Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Overview
This chapter links electricity and magnetism — the field produced by currents (Biot–Savart, Ampère) and the forces magnetic fields exert on currents and moving charges.
1. Magnetic Field Due to Currents
Long straight wire: B = μ₀I/2πr
Centre of circular loop (N turns): B = μ₀NI/2R
Inside a solenoid: B = μ₀nI (n = turns per metre)
μ₀ = 4π × 10⁻⁷ T·m/A
2. Force on Currents and Charges
On a conductor: F = BIL sin θ
On a moving charge (Lorentz): F = qvB sin θ
Radius of circular path: r = mv/qB
3. Torque on a Current Loop
τ = NBIA sin θ (N turns, area A) — principle of the moving-coil galvanometer
Worked Examples
Example 1 — Field of a straight wire
B at 0.05 m from a wire carrying 10 A.
B = μ₀I/2πr = (4π×10⁻⁷ × 10)/(2π × 0.05) = (2×10⁻⁶)/0.05 = 4 × 10⁻⁵ T
Example 2 — Force on a conductor
A 0.2 m wire carrying 5 A is perpendicular to a 0.4 T field. Find the force.
F = BIL sin90° = 0.4 × 5 × 0.2 = 0.4 N
Example 3 — Lorentz force
Force on a 2 μC charge moving at 10⁶ m/s perpendicular to a 0.5 T field.
F = qvB = 2×10⁻⁶ × 10⁶ × 0.5 = 1 N
Example 4 — Solenoid field
Solenoid with 1000 turns/m carrying 2 A.
B = μ₀nI = 4π×10⁻⁷ × 1000 × 2 = 2.51 × 10⁻³ T
Common Mistakes
- Using 2πr for a loop centre (it is 2R) and vice-versa.
- Forgetting sin θ in force formulas.
- Confusing n (turns per metre) with total turns N in a solenoid.
- Wrong direction — use the right-hand / Fleming's left-hand rule.
Exam Tips & Revision Checklist
- Straight wire 2πr; loop centre 2R; solenoid μ₀nI — keep them distinct.
- Magnetic force does no work (always ⟂ to velocity).
- Practised: wire field, conductor force, Lorentz force, solenoid field.
Frequently Asked Questions
B = μ₀I/2πr, where I is the current and r the perpendicular distance. μ₀ = 4π × 10⁻⁷ T·m/A.