Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.1.4

Interactive: Ampère's Law Profiles

Problem 4.1.4

Infinitely long thin wire (Ampère)

By using Ampère's formula, find the magnetic field at a normal distance $ρ$ from an infinitely long thin wire having a current of $I$ flowing along the $+ z$ -direction.

Try it yourself first.

Strategy

The symmetry argument of Section 4.1.8: only $H_{ϕ} (ρ)$ survives, so on a circle round the wire $H$ is constant and tangential, and Ampère's formula gives it in one line. This is the magnetic counterpart of Problem 3.2.1.

1. Symmetry

Translation along $z$ and rotation about $z$ leave the wire unchanged, so $H$ cannot depend on $z$ or $ϕ$ . A radial component would make $\oint B \cdot d S \neq = 0$ over a coaxial cylinder, contradicting $\nabla \cdot B = 0$ . A $z$ -component is excluded by the mirror symmetry of the field of a line current (and, directly, by Biot–Savart's law, whose $a_{z} \times R$ has no $z$ -component). Hence

H = H_{ϕ} (ρ) a_{ϕ} .

2. Current density

$J = I δ (x) δ (y) a_{z}$ $[A/ m^{2}]$ , so $\int J \cdot d S = I$ through any plane $z =$ const.

3. Amperian path and enclosed current

Take $C$ to be the circle $ρ =$ const in a plane $z =$ const, traversed in the $+ a_{ϕ}$ sense (right hand along $+ z$ ). On $C$ , $H_{ϕ}$ is constant and $d l = ρ d ϕ a_{ϕ}$ is parallel to $H$ :

\oint_{C} H \cdot d l = H_{ϕ} \int_{0}^{2 π} ρ d ϕ = 2 π ρ H_{ϕ} = I_{enc} = I .

Answer

H B = \frac{I}{2 π ρ} a_{ϕ} [A/m], = \frac{μ _{0} I}{2 π ρ} a_{ϕ} [T]

(b) The 1/\rho law for I=10 A. Problem 4.1.4. (a) On the circle C the field is tangential and constant, so \oint_C\vec H\cdot d\vec l=2\pi\rho H_\phi. (b) For 10 A the field equals the Earth's field at \rho=4 cm and keeps decaying as 1/\rho: it never vanishes. — **Figure 4.1.21.** (b) The $1/ ρ$ law for $I = 10$ A. Problem 4.1.4. (a) On the circle $C$ the field is tangential and constant, so $\oint_{C} H \cdot d l = 2 π ρ H_{ϕ}$ . (b) For $10$ A the field equals the Earth's field at $ρ = 4$ cm and keeps decaying as $1/ ρ$ : it never vanishes.

Drag the circle and check that $2 π ρ H_{ϕ}$ stays equal to $I$ . Where does 10 A give the Earth's 50 μT?

InteractiveLoading simulation…

Check

The same result followed from Biot–Savart's law integrated along the wire in Problem 4.1.1, with a page of algebra instead of one line: that is what symmetry buys. Direction: $a_{ϕ}$ is given by the right hand, thumb along $+ z$ (the current), fingers curling in the sense of $H$ . Units: A/m. Numbers: $I = 1$ A gives $20$ μT at $1$ cm, and $I = 10$ A matches the Earth's field ( $\approx 50$ μT) at $ρ = 2 \times 1 0^{- 7} \times 10/ (5 \times 1 0^{- 5}) = 4$ cm (Figure 4.1.21b). Read backwards, $I = 2 π ρ H_{ϕ} = \oint_{C} H \cdot d l$ is the principle of the clamp meter and of the Rogowski coil (Application 4.1.3).

Problem 4.1.4: Infinitely long thin wire (Ampère)