Question

A current of 5A is passing through a metallic wire of cross-sectional area \( 4 \times 10^{-6} \, m^2 \). If the density of the charge carriers in the wire is \( 5 \times 10^{22} \, m^{-3} \), the drift speed of the electrons will be:

• A. \( 1.56 \times 10^{-3} \, m/s \)
• B. \( 1.89 \times 10^{-3} \, m/s \)
• C. \( 2.42 \times 10^{-3} \, m/s \)
• D. \( 2.84 \times 10^{-3} \, m/s \)

Hint:

The drift speed can be found using the formula \( v_d = \frac{I}{n A e} \), which involves current, charge carrier density, cross-sectional area, and electron charge.

Answer 1

The Correct Option is A

Step 1: Drift velocity formula
The drift velocity is given by:
\[ v_d = \frac{I}{n A e} \]

Step 2: Substitute values
Given:
\( I = 5\,A,\ n = 5 \times 10^{22}\,m^{-3},\ A = 4 \times 10^{-6}\,m^2,\ e = 1.6 \times 10^{-19}\,C \)

\[ v_d = \frac{5}{(5 \times 10^{22})(4 \times 10^{-6})(1.6 \times 10^{-19})} \]

Step 3: Simplify denominator
\[ (5 \times 4 \times 1.6) = 32 \] \[ 10^{22} \times 10^{-6} \times 10^{-19} = 10^{-3} \] So denominator: \[ 32 \times 10^{-3} = 3.2 \times 10^{-2} \]

Step 4: Final calculation
\[ v_d = \frac{5}{3.2 \times 10^{-2}} = 156.25 \] \[ v_d \approx 1.56 \times 10^{2}\ m/s \]

Final Answer:
\( v_d \approx 1.56 \times 10^{2}\ m/s \)