Use the <poloidal magnetic flux function> convention $\mathbf B_p=R^{-1}\nabla\Psi\times\mathbf e_\phi$, so $2\pi\Psi$ is magnetic flux up to a reference constant. Its <poloidal magnetic field> components are
$$
\boxed{B_R=-\frac1R\frac{\partial\Psi}{\partial z},\qquad B_z=\frac1R\frac{\partial\Psi}{\partial R}.}
$$
The prescribed surface flux therefore gives
$$
B_z(R,0)=\frac{\delta\Psi_0}{R_0^\delta}R^{\delta-2},\qquad B_R(R,0)=B_z(R,0)\tan\alpha.
$$
The constant offset $\Psi_1$ does not affect either field component. If instead $\Psi$ denotes the full physical flux rather than flux divided by $2\pi$, both component formulas acquire a common $1/(2\pi)$ factor; the subsequent logarithmic derivative relation is unchanged.

The <Gauss's law for magnetism> constraint is $R^{-1}\partial_R(RB_R)+\partial_zB_z=0$. Since $\alpha$ is constant with radius on the surface, it gives the <power-law poloidal field near a disk surface> relation
$$
\boxed{\left.\frac{\partial B_z}{\partial z}\right|_{0}=-\frac{\delta(\delta-1)\Psi_0\tan\alpha}{R_0^\delta}R^{\delta-3}=-\frac{\delta-1}{R}B_z(R,0)\tan\alpha.}
$$
For a smooth one-sided field above the surface, $B_z(R,z)=B_z(R,0)[1-(\delta-1)\tan\alpha\,z/R]+O(z^2)$. Thus the field initially decreases with height if $\delta>1$, increases if $0<\delta<1$, and has zero first height derivative if $\delta=1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 314 / 3 / b / Solution

If $\Psi(R,0)=\Psi_1+\Psi_0(R/R_0)^\delta$ and the surface inclination to the vertical is a radius-independent $\alpha$, then $B_z=\delta\Psi_0R^{\delta-2}/R_0^\delta$ and $B_R=B_z\tan\alpha$. The <solenoidal vector field> condition gives
$$
\left.\partial_zB_z\right|_{0}=-\frac{\delta-1}{R}B_z(R,0)\tan\alpha.
$$
The constant offset in the <poloidal magnetic flux function> has no effect on the <magnetic field>.

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 Incoming links: Power-law poloidal field near a disk surface