A) 5 A
B) 80 A
C) 8 A
D) 0.8 A
Correct Answer: D
Solution :
Drift velocity is defined as the average velocity that a particle attains due to an electric field. \[{{v}_{d}}=\frac{i}{neA}\] where i is current, n the number of electrons, e the electron charge and A the area of cross- section of wire. Number of electrons per unit volume \[n=\frac{2\times {{10}^{21}}}{A\times 100}\] Hence, current in the wire \[i=neA{{v}_{d}}\] \[i=\frac{2\times {{10}^{21}}}{A\times 100}\times 1.6\times {{10}^{-19}}\times A\times 0.25\] \[i=0.8\text{ }A\] Note: Since, particles can accelerate arbitrarily close to the speed of light in the absence of other forces the term drift velocity can only really apply to carriers in materials and not to particles in vacuum.
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