Question

A silver wire 2.7 mm in diameter transfers a charge of 431 C in 88 min. Silver contains 5.8 × 10²⁸ free electrons per cubic meter. What is the magnitude of the drift velocity of the electrons in the wire? (Give your answer in scientific notation using m/s as unit)

Answer 1

Given : d = 2.7 mm , Q= 431 C, t= 88 min., n= 5.8×1028 m-3

Solution:

Step:1

We know that , current through wire is given by : I = Q/t

I= 431 C/(88×60 s)= 0.0816 A

Step:2

Using the equation : I = nqvdA

Where vd is the drift velocity and A is the cross sectional area.

A = (\pi/4)d2 = (\pi/4)(2.7×10-3)2 = 5.72×10-6 m2

Now ,

0.0816A = (5.8×1028 m-3)(1.6×10-19C)(vd)(5.72×10-6 m2)

vd = 0.0816/(5.308×104) = 1.54×10-6 m/s

Answer: vd = 1.54×10-6 m/s

0.0816A = (5.8×1028 m-3)(1.6×10-19C)(vd)(5.72×10-6 m2)

vd = 0.0816/(5.308×104) = 1.54×10-6 m/s

Answer: vd = 1.54×10-6 m/s