Question

There are $8.4\times {10}^{22}$ free electrons per $c{m}^3$ in copper, the current in the wire is $0.21A$ ($e=1.6\times {10}^{-19}C$). Then the drifts velocity of electrons in a copper wire of $1m{m}^2$ cross section will be:-

• A. $1.56\times {10}^{-5}m/s$
• B. $2.12\times {10}^{-5}m/s$
• C. $0.78\times {10}^{-5}m/s$
• D. none of these

Answer 1

Answer: (A)

Number of free electrons $n=8.4\times {10}^{22}\times {10}^6$ per $m^3=8.4\times {10}^{28}$ per $m^3$

Current in the wire $I=0.21$ A

Cross-section of the wire $A=1m{m}^2={10}^{-6}{m}^2$

Thus drift velocity of electrons $v=\frac{I}{neA}=\frac{0.21}{(8.4\times {10}^{28})\times (1.6\times {10}^{-19})\times \left({10}^{-6}\right)}=1.56\times {10}^{-5}m/s$