Short notes

Magnetostatics Short Notes

A long straight conductor carrying 10 A of DC current produces a magnetic field H = I/(2πr) at radius r — at r = 5 cm, that's H = 10/(2π × 0.05) = 31.83 A/m and B = µ₀H = 4π×10⁻⁷ × 31.83 ≈ 40 µT. That calculation using Ampere's law in integral form is the magnetostatics equivalent of Gauss's law in electrostatics, and it appears in the first ten minutes of every magnetostatics exam paper.

EEE, ECE, EI

How it works

The Biot-Savart law gives the differential magnetic field contribution dH = I(dl × r̂)/(4πr²), used for arbitrary current geometries. For a finite straight wire of length L at perpendicular distance d, H = I(sinα₂ − sinα₁)/(4πd). Ampere's circuital law in integral form: ∮H·dl = I_enclosed. For an infinite straight wire: H = I/(2πρ) φ̂ in cylindrical coordinates. A toroidal coil with N turns and mean radius R has H = NI/(2πR) inside the core, zero outside — a key result for transformer analysis.

Key points to remember

Magnetic flux Φ = ∫B·dS, measured in Webers. The boundary conditions: normal B is continuous across interfaces (B₁ₙ = B₂ₙ); tangential H is discontinuous by the surface current density (H₁ₜ − H₂ₜ = K). Magnetic vector potential A is defined by B = ∇×A, and satisfies Poisson's equation ∇²A = −µJ. For a solenoid with n turns/meter carrying current I: B = µ₀µr·nI inside, zero outside. Energy stored in magnetic field: W_m = ½µ₀∫|H|² dV. Inductance L = NΦ/I, and mutual inductance M = N₂Φ₁₂/I₁.

Exam tip

The examiner always asks you to find H inside and outside a coaxial cable or toroid using Ampere's law — define your Amperian loop clearly, calculate enclosed current for each region, and present results in a piecewise expression.

More Electromagnetic Theory notes