Electromagnetic Fields

Syllabus

Coulomb's law, electric field intensity and flux density, Gauss's law, divergence, fields and potentials of point, line, plane and spherical charges, dielectrics, capacitance of simple configurations; Biot-Savart law, Ampere's law, curl, Faraday's law, Lorentz force, inductance, magnetomotive force, reluctance, magnetic circuits, self and mutual inductance.

Key concepts and formulas

Electrostatics (ε₀ = 8.854 × 10⁻¹² F/m, 1/(4πε₀) ≈ 9 × 10⁹):

• Coulomb: F = q₁q₂/(4πεr²). Field of a point charge E = q/(4πεr²); potential V = q/(4πεr).
• Gauss's law: ∮ D·dS = Q_enc, or ∇·D = ρ_v.
• Infinite line charge: E = ρ_L/(2πεr). Infinite sheet: E = ρ_s/(2ε), independent of distance.
• E = −∇V. Energy stored in a capacitor W = ½CV².

Capacitance. Parallel plate C = εA/d; coaxial cable C = 2πεL/ln(b/a); isolated sphere C = 4πεa.

Magnetostatics (μ₀ = 4π × 10⁻⁷ H/m):

• Biot-Savart law; field of a long straight wire B = μ₀I/(2πr); at the centre of a circular loop B = μ₀I/(2a); inside a long solenoid B = μ₀nI (n = turns per metre).
• Ampere's law: ∮ H·dl = I_enc, or ∇ × H = J.
• Lorentz force F = q(E + v × B); force on a conductor F = BIL sin θ. Force per unit length between parallel wires μ₀I₁I₂/(2πd).

Faraday's law. emf = −dΦ/dt = −N dΦ/dt for N turns; motional emf = Blv.

Magnetic circuits. mmf = NI; reluctance S = l/(μA); flux Φ = mmf/S. Inductance L = N²/S = μN²A/l; energy ½LI². Mutual inductance M = k√(L₁L₂).

Maxwell's equations (differential form): ∇·D = ρ_v, ∇·B = 0, ∇ × E = −∂B/∂t, ∇ × H = J + ∂D/∂t.

Worked example

Find the capacitance of a parallel plate capacitor with plates of 0.02 m² separated by 1 mm of dielectric with ε_r = 5. Find the field 5 cm from a long wire carrying 10 A. Then find the inductance of a toroid with 500 turns on an iron core (μ_r = 1000), mean length 0.4 m and cross-section 4 cm².

import math

e0, mu0 = 8.854e-12, 4 * math.pi * 1e-7
C = e0 * 5 * 0.02 / 1e-3
print(f"C = {C * 1e12:.2f} pF")
print(f"B at 5 cm from 10 A wire = {mu0 * 10 / (2 * math.pi * 0.05) * 1e6:.1f} uT")
N, mur, l, A = 500, 1000, 0.4, 4e-4
S = l / (mu0 * mur * A)
print(f"reluctance = {S:.4e} A/Wb, L = {N ** 2 / S:.4f} H")

C = 885.40 pF
B at 5 cm from 10 A wire = 40.0 uT
reluctance = 7.9577e+05 A/Wb, L = 0.3142 H

Practice questions