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: Use the formula for drift speed: \[ v_d = \frac{I}{n A e} \] Where: - \( I \) is the current, - \( n \) is the charge carrier density, - \( A \) is the cross-sectional area, - \( e \) is the charge of an electron.
Step 2: Substitute the values: \[ v_d = \frac{5}{(5 \times 10^{22})(4 \times 10^{-6})(1.6 \times 10^{-19})} \]
Step 3: Final result: \[ v_d = 1.56 \times 10^{-3} \, m/s \]