Formulas/physics/Moving Charges/Field Due to Infinite Straight Wire

Field Due to Infinite Straight Wire

Limiting case of the finite wire formula (φ₁ = φ₂ = 90°). Also derived directly via Ampere's law. Field lines are concentric circles; direction by right-hand thumb rule.

Class 12

Derivation

Symmetry Argument

An infinite straight wire along the $z$ -axis carrying current $I$ has:

Translational symmetry along $z$ : $B$ cannot depend on $z$ .
Azimuthal symmetry: $B$ cannot depend on azimuthal angle $\phi$ .
Direction: $\vec{B}$ has no radial or axial component by symmetry. So $\vec{B} = B(d)\,\hat{\phi}$ .

Ampere's Law

Choose a circular Amperian loop of radius $d$ centred on the wire. Since $\vec{B} \parallel d\vec{l}$ and $|\vec{B}|$ is constant:

\oint \vec{B} \cdot d\vec{l} = B \cdot 2\pi d = \mu_0 I

\boxed{B = \frac{\mu_0 I}{2\pi d}}

Direction: azimuthal, given by the right-hand thumb rule.

Field Lines

Concentric circles centred on the wire. The field falls as $1/d$ — slower than the $1/d^2$ falloff of a point charge, reflecting the infinite extent of the source.

Note

This form of Ampere's law applies to steady currents only. For time-varying fields, the displacement current term $\mu_0\varepsilon_0\,\partial\vec{E}/\partial t$ must be added.

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