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A cylindrical magnetostatic equilibrium is an axisymmetric vector field and pressure configuration independent of the axial coordinate. Its radial Lorentz force density balances the pressure gradient, as expressed by cylindrical magnetostatic pressure balance.

With no axial electric current, cylindrical magnetostatic pressure balance reduces to constant $p+B_z^2/(8\pi )$. For $p=p_0{e}^{-R^2/a^2}$ and $B_z(0)=0$, its radial solutions areAs $R\downarrow 0$, $B_z\sim \sigma \sqrt{8\pi p_0}R/a$ and $j_{\varphi }\to -\sigma c\sqrt{8\pi p_0}/(4\pi a)$. These finite scalar limits do not give a smooth function of Cartesian position: $R=\sqrt{x^2+y^2}$ has a cusp at the axis, and the azimuthal unit vector has no unique limit there. Thus these data describe radial equilibrium away from the axis, but cannot satisfy smooth vector-field regularity on the axis without modifying the data.

With no dependence on the azimuthal or axial coordinates, Gauss's law for magnetism in cylindrical coordinates givesRegularity at the axis forces $C=0$, so $B_R=0$. The radial component of the magnetostatic Ampère-Maxwell equation isConsequentlyThis is also the starting point of cylindrical magnetostatic pressure balance.

The remaining magnetic field has $\mathbf{B}=B_{\varphi }(R){\mathbf{e}}_{\varphi }+B_z(R){\mathbf{e}}_z$. The radial Lorentz force density separates into a magnetic pressure gradient and inward magnetic tension:Here $(\mathbf{B}\cdot \nabla )\mathbf{B}$ has radial component $-B_{\varphi }^2/R$ because ${\partial }_{\varphi }{\mathbf{e}}_{\varphi }=-{\mathbf{e}}_R$. Thus magnetostatic equilibrium givesIntegrating the axial component of the magnetostatic Ampère-Maxwell equation gives $I(R)=cR{B}_{\varphi }/2$, since regularity removes the integration constant. Substituting $B_{\varphi }=2I/(cR)$ combines the toroidal terms intoThis is the cylindrical magnetostatic pressure balance equation in terms of enclosed electric current.

For constant $B_z$, cylindrical magnetostatic pressure balance givesRegularity gives $I(0)=0$. With $x=R^2/a^2$, integration yields the power-law pressure-supported axial currentThe 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 whenAt $k=1$ 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.

Setting $I=0$ in cylindrical magnetostatic pressure balance gives $p+B_z^2/(8\pi )=\mathrm{constant}$. The axial boundary value sets this constant to $p_0$. Thus the magnetic pressure support with vanishing axial field solution isThe azimuthal component of the magnetostatic Ampère-Maxwell equation givesSince $1-e^{-R^2/a^2}=R^2/a^2+O(R^4)$, the radial limit isThere is a regularity subtlety: $B_z\sim \sigma \sqrt{8\pi p_0}R/a$ is not a smooth function of Cartesian position at the axis, and the nonzero limiting $j_{\varphi }$ multiplies an azimuthal unit vector with no unique direction there. These formulas solve the radial problem for $R>0$ and give the requested scalar limit, but the data in this part do not admit the fully smooth vector-field regularity assumed in part (a).

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 $x=R^2/a^2$,The nontrivial electric current is finite at infinity exactly when $k>1$, in which case $I^2(\infty )=2\pi c^2{p}_0{a}^2/(k-1)$. Either sign of $I$ is possible because the pressure balance fixes its square.