Question

There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is 1 $m m ^ { 2 }$ . If the number of free electrons per $c m ^ { 3 }$ is $8.4 \times 10 ^ { 22 }$ , then the drift velocity would be

• A. 1.0
• B. 1.0
• C. 0.1
• D. 0.01

Answer 1

Answer: (C)

0.1

Answer 2

$v _ { d } = \frac { i } { n A e } = \frac { 1.344 } { 10 ^ { - 6 } \times 1.6 \times 10 ^ { - 19 } \times 8.4 \times 10 ^ { 22 } }$

$= \frac { 1.344 } { 10 \times 1.6 \times 8.4 } = 0.01 \mathrm {~cm} / \mathrm { s } = 0.1 \mathrm {~mm} / \mathrm { s }$