Magnetic circuits and Ampere’s law

Ampere's circuital law is the starting point. For any closed path that links a coil of $N$ turns carrying current $I$ — in practice, the mean flux path through a magnetic core —

where $H$ is the magnetic field intensity (A/m) along the path and $dl$ is an element of path length. The right-hand side $NI$ is the magnetomotive force (mmf), in ampere-turns (AT): it is the "cause" that drives flux around the circuit, exactly as EMF drives current around an electric circuit.

Flux, flux density and permeability. The flux $\Phi$ (weber, Wb) through a cross-section of area $A$ gives flux density $B = \Phi/A$ (tesla, T). The material connects $B$ to $H$ through permeability,

where: $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$ is the permeability of free space and $\mu_r$ is the relative permeability of the core material — typically hundreds to a few thousand for magnetic steel, and exactly 1 for air or vacuum. A high-$\mu_r$ core needs far less $H$ (hence less current) to push the same flux density than air does.

Reluctance and the circuit analogy. For a uniform path of mean length $l$ and cross-section $A$, define reluctance

Substituting into Ampere's law for a simple loop gives the working equation of every magnetic-circuit problem:

where: $N$ is the number of coil turns, $I$ the coil current, $\Phi$ the flux and $S$ the reluctance of the complete flux path. This is the exact analogue of $V = IR$: mmf $\leftrightarrow$ voltage, flux $\leftrightarrow$ current, reluctance $\leftrightarrow$ resistance. Just as series resistors carrying the same current add, series segments of a magnetic path carrying the same flux add their reluctances:

Because air has vastly higher reluctance per unit length than iron, even a millimetre-scale gap can dominate the mmf a coil must supply — the theme of the worked example below.

A ring (toroidal) core of mean length $l = 0.5\,\text{m}$, cross-sectional area $A = 4\times10^{-4}\,\text{m}^2$ and relative permeability $\mu_r = 1000$ is wound with $N = 500$ turns. Find the coil current needed to establish a flux $\Phi = 8\times10^{-4}\,\text{Wb}$, then find the extra mmf needed if a $1\,\text{mm}$ air gap is cut into the core (rest of core now $l_i = 0.499\,\text{m}$, same $A$ and flux, no fringing).

Step 1 — flux density and field intensity. $B = \Phi/A = (8\times10^{-4})/(4\times10^{-4}) = 2\ \text{T}$. Then $H = B/(\mu_0\mu_r) = 2/(4\pi\times10^{-7}\times1000) \approx 1591.5\ \text{A/m}$.

Step 2 — solid core current. $NI = Hl = 1591.5\times0.5 = 795.8\ \text{AT}$, so $I = 795.8/500 \approx 1.59\ \text{A}$. Cross-check via reluctance: $S = l/(\mu_0\mu_r A) \approx 9.95\times10^{5}\ \text{AT/Wb}$, giving $NI = \Phi S \approx 796\ \text{AT}$ — matches.

Step 3 — add the gap: two reluctances in series. Iron path (now $l_i = 0.499\,\text{m}$): $S_i = l_i/(\mu_0\mu_r A) \approx 9.93\times10^{5}\ \text{AT/Wb}$. Air gap ($\mu_r=1$, $l_g = 1\times10^{-3}\,\text{m}$): $S_g = l_g/(\mu_0 A) \approx 1.989\times10^{6}\ \text{AT/Wb}$. Total $S = S_i + S_g \approx 2.982\times10^{6}\ \text{AT/Wb}$.

Step 4 — new mmf and current. $NI = \Phi S = (8\times10^{-4})(2.982\times10^{6}) \approx 2386\ \text{AT}$, so $I = 2386/500 \approx 4.77\ \text{A}$. The current jumps from about 1.59 A to about 4.77 A — roughly triple — because a gap only 0.2% of the mean path length has almost twice the reluctance of the entire iron path.

Competitive stretch

If the same 1 mm gap were doubled to 2 mm (same $A$, same flux), $S_g$ doubles to about $3.98\times10^{6}\ \text{AT/Wb}$, total $S \approx 4.97\times10^{6}\ \text{AT/Wb}$, and $NI \approx 3978\ \text{AT}$, so $I \approx 7.96\ \text{A}$ — the current has grown roughly linearly with the gap length because $S_g$ dominates $S_i$ once the gap is present. This is why designers treat air-gap length as one of the most sensitive parameters in a magnetic circuit.

• Confusing $B$ and $H$: $B = \mu H$ depends on the material, while $H = NI/l$ depends only on current and geometry — never treat their numerical values as interchangeable.
• Using $\mu_0\mu_r$ for the gap reluctance instead of plain $\mu_0$; the gap is air, so $\mu_r = 1$ there even though the rest of the core uses the iron's $\mu_r$.
• Adding fluxes instead of reluctances (or vice versa) in a series path — series segments carry the same flux and their reluctances add, just like series resistors carrying the same current.
• Using the physical/outer core length instead of the mean (centre-line) flux path length in $l$.

Ask the tutor: If the jump from a solid core's 1.59 A to the gapped core's 4.77 A still feels surprising, ask the AI tutor to walk through why the gap reluctance $S_g$ ends up nearly twice $S_i$ despite being a thousand times shorter.