physics question...?

needhelp!

2007-09-26 6:19 am

An electrical conductor designed to carry large currents has a square cross section 2.00 mm on a side, and is 12.0 cm long. The resistance between its ends is 0.0625 ohms. What is the resistivity of the material? If the electric field magnitude in the conductor is 1.28 V/m, what is the total current? If the material has 8.5×10^16 free electrons/m3, find the average drift speed

回答 (1)

[匿名]

2007-09-26 6:45 am

✔ 最佳答案

You divide the resistance by the length and multiply by the cross-sectional area to get the resistivity in ohm-cm. Don't forget to convert mm into cm. The voltage is 1.28V/m x 0.12m = 0.1536 volts. Ohm's law says that the current I= V/R= 0.1536/0.0625= 2.4576 amps. I amp is 6E21 electrons/sec, so 2.4576 amps = 1.47456E22 electrons/ sec. Your conductor has a volume of 0.002x0.002x0.12=4.8E-7 cu.m., so it contains 4.08E10 electrons. So it will take 4.08E10/1.47456E22=2.76 E-12 secs for all of them to move out of the conductor, so the drift velocity is 0.12/2.76E-12=4E10 m/s. Something's wrong here. Can you spot any errors I've made? I can't.

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