Question

A copper wire with a cross-sectional area of $2 \times 10^{-6}$ m$^2$ has a free electron density equal to $5 \times 10^{22}$/cm$^3$. If this wire carries a current of $16$ A, the drift velocity of the electron is}

• A. $1$ m/s
• B. $0.1$ m/s
• C. $0.01$ m/s
• D. $0.001$ m/s
• E. $0.0001$ m/s

Hint:

Always convert number density from cm$^{-3}$ to m$^{-3}$ by multiplying by $10^6$.

Answer 1

Answer: (E)

Concept:
Drift velocity: \[ v_d = \frac{I}{nqA} \]

Step 1: Convert number density to SI units.
\[ n = 5 \times 10^{22} \text{ cm}^{-3} = 5 \times 10^{28} \text{ m}^{-3} \]

Step 2: Substitute values.
\[ I = 16,\quad q = 1.6 \times 10^{-19},\quad A = 2 \times 10^{-6} \] \[ v_d = \frac{16}{(5 \times 10^{28})(1.6 \times 10^{-19})(2 \times 10^{-6})} \]

Step 3: Calculate denominator.
\[ 5 \times 1.6 \times 2 = 16 \] \[ 10^{28} \times 10^{-19} \times 10^{-6} = 10^3 \] \[ v_d = \frac{16}{16 \times 10^3} = \frac{1}{10^3} = 10^{-3} \text{ m/s} \]