Magnetostatics

Magnetostatics describes steady magnetic fields from currents — inductance, forces and boundary conditions. GATE EC tests Ampere's/Biot-Savart laws and B-field boundary conditions (1–2 marks).

Key results

• Biot-Savart: dB = (μI dl × r̂)/(4π r²). Ampere's law: ∮ H·dl = I_enclosed; ∇×H = J.
• No magnetic monopoles: ∇·B = 0; ∮B·dS = 0. B = μH.
• Infinite wire: H = I/(2πr). Solenoid: B = μnI (n turns/length). Toroid: H = NI/(2πr).
• Force: F = qv×B (on charge); F = IL×B (on wire). Torque on loop = m×B (m = NIA).
• Boundary: normal B continuous (B_n1 = B_n2); tangential H differs by surface current (H_t1 − H_t2 = K).
• Inductance L = NΦ/I; energy = ½LI²; energy density = ½μH² = B²/2μ.

Quantity	Expression
Ampere's law	∮H·dl = I_enc
Infinite wire H	I/(2πr)
Solenoid B	μnI
No monopoles	∇·B = 0
Energy density	½μH²

Exam Tricks & Tips

• 🎯 Use Ampere's law with symmetry (∮H·dl = I_enc) to get H around wires/solenoids/toroids without Biot-Savart.
• 🎯 Normal B is continuous across a boundary; tangential H jumps by the surface current density K.
• 🎯 ∇·B = 0 always (no magnetic monopoles) — magnetic flux lines close on themselves.
• 🎯 Solenoid field B = μnI is uniform inside and ~0 outside (n = turns per unit length).
• 🎯 Magnetic energy density = ½μH² = B²/(2μ) — analogous to ½εE² in electrostatics.
• ❌ Common mistake: swapping the B and E boundary conditions — for B it's the NORMAL component that's continuous (opposite of D's normal jump only via charge).

Expected exam pattern

1–2 marks: compute H/B from Ampere's law, force on a charge/wire, inductance/energy, or apply magnetic boundary conditions.

Quick recap

Ampere: ∮H·dl=I_enc, B=μH. Wire H=I/2πr; solenoid B=μnI. ∇·B=0 (no monopoles). Force F=qv×B, IL×B. Boundary: B_n continuous, H_t jumps by K. Energy density ½μH².

Cover the answer, recall, then check. 12 cards on magnetostatics for GATE EC.

Q1. State Ampere's circuital law.
A1. ∮ H·dl = I_enclosed; differentially ∇×H = J.

Q2. State the Biot-Savart law.
A2. dB = (μ·I dl × r̂)/(4π r²).

Q3. Why is ∇·B = 0?
A3. There are no magnetic monopoles; magnetic flux lines are closed loops.

Q4. Magnetic field of an infinite straight wire?
A4. H = I/(2πr) (B = μI/(2πr)), circling the wire.

Q5. Magnetic field inside a long solenoid?
A5. B = μnI (n = turns per unit length), uniform inside, ~0 outside.

Q6. Force on a moving charge and on a current-carrying wire?
A6. F = qv × B (charge); F = IL × B (wire).

Q7. Boundary condition on normal B?
A7. It is continuous: B_n1 = B_n2.

Q8. Boundary condition on tangential H?
A8. H_t1 − H_t2 = K (the surface current density).

Q9. Magnetic energy density?
A9. w = ½μH² = B²/(2μ) (J/m³).

Q10. Inductance in terms of flux linkage?
A10. L = NΦ/I; stored energy = ½LI².

Q11. Field of a toroid with N turns?
A11. H = NI/(2πr) inside the core, ~0 outside.

Q12. Torque on a current loop of moment m = NIA in field B?
A12. τ = m × B.

Magnetostatics describes the fields of steady currents — the basis of inductance, transformers, and the magnetic side of Maxwell's equations. GATE tests Ampère's law for symmetric currents, the Biot-Savart field, inductance, and magnetic boundary conditions, mirroring the electrostatics toolkit.

Core concept: Steady currents produce a magnetic field H (and flux density B); Ampère's and Biot-Savart laws give the field, and energy is stored in inductance.

The theory

Beginner — Biot-Savart and force

• Biot-Savart law: dH = (I dl × r̂)/(4π r²) — the field contribution of a current element.
• Magnetic flux density B = µH, with µ = µ0µr and µ0 = 4π×10⁻⁷ H/m.
• Lorentz force on a charge: F = q(E + v × B); on a current element: dF = I dl × B.

Intermediate — Ampère's law and standard fields

Ampère's circuital law: ∮ H·dl = I_enclosed. For symmetric currents it gives H directly:

• Infinite straight wire: H = I/(2π r) (φ-directed).
• Infinite solenoid: H = nI (n = turns/length) inside, ~0 outside.
• Toroid: H = NI/(2π r) inside.
  Magnetic flux Φ = ∫B·dS; there are no magnetic monopoles, so ∮B·dS = 0 (flux lines close on themselves).

Advanced — inductance, energy, and boundaries

• Inductance L = Φ_linkage/I = NΦ/I; energy stored W = ½LI² = ½∫(B²/µ)dV (energy density B²/2µ).
• Solenoid inductance L = µN²A/l; coaxial and parallel-wire inductances from flux integration.
• Magnetic boundary conditions:
  • Normal B is continuous: Bn1 = Bn2 (no monopoles).
  • Tangential H differs by surface current: Ht1 − Ht2 = K (continuous if K = 0).
• Magnetic vector potential A: B = ∇×A, with ∇²A = −µJ (analogous to Poisson's equation). Magnetic materials (dia/para/ferromagnetic) set µr; ferromagnets have µr ≫ 1 and hysteresis.

Worked example

An air-cored solenoid has N = 500 turns, length l = 25 cm, cross-section A = 4 cm². Find its inductance.
n = N/l = 500/0.25 = 2000 turns/m.
L = µ0·N²·A/l = (4π×10⁻⁷)(500²)(4×10⁻⁴)/(0.25)
= (1.2566×10⁻⁶)(250000)(4×10⁻⁴)/0.25 = (1.2566×10⁻⁶)(250000)(1.6×10⁻³)
= 1.2566×10⁻⁶ × 400 = 0.503 mH.

GATE relevance

1–2 mark NAT/MCQ: field from a symmetric current via Ampère's law, Biot-Savart field, force on charges/currents, inductance and stored energy, and magnetic boundary conditions. Ampère-symmetry and inductance problems recur.

Exam tricks and shortcuts

• Ampère's law for symmetry: wire H = I/2πr, solenoid H = nI, toroid H = NI/2πr.
• No monopoles ⇒ ∮B·dS = 0, Bn continuous across boundaries.
• Inductance L = NΦ/I; energy density = B²/2µ; W = ½LI².
• Boundary: Bn continuous, Ht jumps by surface current K.
• Mnemonic: "Normal B and tangential H are the well-behaved ones" (dual to electrostatics).

Key formulas

• Biot–Savart: dH = (I dl × â_R)/(4πR²).
• Ampère's law: ∮H·dl = I_enc; B = μH.
• Straight wire: H = I/(2πr); solenoid H = nI.
• Force: F = qv × B; on wire F = IL × B.
• Energy density: w_m = ½μH² = B²/2μ.
• Boundary: H_tan difference = K (surface current); B_normal continuous.

Ampère's law gives fields of symmetric currents; magnetic energy is stored in the field (B²/2μ).

Worked Example

Problem: (a) A long straight wire carries I = 10 A. Find the magnetic flux density B at a radial distance of 5 cm. (b) A second parallel wire 10 cm away also carries 10 A in the same direction. Find the force per unit length between the wires. (µ₀ = 4π × 10⁻⁷ H/m.)

Solution:
(a) Ampère's law for an infinite straight wire:
B = µ₀I/(2πr) = (4π × 10⁻⁷ × 10)/(2π × 0.05).
Cancel π: B = (4 × 10⁻⁷ × 10)/(2 × 0.05) = (4 × 10⁻⁶)/(0.1) = 4 × 10⁻⁵ T = 40 µT.

(b) Force per unit length between two parallel currents:
F/ℓ = µ₀I₁I₂/(2πd) = (4π × 10⁻⁷ × 10 × 10)/(2π × 0.10).
Cancel π: F/ℓ = (4 × 10⁻⁷ × 100)/(2 × 0.10) = (4 × 10⁻⁵)/(0.2) = 2 × 10⁻⁴ N/m.
Because both currents flow the same way, the force is attractive.

Answer: (a) B = 40 µT; (b) F/ℓ = 2 × 10⁻⁴ N/m, attractive.