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

Home exercise 4.1.7

Infinitely long solenoid

An infinitely long solenoid has $n$ closely wound turns per metre, each carrying a current $I$ . Apply Ampère's formula to the rectangular path $C$ shown in the figure (one side of length $ℓ$ inside the solenoid, the opposite side outside) and show that

H = n I a_{z} inside, H = 0 outside,

independently of the radius of the solenoid and of the position of the path.

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 6-3.

Try it yourself first.

Strategy

Symmetry leaves only $H_{z} (ρ)$ . A rectangle with both long sides outside shows that $H_{z}$ is the same everywhere outside, hence zero, and a rectangle with both sides inside shows that $H_{z}$ is uniform inside. The straddling rectangle then threads $n ℓ$ turns and gives the value.

1. Symmetry

The solenoid is invariant under translation along $z$ and rotation about $z$ , so $H$ depends on $ρ$ only. For closely wound turns the current is a cylindrical sheet $J_{S} = n I a_{ϕ}$ at $ρ = b$ . Each turn is a loop whose field is along $\pm a_{z}$ on the axis. By the symmetry of the sheet, $H_{ϕ} = 0$ everywhere (no net axial current is enclosed by any coaxial circle) and $H_{ρ} = 0$ (it would violate $\nabla \cdot B = 0$ ). Hence $H = H_{z} (ρ) a_{z}$ . Far from an infinitely long sheet the field must vanish, and since $H_{z}$ cannot vary with $ρ$ outside (next step) it is zero everywhere outside.

2. A path entirely outside, or entirely inside

Take a rectangle with both long sides outside, at $ρ_{1}$ and $ρ_{2} > ρ_{1}$ . No current threads it, the short sides are perpendicular to $H$ , so $ℓ [H_{z} (ρ_{1}) - H_{z} (ρ_{2})] = 0$ : $H_{z}$ is the same at every $ρ > b$ , hence equal to its value at infinity, zero. The same argument with both sides inside shows that $H_{z}$ is uniform inside.

3. The path straddling the wall

Now one side of $C$ , of length $ℓ$ , is inside and the other outside. Traverse it so that the inside side runs along $+ a_{z}$ . The two short sides contribute nothing ( $H ⊥ d l$ ), the outside side contributes nothing ( $H = 0$ ), so

\oint_{C} H \cdot d l = H_{z} ℓ = I_{enc} = (n ℓ) I,

because the rectangle is pierced by $n ℓ$ turns, each carrying $I$ in the same sense.

Answer

H = n I a_{z} inside, H = 0 outside

The result contains neither the radius $b$ nor the distance of the path from the wall.

Check

Direction: with the current circulating counter-clockwise when seen from $+ z$ , the right hand gives $H$ along $+ a_{z}$ inside. Units: (turns/m) $\times$ A $=$ A/m. Numbers: $n = 2000$ m $^{- 1}$ and $I = 3$ A give $H = 6$ kA/m and $B = μ_{0} H = 7.5$ mT in air. The uniform interior field is why the solenoid is the standard source of a known $B$ , and $Φ = μ_{0} n I π b^{2}$ per turn gives its inductance per unit length, $L^{'} = μ_{0} n^{2} π b^{2}$ , in the next chapter. A finite solenoid differs only near its ends, where field lines leak out (Home exercise 4.1.2): at the centre of a solenoid whose length is five times its diameter the field is already within $2 %$ of $n I$ .

Home exercise 4.1.7: Infinitely long solenoid