At $\kappa=\kappa_c=1/2$, the <indicial equation> has the <multiple root> $s=-1/2$. The general axial component is
$$
\boxed{B_z=R^{-1/2}\left(C+D\ln\frac R{R_0}\right)}.
$$
Since $B_\phi=-2R B_z'$, the azimuthal component is
$$
\boxed{B_\phi=R^{-1/2}
\left(C+D\ln\frac R{R_0}-2D\right)},
\qquad B_R=0.
$$
Consequently
$$
\frac{B_\phi}{B_z}
=1-\frac{2D}{C+D\ln(R/R_0)}\longrightarrow1
$$
as $R\to\infty$; the same ratio is identically one when $D=0$. A <magnetic field line> has tangent parallel to $\mathbf B$, so its asymptotic angle $\vartheta$ with the $z$-axis satisfies
$$
\tan\vartheta=\left|\frac{B_\phi}{B_z}\right|\longrightarrow1.
$$
Hence the field lines become helices making the constant angle
$$
\boxed{\vartheta=\frac\pi4=45^\circ}
$$
with the axis.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-314/1/d/solution