Problem 4.2.3: The infinitely long solenoid

Nothing depends on $z$ or $\varphi$, and $\nabla \cdot B=0$ with no $z$ dependence forbids a radial component (a radial $B_{\rho }(\rho )$ would give $\frac{1}{\rho }\frac{d}{d\rho }(\rho B_{\rho })\ne 0$ unless $B_{\rho }\propto 1/\rho$, which is singular on the axis), so $B=B_z(\rho )a_z$. Outside, $B_z$ must vanish as $\rho \to \infty$, and Ampère's law on a rectangle lying wholly outside encloses no current, so $B_z$ has the same value at every $\rho$ outside: $B_z=0$ everywhere outside.

Apply ${\oint }_{C}H\cdot dl={\int }_{S}J\cdot dS$ to the rectangle $C$ of length $L$ with one long side inside and one outside. The two radial sides cancel each other ($H\perp dl$), the outer side contributes nothing, and the contour links $NL$ turns:

With the cross-section $S=\pi a^2$ each turn links $\Phi ={\mu }_{0}NIS$, and a length $l$ of the coil contains $Nl$ turns:

$B$ is independent of $\rho$ inside and of the coil radius: the same for a thin or a fat solenoid. Units: ${\mu }_{0}NI$ is (H/m)(1/m)(A) $=$ (V$\cdot$s/A/m)(A/m) $=$ V$\cdot$s/m${}^2$ $=$ T. The jump of $H_z$ from $NI$ inside to $0$ outside is exactly the first boundary condition, Eq. (4.2.11), with the sheet current $J_{nS}=NI$ of the winding. With a core of permeability $\mu$ both $B$ and $L$ are multiplied by ${\mu }_r$, Eq. (4.2.21). For a finite solenoid the result is an upper bound: Figure 4.2.18 shows that $B_z$ falls to half the ideal value at each end, and the inductance of a finite coil is ${\mu }_0{N}^{2}Sl$ times the Nagaoka coefficient, which tends to $1$ only when the length is much larger than the radius. This is why Worked example 4.2.2 uses a solenoid ten times longer than its radius.