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

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).