-sectional area 1 mm 2 1 \text{ mm}^2 carrying a current of 2 A. If each cubic meter of the material contains 1 0 29 10^{29} free electrons, find the average time taken by an electron to cross the length of the wire.

Note: Put your answer in hours.


The answer is 8.9.

Rishabh Bhardwaj by Rishabh Bhardwaj , 25, India

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1 solution

Michael Ng
Sep 14, 2015

Use the electron flow equation I = n A v e I=nAve .

Therefore v = I n A e = 2 1 0 29 × 1 × 1 0 − 6 × 1.6 × 1 0 − 19 = 1 8000 v=\frac{I}{nAe}=\frac{2}{10^{29}\times1\times10^{-6}\times1.6\times10^{-19}}=\frac{1}{8000}

And so in order to travel 4 4 m, it takes 32000 32000 seconds and therefore 8.9 \boxed{8.9} hours.

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