Question

The drift velocity of electrons in a silver wire with cross-sectional area 3.14×10⁻6m² carrying a current of 20A is. Atomic weight of Ag = 108, density of silver = 10.5×10³kg m⁻3.

• A. \(2.798\times10^{-4}\,\text{m/s}\)
• B. \(67.98\times10^{-4}\,\text{m/s}\)
• C. \(0.67\times10^{-4}\,\text{m/s}\)
• D. 6.798×10⁻4m/s

Hint:

Drift velocity is very small because: vd=(I)/(nqA) Large number density of electrons reduces vd.

Answer 1

The Correct Option is D

Step 1: Number density of free electrons (Ag is monovalent): n=(ρ NA)/(M) =(10.5×10³)/(0.108)×6.02×10²3 ≈5.85×10²8m⁻3
Step 2: Drift velocity formula: vd=(I)/(nqA)
Step 3: Substituting values: vd=\frac20(5.85×10²8)(1.6×10⁻19)(3.14×10⁻6) ≈6.8×10⁻4m/s