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In SI units, the source-free Maxwell equations in vacuum areUse complex amplitudes with phase $e^{i(\omega t-k\cdot x)}$ and take real parts at the end. The electric divergence equation gives $k\cdot E_0=0$. Faraday's equation gives $k\times E_0=\omega B_0$, soThe magnetic divergence equation follows because $k\cdot (k\times E_0)=0$. Ampere's equation becomes $k\times B_0=-{\mu }_0{ϵ}_0\omega E_0$. Since $k\times (k\times E_0)=-|k{|}^2{E}_0$, a nonzero wave satisfiesThus both fields are transverse, mutually perpendicular for linear polarization, and obey the vacuum plane-wave dispersion relation.

For the interface let $n=e_z$ point from the minus side to the plus side and write $[E]=E_{+}-E_{-}$, similarly for $B$. Applying the integral Maxwell equations to a thin pillbox and a thin loop gives the electromagnetic boundary conditionsHere $j$ is the tangential surface current. The jump notation fixes the signs, including the direction of the normal.

For the perfect conductor, take the time-varying interior fields to vanish. Choose positive $k$ with $\omega =ck$. The incident electric field has complex form $E_0{e}_x{e}^{i\omega t+ikz}$. To cancel its tangential value at the surface, use the reflected field $-E_0{e}_x{e}^{i\omega t-ikz}$, whose wavevector points upward. The total vacuum fields are thereforeThe sign of each magnetic contribution follows from $B_0=k\times E_0/\omega$. At $z=0$, the electric field is zero and the magnetic field is tangential. The jump conditions yieldFor a real incident amplitude this is $j=(2E_0/{\mu }_{0}c)\cos (\omega t)e_x$. It supplies the reflected wave while satisfying the normal-field and tangential-electric conditions. Any independently imposed static interior field is separate from this oscillatory response.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. The resistive induction equation, together with the solenoidal conditions on both fields, isSince $D_t\mathbf{x}=\mathbf{u}$, the product rule givesAlso $\Delta P=\mathbf{x}\cdot \Delta \mathbf{B}+2\nabla \cdot \mathbf{B}=\mathbf{x}\cdot \Delta \mathbf{B}$. Substitution proves the radial magnetic induction scalar equation:

In the insulating exterior the quasistatic magnetic field is current-free, so $\nabla \times \mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$. Thus $\Delta \mathbf{B}=0$ and $\Delta P=0$ for $r>a$, with decay at infinity for an isolated field. This is an instantaneous exterior Laplace equation, rather than a diffusion equation with the interior value of $\eta$.

Use the usual nonmagnetic-interface assumptions: equal magnetic permeability on both sides and no surface current. The electromagnetic boundary conditions then make $B_r$ and the tangential magnetic field continuous. Hence $P=r{B}_r$ is continuous. In spherical coordinates, the solenoidal magnetic-field constraint readswhere the last term is the angular divergence on the unit sphere. Its right-hand side is also continuous, giving the insulating boundary condition for the radial magnetic scalar:Equivalently, the decaying exterior degree-$l$ spherical harmonic is proportional to $r^{-(l+1)}$, so its boundary amplitude satisfies ${\partial }_r{P}_{lm}(a)=-(l+1)P_{lm}(a)/a$.

Inductance and interface conditions. For a linear coil, self-inductance is the ratio $L=\Lambda /I$ of its own magnetic flux linkage $\Lambda =N\Phi$ to current. Its induced electromotive force is $-LdI/dt$ for fixed $L$. Across an interface with unit normal $\mathbf{n}$ from medium 1 to 2,These electromagnetic boundary conditions follow from Gauss's law for magnetism and Ampère's law. Here ${\mathbf{K}}_f$ is free surface current. At the iron–air faces there is no free surface current, so normal magnetic flux density and tangential magnetic field intensity are continuous. In each isotropic linear medium, $\mathbf{B}={\mu }_0{\mu }_r\mathbf{H}$.

Complete core. A slender toroid has approximately uniform $H$ along mean path length $\ell$. Ampère's law gives $H\ell =NI$, so $B=\mu {\mu }_{0}NI/\ell$. The flux per turn is $\Phi =BA$, yielding the inductance of a toroidal winding,This uses constant relative magnetic permeability $\mu$ and neglects leakage, fringing and saturation.

Why the gap lowers inductance. The gap replaces an easy magnetic path in iron with air. Maintaining the same flux now requires a much larger magnetic field intensity in part of the path and therefore more ampere-turns. Equivalently, magnetic reluctance increases, so the flux linkage for a fixed current and hence self-inductance decrease. A narrow gap can matter greatly because its permeability is smaller by the factor $\mu$.

Oblique-gap calculation. In the original diagram $t$ is the perpendicular separation of the cut faces. Let $\mathbf{s}$ point along the core, and let $\mathbf{n},\mathbf{q}$ be normal and tangential to the gap faces. With the face at angle $\theta$ to the core direction, $\mathbf{s}=\sin \theta \mathbf{n}+\cos \theta \mathbf{q}$. Write the core flux density as $B\mathbf{s}$. The boundary conditions givesoThe gap area normal to its nearly normal field is $A/\sin \theta$, not $A$. Its leading reluctance is consequently $t\sin \theta /({\mu }_{0}A)$. Adding the core reluctance gives the inductance of a toroid with an oblique narrow air gap,Thus at $\theta =45^{\circ }$,The approximation retains the possibly order-one ratio $\mu t/\ell$ while neglecting $t/\ell$, fringing and the small tangential gap contribution. More precisely, a local uniform-field path along the core, with missing iron length $t/\sin \theta$ subtracted, gives denominator $\ell /\mu +t\sin \theta (1-1/\mu )$. Its correction is beyond the stipulated large-$\mu$, narrow-gap accuracy; simply neglecting $\mu t$ would not be justified.

Field in the slot. The component expression above givesFor the $45^{\circ }$ gap and $\mu \gg 1$,Its units are $\mathrm{A}{\mathrm{m}}^{-1}$. The field is almost perpendicular to the cut faces, rather than following the iron direction straight across the cut. Its small angle away from the face normal satisfies $\tan \alpha =1/\mu$ at $45^{\circ }$; the sense reverses if the current reverses.