Lesson 4 of 12 · 7 min
Biot–Savart law
NCERT §4.4
Kabir's long straight lead carries 10 A. Take just a 1 mm piece of it. How much field does that little piece make at a point 2 cm away?
The lesson in notes
In short
Every known magnetic field comes from currents (moving charges) or from the intrinsic magnetic moments of particles. The Biot–Savart law gives the field of a small piece of current.
A current element I dl produces at a point a displacement r away the field dB = (μ₀/4π) I dl × r/r³. Its size is dB = (μ₀/4π) I dl sin θ/r², with θ the angle between dl and r.
dB is perpendicular to the plane containing dl and r; its sense follows the right-hand screw rule for dl × r.
μ₀ is the permeability of free space, with μ₀/4π = 10⁻⁷ T m A⁻¹ (μ₀ = 4π × 10⁻⁷ T m A⁻¹ in SI).
Like Coulomb's law, it is long range (1/r²) and obeys superposition, and the field is linear in its source. Unlike it, the source I dl is a vector, the field is perpendicular to r rather than along it, and there is an angle factor sin θ.
On the line of the element itself (θ = 0) the element makes no field at all.
The constants are linked to the speed of light: ε₀μ₀ = 1/c², with c = 3 × 10⁸ m/s. Fixing μ₀ fixes ε₀.
Example 4.4: a 1 cm element carrying 10 A along x, seen from 0.5 m along y (θ = 90°), gives dB = 10⁻⁷ × 10 × 10⁻²/0.25 = 4 × 10⁻⁸ T, pointing along +z.