Moving Charges and Magnetism is Chapter 4 of CBSE Class 12 Physics. This chapter explores the magnetic effects of electric current — the Biot-Savart law, Ampere’s circuital law, force on a moving charge in a magnetic field, and the force between current-carrying conductors. It forms the foundation for understanding electromagnets, motors, and many modern devices.
This chapter carries 7–8 marks. Biot-Savart law derivations, force on a current-carrying conductor, and cyclotron numericals are heavily tested.
Key Concepts
1. Magnetic Force on a Moving Charge
Lorentz Force: F = qv × B = qvB sin θ
- F = force on charge (N)
- q = charge (C)
- v = velocity of charge (m/s)
- B = magnetic field (Tesla, T)
- θ = angle between v and B
Force is maximum when θ = 90° and zero when θ = 0° or 180° (charge moving parallel to field).
Direction: given by Fleming’s Left-Hand Rule or the right-hand cross product rule.
Important: Magnetic force does no work on the charge (F ⊥ v always). It only changes direction, not speed.
Motion of a Charged Particle in Magnetic Field
| Angle (θ) | Path |
|---|---|
| 0° or 180° | Straight line (no force) |
| 90° | Circle (radius r = mv/(qB)) |
| Between 0° and 90° | Helix (spiral) |
Radius of circular motion: r = mv/(qB)
Time period: T = 2πm/(qB) — independent of velocity!
2. Biot-Savart Law
The magnetic field dB due to a small current element Idl at a point P at distance r is:
dB = (μ₀/4π) × (Idl × r̂)/r²
where μ₀ = 4π × 10⁻⁷ T·m/A (permeability of free space)
Magnetic Field Due to Common Configurations
| Configuration | Magnetic Field |
|---|---|
| Centre of circular loop (radius R, current I) | B = μ₀I/(2R) |
| On axis of circular loop (at distance x) | B = μ₀IR²/[2(R² + x²)^(3/2)] |
| Infinite straight wire (at distance r) | B = μ₀I/(2πr) |
| Inside a solenoid (n turns/length) | B = μ₀nI |
| Inside a toroid (N total turns, radius r) | B = μ₀NI/(2πr) |
3. Ampere’s Circuital Law
The line integral of B around any closed loop equals μ₀ times the total current enclosed:
∮ B · dl = μ₀I_enclosed
Useful when there is symmetry (infinite wire, solenoid, toroid).
4. Force on a Current-Carrying Conductor
F = Il × B = BIl sin θ
Direction: Fleming’s Left-Hand Rule
Force Between Two Parallel Current-Carrying Conductors
F/l = μ₀I₁I₂/(2πd)
- Parallel currents (same direction): attract
- Anti-parallel currents (opposite direction): repel
Definition of Ampere: 1 Ampere is the current which, when flowing through two infinite parallel conductors 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length.
5. Cyclotron
A device that accelerates charged particles to high energies using electric and magnetic fields.
- Particle moves in semicircles inside two D-shaped electrodes (dees)
- Magnetic field provides circular motion; Electric field accelerates at each half-turn
- Cyclotron frequency: ν = qB/(2πm) — independent of radius and speed
- Maximum energy: E = q²B²R²/(2m), where R is the radius of the dee
- Cannot accelerate electrons (too light — relativistic effects) or neutral particles
Important Definitions
| Term | Definition |
|---|---|
| Magnetic field (B) | Region where a moving charge or current experiences a force; unit: Tesla (T) |
| Lorentz force | Total force on a charge in E and B fields: F = qE + qv × B |
| Biot-Savart law | Gives the magnetic field due to a small current element |
| Ampere’s circuital law | ∮B·dl = μ₀I_enclosed around a closed loop |
| Cyclotron | Device that accelerates charged particles using crossed E and B fields |
Solved Examples
Example 1
A proton moves with velocity 5 × 10⁶ m/s perpendicular to a magnetic field of 0.2 T. Find the radius of the circular path. (mp = 1.67 × 10⁻²⁷ kg)
Answer: r = mv/(qB) = (1.67 × 10⁻²⁷ × 5 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.2) = 8.35 × 10⁻²¹/3.2 × 10⁻²⁰ = 0.26 m
Example 2
Find the magnetic field at the centre of a circular loop of radius 5 cm carrying 2 A current.
Answer: B = μ₀I/(2R) = (4π × 10⁻⁷ × 2)/(2 × 0.05) = 8π × 10⁻⁷/0.1 = 2.51 × 10⁻⁵ T
Example 3
Two parallel wires 10 cm apart carry currents of 5 A and 10 A in the same direction. Find the force per unit length.
Answer: F/l = μ₀I₁I₂/(2πd) = (4π × 10⁻⁷ × 5 × 10)/(2π × 0.1) = (200π × 10⁻⁷)/(0.2π) = 10⁻⁴ N/m (attractive)
Example 4
A solenoid of length 50 cm has 500 turns and carries 3 A. Find B inside.
Answer: n = 500/0.5 = 1000 turns/m. B = μ₀nI = 4π × 10⁻⁷ × 1000 × 3 = 3.77 × 10⁻³ T ≈ 3.8 mT
Important Questions for Board Exams
1-Mark
- State Biot-Savart law.
- What is the force on a charge moving parallel to a magnetic field?
3-Mark
- Derive the expression for magnetic field at the centre of a current-carrying circular loop.
- Using Ampere’s law, derive B inside a solenoid.
- Explain the working principle of a cyclotron.
5-Mark
- State Biot-Savart law. Derive the expression for B on the axis of a circular current loop.
- Derive the force per unit length between two parallel current-carrying conductors. Define the Ampere.
Quick Revision Points
- F = qvB sin θ; magnetic force ⊥ velocity → does no work
- Circular path: r = mv/(qB); T = 2πm/(qB) — independent of v
- Biot-Savart: dB = (μ₀/4π)(Idl sin θ)/r²
- Centre of loop: B = μ₀I/(2R); Straight wire: B = μ₀I/(2πr)
- Solenoid: B = μ₀nI; Toroid: B = μ₀NI/(2πr)
- Ampere’s law: ∮B·dl = μ₀I_enc
- F between wires: F/l = μ₀I₁I₂/(2πd); same direction → attract
- Cyclotron: ν = qB/(2πm); can’t accelerate electrons or neutrals
Previous: Ch 3 — Current Electricity
Next: Ch 5 — Magnetism and Matter
Chapter Navigation
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Next: Magnetism and Matter Class 12 Notes
Related Chapters in Class 12 Physics
- Current Electricity Class 12 Notes
- Magnetism and Matter Class 12 Notes
- Electromagnetic Induction Class 12 Notes
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