Drift velocity: Class 12 practice
Current I = neAv_d, where n is the free-electron density, e = 1.6 × 10⁻¹⁹ C, A the cross-section and v_d the drift velocity. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Charge and current
The idea behind drift velocity
Current I = neAv_d, where n is the free-electron density, e = 1.6 × 10⁻¹⁹ C, A the cross-section and v_d the drift velocity. Current density J = I/A = nev_d.
Step-by-step method
- Convert area to m².
- v_d = I/(neA).
- J = I/A.
Worked example 1 (Basic)
A 3 A current flows in a wire of cross-section 2 mm². Find the current density.
- J = 3/0.000002
- Answer: 1.50e+6 A/m²
Worked example 2 (Standard)
A copper wire (n = 8.5 × 10²⁸ m⁻³) of cross-section 1 mm² carries 4 A. Find the drift velocity.
- v_d = 4/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.000001)
- = 2.941e-4 m/s
- Answer: 2.941e-4 m/s
Worked example 3 (Challenge)
A copper wire (n = 8.5 × 10²⁸ m⁻³) of cross-section 3 mm² carries 2 A. Find the drift velocity and the time for an electron to drift 1 m.
- v_d = 2/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.000003)
- = 4.902e-5 m/s
- t = 1/v_d = 5.67 h
- Answer: 4.902e-5 m/s; 5.67 h
Common mistakes and misconceptions
- Using mm² as m²
- Confusing drift speed with signal speed
Practice questions
- (Basic) A 4 A current flows in a wire of cross-section 5 mm². Find the current density.
- (Basic) A 6 A current flows in a wire of cross-section 4 mm². Find the current density.
- (Standard) A copper wire (n = 8.5 × 10²⁸ m⁻³) of cross-section 3 mm² carries 3 A. Find the drift velocity.
- (Standard) A copper wire (n = 8.5 × 10²⁸ m⁻³) of cross-section 1 mm² carries 3 A. Find the drift velocity.
- (Challenge) A copper wire (n = 8.5 × 10²⁸ m⁻³) of cross-section 1 mm² carries 3 A. Find the drift velocity and the time for an electron to drift 1 m.
- (Challenge) A copper wire (n = 8.5 × 10²⁸ m⁻³) of cross-section 1 mm² carries 4 A. Find the drift velocity and the time for an electron to drift 1 m.
Answer key with full working
- Q1: J = 4/0.0000049999999999999996 → Answer: 8.00e+5 A/m²
- Q2: J = 6/0.000004 → Answer: 1.50e+6 A/m²
- Q3: v_d = 3/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.000003) → = 7.353e-5 m/s → Answer: 7.353e-5 m/s
- Q4: v_d = 3/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.000001) → = 2.206e-4 m/s → Answer: 2.206e-4 m/s
- Q5: v_d = 3/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.000001) → = 2.206e-4 m/s → t = 1/v_d = 1.26 h → Answer: 2.206e-4 m/s; 1.26 h
- Q6: v_d = 4/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.000001) → = 2.941e-4 m/s → t = 1/v_d = 0.94 h → Answer: 2.941e-4 m/s; 0.94 h
Syllabus note
This skill is listed under "Current Electricity" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is drift velocity in?
"Current Electricity", Class 12 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.