Magnetic field outside an infinite solenoid

The argument below is paraphrased from Zangwill, Section 10.2.2. Set the origin of our coordinates at the spot we're trying to find the field, and let the axis of the solenoid be the $z$-axis. Consider a horizontal "slice" of the solenoid of width dz at a height $z$ above the xy-plane. By the Biot-Savart Law, the magnetic field due to this slice is just that of a current loop: $$ dvec{B} = - frac{mu_0 K dz}{4 pi} oint frac{dvec{ell} times hat{r}}{r^2} = - frac{mu_0 K dz}{4 pi} oint frac{dvec{ell} times vec{r}}{r^3}. $$ (The minus sign is there because $vec{r}$ in this formula denotes the source point, not the field point.) The slice can be thought of as a parametric curve $vec{ell}(lambda)$, and it can be related to $vec{r}$ by $vec{r} = vec{ell} + z hat{z}$, where $vec{ell}$ is in the xy-plane. Since $vec{ell}$ is perpendicular to $hat{z}$ by construction, we have $r = sqrt{ell^2 + z^2}$, and so we can write the full integral as begin{equation} dvec{B} = - frac{mu_0 K dz}{4 pi} oint frac{dvec{ell} times (vec{ell} + z hat{z})}{(ell^2 + z^2)^{3/2}} = - frac{mu_0 K dz}{4 pi} left[ oint frac{dvec{ell} times vec{ell} }{(ell^2 + z^2)^{3/2}} + z oint frac{dvec{ell} times hat{z} }{(ell^2 + z^2)^{3/2}} right ] end{equation} When we integrate from $z = -infty$ to $infty$, the second term will vanish (because it is odd with respect to $z$, while the first term evaluates to $2/ell^2$. Thus, we now have $$ vec{B} = -frac{mu_0 K}{2 pi} oint frac{dvec{ell} times vec{ell} }{ell^2}. $$ Writing $vec{ell}$ in cylindrical coordinates, we have $vec{ell} = s hat{s}$ and $dvec{ell} = ds hat{s} + s , dphi hat{phi}$; thus, $-dvec{ell} times vec{ell} = -s^2 , dphi hat{z}$, and we have $$ vec{B} = frac{mu_0 K}{2 pi} hat{z} oint , dphi. $$ The integral is now the net solid angle traversed when we go around the loop. If the curve $vec{ell}(lambda)$ encloses the origin (i.e., we are inside the solenoid), this will be $2pi$; otherwise, it will be zero. Thus, $$ vec{B} = begin{cases} mu_0 K hat{z} & text{inside} 0 & text{outside}.end{cases} $$ Note that this derivation did not assume any particular shape for the cross-section of the solenoid (i.e., the shape of the curve $vec{ell}$.)

To preemptively address any concerns about infinite solenoids being unrealistic: how would this derivation be modified if we had a non-infinite (but long) solenoid? Return to the step where we had written $dvec{B}$ in terms of $dz$ and two integrals. If we assume that the solenoid now stretches from $z_1$ to $z_2$ instead, the expression becomes: $$ vec{B} = - frac{mu_0 K}{4 pi} left[ oint frac{sigma(z) , dvec{ell} times vec{ell} }{ell^2 sqrt{(ell/z)^2 + 1}} - oint frac{dvec{ell} times hat{z} }{|z| sqrt{(ell/z)^2 + 1}} right ]_{z = z_1}^{z_2}. $$ where $sigma(z)$ is the sign of $z$. While we can't evaluate these integrals in as much generality, we still retain some important features. The first term will still point in the z-direction regardless of $z_1$ and $z_2$, and the second term will always point in the xy-plane. Moreover, the second term still vanishes so long as $z_1 = - z_2$. Thus, at the midpoint of a long solenoid, the external field still points along the axis of the solenoid (as it must by symmetry.)

If we wanted, we could expand these integrals in a power series in $ell/z_1$ and $ell/z_2$, assuming this quantity is small; this is saying that the perpendicular distance from the field point to the solenoid is much smaller than the distance to its ends. In this limit, we would recover the infinite-solenoid result plus corrections due to the finite length of the solenoid. The leading-order correction to the horizontal field would be $mathcal{O}(ell/z)$, while the leading-order correction to the vertical field will be $mathcal{O}(ell/z)^2$.