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A 3.0×10−4 V/m electric field creates a 2.1×1017 electrons/s current in a 2.0-mm-diameter aluminum wire. Part...

A 3.0×10−4 V/m electric field creates a 2.1×1017 electrons/s current in a 2.0-mm-diameter aluminum wire.

Part A

What is the drift speed?

Part B

What is the mean time between collisions for electrons in this wire?

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Answer #1

The expression for the drift velocity is as follows nA Now the radius becomes: 2.00 mm 2.00 mm 107 m 2 1.00 mm -1x10 m And area is, A (3.14) (1x10 m) 3.14x10-6 m2 Therefore, the drift velocity is, 2.1x10 electrons/s 6.00×1028 m-3)(3.14×10-6 m2 -0.112 x10-3 m/s -1.12x106 m/s The expression for the mean time between the Collison is as follows: eE 9.1x10-31 kg) (1.12x10-* m/s ( 1.6×10-19 C)(3.0×10-4 V/m ) 21.23x10-15 s

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