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

Problem 4.1.5

Interactive: Ampère's Law Profiles

Problem 4.1.5

Solid wire of radius b

By using Ampère's formula, find the magnetic field inside and outside a wire with a radius of $b$ having a current of $I$ .

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 6-1.

Try it yourself first.

Strategy

Exactly as in Problem 4.1.4: $H = H_{ϕ} (ρ) a_{ϕ}$ , so the same circular path $C$ works inside and outside the wire, and only $I_{enc}$ changes. Inside, the circle encloses only the part of the current that flows through it.

1. Symmetry

The thick wire has the same symmetries as the thin one (translation along $z$ , rotation about $z$ , the mirror), so again $H = H_{ϕ} (ρ) a_{ϕ}$ , and on a coaxial circle of radius $ρ$ Ampère's formula reads $2 π ρ H_{ϕ} = I_{enc} (ρ)$ .

2. Current density

Uniform over the cross-section $π b^{2}$ :

J = \frac{I}{π b ^{2}} a_{z} [A/ m^{2}], ρ \leq b, J = 0 outside .

3. Inside, ρ ≤ b

The circle encloses only the part of the current that crosses the disk of radius $ρ$ (Figure 4.1.22a):

I_{enc} 2 π ρ H_{ϕ} = \frac{I}{π b ^{2}} π ρ^{2} = I \frac{ρ ^{2}}{b ^{2}}, = I \frac{ρ ^{2}}{b ^{2}} ⟹ H_{ϕ} = \frac{I ρ}{2 π b ^{2}} .

4. Outside, ρ ≥ b

The circle encloses all of it, $2 π ρ H_{ϕ} = I$ .

Answer

H H = \frac{I ρ}{2 π b ^{2}} a_{ϕ} (ρ \leq b), = \frac{I}{2 π ρ} a_{ϕ} (ρ \geq b) [A/m]

(a) Cross-section with the two Amperian circles.

$(b) The field profile. Problem 4.1.5. (a) Inside the wire the circle encloses only the fraction \rho^2/b^2 of the current, outside it encloses all of it. (b) H_\phi rises linearly in the metal, peaks at the surface with I/2\pi b, and falls as 1/\rho in the air: outside, the wire is indistinguishable from the thin wire of Problem 4.1.4.$

Figure 4.1.22. (b) The field profile. Problem 4.1.5. (a) Inside the wire the circle encloses only the fraction

ρ^{2} / b^{2}

of the current, outside it encloses all of it. (b)

H_{ϕ}

rises linearly in the metal, peaks at the surface with

I /2 π b

, and falls as

1/ ρ

in the air: outside, the wire is indistinguishable from the thin wire of Problem 4.1.4.

Move the circle inside the wire and see $I_{enc}$ grow as $ρ^{2}$ , then check that $H_{ϕ}$ is continuous at $ρ = b$ .

InteractiveLoading simulation…

Check

$H_{ϕ} \to 0$ on the axis (no current enclosed, and $a_{ϕ}$ is not even defined there). The two branches agree at $ρ = b$ , giving the maximum $H_{m a x} = I /2 π b$ : no jump, as it must be without a surface current. Outside, the wire behaves exactly like the thin wire of Problem 4.1.4, its radius has disappeared (Figure 4.1.22b). Units: A/m. The field inside the copper stores energy: $\int \frac{1}{2} μ_{0} H^{2} d S$ per unit length is $μ_{0} I^{2} /16 π$ , an internal inductance of $μ_{0} /8 π = 50$ nH/m, independent of $b$ . At high frequency the current crowds into a thin skin and this interior field collapses.

Problem 4.1.5: Solid wire of radius b