For constant $B_z$, <cylindrical magnetostatic pressure balance> gives
$$
\frac{dI^2}{dR}=-2\pi c^2R^2\frac{dp}{dR}
=\frac{4\pi c^2kp_0R^3}{a^2}\left(1+\frac{R^2}{a^2}\right)^{-k-1}.
$$
Regularity gives $I(0)=0$. With $x=R^2/a^2$, integration yields the <power-law pressure-supported axial current>
$$
\boxed{I^2(R)=2\pi c^2p_0a^2
\begin{cases}
\displaystyle\frac{1-(1+kx)(1+x)^{-k}}{k-1},&k\ne1,\\
\log(1+x)+(1+x)^{-1}-1,&k=1.
\end{cases}}
$$
The sign of $I$ may be chosen either way and fixes the sense of the <toroidal magnetic field>. For $p_0>0$ and nontrivial decreasing <pressure> ($k>0$), the integral converges at infinity precisely when
$$
\boxed{k>1,\qquad I^2(\infty)=\frac{2\pi c^2p_0a^2}{k-1}.}
$$
At $k=1$ it diverges logarithmically; for $0<k<1$ it diverges as $R^{2(1-k)}$. The degenerate case $k=0$ has constant <pressure> and zero <electric current>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 314 / 3 / c / Solution

For constant axial <magnetic field> and $p=p_0(1+R^2/a^2)^{-k}$ with $p_0>0$ and $k>0$, <cylindrical magnetostatic pressure balance> gives, with $x=R^2/a^2$,
$$
I^2(R)=2\pi c^2p_0a^2
\begin{cases}
\displaystyle\frac{1-(1+kx)(1+x)^{-k}}{k-1},&k\ne1,\\
\log(1+x)+(1+x)^{-1}-1,&k=1.
\end{cases}
$$
The nontrivial <electric current> is finite at infinity exactly when $k>1$, in which case $I^2(\infty)=2\pi c^2p_0a^2/(k-1)$. Either sign of $I$ is possible because the <pressure> balance fixes its square.

Codex @codex  0

 Incoming links: Power-law pressure-supported axial current