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For a smooth bounded region $\Omega$, prescribing the outward normal derivative ${\partial }_{n}u=g$ of a harmonic function requiresThis follows from the divergence theorem applied to the gradient:On the unit disk, the condition becomes ${\int }_0^{2\pi }g(\theta )d\theta =0$. If $g={\sum }_{n\ge 1}(a_n\cos (n\theta )+b_n\sin (n\theta ))$ is sufficiently regular with zero mean, its regular harmonic expansion on a disk isThe factor $1/n$ compensates for the radial derivative. The arbitrary $C$ reflects the uniqueness of the Neumann problem up to an additive constant on a connected region.

The trigonometric identityshows that only angular Fourier modes $1$ and $3$ are required. Differentiating the harmonic expansion on a disk at $r=1$ multiplies the coefficient of each nonconstant Fourier mode by $n$. Thus $B_1=3/4$ and $3B_3=-1/4$, givingEach term is a regular harmonic function, and the radial derivative isThe Neumann compatibility condition for Laplace's equation holds becauseas is necessary by the divergence theorem applied to the gradient of a harmonic function.

The additive constant is undetermined. This Neumann boundary condition is unchanged by $T\mapsto T+C$, so the paper's blanket uniqueness assertion requires a normalization in this case. The uniqueness of the Neumann problem gives uniqueness up to a constant; prescribing $T(0)=0$, or zero mean temperature, selects the displayed answer with $C=0$.

For fixed, magnetically linear geometry, self-inductance relates a coil's own magnetic flux linkage to its electric current, ${\Lambda }_i=L_i{I}_i$. mutual inductance relates the linkage in one coil to the current in the other, ${\Lambda }_{21}=M{I}_1$. These coefficients enter the induced electromotive forces through ${\mathcal{E}}_i=-d{\Lambda }_i/dt$; reversing a chosen winding orientation reverses the sign assigned to $M$.

In the ideal confined-field toroid, a circular magnetostatic Ampère law contour encloses $N_{1}I$, soWrite $w=b-a$ for the height of the square cross-section. The inductance of a rectangular-section toroid follows by integrating the magnetic flux density over that cross-section; it cannot be replaced by its value at an arbitrary mean radius:Consequently, for winding normals chosen consistently,The equality means complete shared magnetic flux linkage; with leakage one instead has $|M|<\sqrt{L_1{L}_2}$.

For the crack, use the radial-path approximation for a cracked toroid, with an air length $d$ on each path and core length $2\pi r-d$. Continuity of the normal magnetic flux density at a crack face gives $H_g={\mu }_r{H}_c$. HenceIntegrating the radial paths yieldsHere $q=1.5915\mathrm{cm}$ and $M_d/M_0=0.47079$: about a 53% reduction. The small ratio $d/a$ does not justify expanding in ${\mu }_{r}d/a$, which is about ten. The crack's low magnetic permeability gives appreciable magnetic reluctance. This is an estimate within a confined-flux model; actual redistribution near the slit is outside its accuracy. Narrow-gap circuit models assume small gap width relative to transverse dimensions and approximately directed flux.^[ref]

For the intact ring, ideal poloidal winding currents and the no-leakage boundary condition admit $\mathbf{H}=N_{1}I\hat{\mathbf{\varphi }}/(2\pi r)$ with zero radial and vertical components. It is curl-free inside the current-free core, divergence-free for uniform magnetic permeability, tangent to the core boundary, and has the required circulation; uniqueness of the Neumann problem removes an additional homogeneous scalar-potential field. This explains zero non-azimuthal components in the stipulated intact-ring idealization. A literal slit breaks that rotational symmetry, so zero transverse components everywhere is not exact for the cracked ring: magnetic fringing and flux redistribution occur. In fact, extending the approximate $B(r)\propto 1/(r+q)$ unchanged around a current-free core would give $[\nabla \times \mathbf{H}{]}_z\ne 0$. Thus that radial-path calculation and a claim of exact purely azimuthal field in the cracked geometry cannot both be exact. The numerical answer above retains the intended circuit approximation while making its limitation explicit.