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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.

For an ideal electrical transformer, let $n=N_p/N_s$. The reflected secondary load is $R_{\mathrm{in}}=n^2{R}_L$. Matching a $4\Omega$ load to an $8\Omega$ source therefore requiresWith the generator replaced by its internal resistance, its resistance referred to the secondary isThis is the source output resistance relevant to matching and maximum power transfer.

To find the windings, use the inductance of a toroidal winding. The mean magnetic path is $\ell =2\pi (0.10)\mathrm{m}$, the core area is $A=30\times 10^{-4}=0.003{\mathrm{m}}^2$ and $\mu =500{\mu }_0$. ThusThe ideal secondary turn count is $N_s=500/\sqrt{2}=353.55$, giving the practical nominal designWith integer turns the irrational ratio cannot be exact: $500:354$ reflects the $4\Omega$ load as $7.98\Omega$, an approximately $0.25\%$ resistance mismatch. These turn counts use the stated $0.75\mathrm{H}$ self-inductance; no operating frequency was specified from which to set a separate minimum inductance.

The $200\Omega$ condition must refer to the unloaded magnetizing reactance, not to the matched loaded input resistance. With $L_p=0.75\mathrm{H}$,The secondary self-inductance is $L_s=L_p/n^2=0.375\mathrm{H}$, so its corresponding unloaded coil impedance satisfiesIf output impedance means this transformed coil reactance, $100\Omega$ is its minimum magnitude. It is distinct from the $4\Omega$ transformed source resistance. This is the magnetizing reactance and reflected resistance in a transformer distinction. A finite magnetizing inductance adds a parallel branch, $Z_{\mathrm{in}}=(8\Omega )\parallel (i\omega L_p)$, so exact purely resistive matching is an ideal large-reactance approximation; interpreting the loaded input itself as both $8\Omega$ and at least $200\Omega$ would be impossible.