A wire of diameter 0.02 meter contains 1028 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 × 10−29 m/s
B. 1 × 10−19 m/s
C. 5 × 10−10 m/s
D. 2 × 10−4 m/s
E. 8 × 103 m/s
(GR8677 #09)
Solution:
Drift velocity is the average velocity of a carrier that is moving under the influence of an electric field.
Velocity: v = L/t
In a wire with length L and cross sectional area A, there are n electrons with charge qe per cubic meter.
Total number of mobile electrons in the wire, Q = nqeLA
Current: I = Q/t = nqeLA/t = nqevA
v = I/nqeA
qe = charge of an electron = 1.6 × 10−19 C
n = 1028 electrons/m³
A = πr² = 0.5π × 10−4
I =100 Ampere
v = 102 / (1028× 1.6 × 10−19× 0.5π × 10−4)
= 102−28+19+4 / (1.6 × 0.5π)
≈ 10−4
Answer: D
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