Solving a Magnetic Circuit Magnetism Example: Air Gap and Fringing Walkthrough

When you are designing relays, contactors, or custom inductors, magnetic circuit analysis is just as critical as Ohm's law for standard wiring. Yet, most textbook problems gloss over the physical realities that cause bench prototypes to fail. In this guide, we will dissect a comprehensive magnetism example that mirrors real-world electromechanical design, focusing on the non-linear B-H characteristics of silicon steel and the often-missed fringing effects in air gaps.

The Magnetism Example: Problem Statement and Core Parameters

To solve this, we apply Hopkinson’s Law (the magnetic equivalent of Ohm’s Law, $\mathcal{F} = \Phi \mathcal{R}$) derived from Ampere’s Circuital Law. The total magnetomotive force (MMF, $\mathcal{F}$) supplied by the coil must equal the sum of the MMF drops across the steel core and the air gap.

Reference Data: M-6 Silicon Steel B-H Characteristics

You cannot use a constant permeability ($\mu$) for ferromagnetic materials. As flux density ($B$) increases, the steel saturates, requiring disproportionately more magnetic field intensity ($H$). Below is the empirical B-H data for M-6 silicon steel, a standard material used in transformer and inductor cores. We will interpolate from this table during our solution.

Source data adapted from standard magnetics reference tables for M-6 grain-oriented electrical steel. For deeper theoretical background on magnetic circuits, refer to the All About Circuits magnetic circuits chapter or MIT OpenCourseWare's lecture on magnetic circuits.

Step-by-Step Algebraic Solution and Common Traps

Before calculating, we must define our geometric parameters in standard SI units (meters and square meters).

• Total path length ($l_{total}$): $40 \text{ cm} = 0.4 \text{ m}$
• Air gap length ($l_g$): $2 \text{ mm} = 0.002 \text{ m}$
• Net core length ($l_c$): $l_{total} - l_g = 0.4 - 0.002 = 0.398 \text{ m}$
• Core Area ($A_c$): $10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2$
• Desired Flux ($\Phi$): $1.2 \text{ mWb} = 1.2 \times 10^{-3} \text{ Wb}$

Step 1: Calculate Flux Density in the Core ($B_c$)

$$B_c = \frac{\Phi}{A_c} = \frac{1.2 \times 10^{-3} \text{ Wb}}{10 \times 10^{-4} \text{ m}^2} = 1.2 \text{ T}$$

Step 2: Determine Field Intensity in the Core ($H_c$)

Looking at our B-H table above, at $B_c = 1.2 \text{ T}$, the corresponding field intensity is exactly:

$$H_c = 130 \text{ A/m}$$

Step 3: Calculate MMF Drop Across the Steel Core ($\mathcal{F}_c$)

$$\mathcal{F}_c = H_c \times l_c = 130 \text{ A/m} \times 0.398 \text{ m} = 51.74 \text{ A}\cdot\text{t}$$

Step 4: Calculate Flux Density in the Air Gap ($B_g$)

First, apply the 5% fringing factor to the gap area:

$$A_g = 1.05 \times A_c = 1.05 \times 10 \times 10^{-4} \text{ m}^2 = 1.05 \times 10^{-3} \text{ m}^2$$

Now, find $B_g$:

$$B_g = \frac{\Phi}{A_g} = \frac{1.2 \times 10^{-3} \text{ Wb}}{1.05 \times 10^{-3} \text{ m}^2} \approx 1.1428 \text{ T}$$

Step 5: Determine Field Intensity in the Air Gap ($H_g$)

The permeability of free space ($\mu_0$) is $4\pi \times 10^{-7} \text{ H/m}$ (approx. $1.2566 \times 10^{-6} \text{ H/m}$).

$$H_g = \frac{B_g}{\mu_0} = \frac{1.1428}{4\pi \times 10^{-7}} \approx 909,465 \text{ A/m}$$

Step 6: Calculate MMF Drop Across the Air Gap ($\mathcal{F}_g$)

$$\mathcal{F}_g = H_g \times l_g = 909,465 \text{ A/m} \times 0.002 \text{ m} = 1818.93 \text{ A}\cdot\text{t}$$

Step 7: Calculate Total Required Current ($I$)

Total MMF ($\mathcal{F}_{total}$) is the sum of the drops. Since $\mathcal{F}_{total} = N \times I$:

$$N \cdot I = \mathcal{F}_c + \mathcal{F}_g$$

$$500 \cdot I = 51.74 + 1818.93 = 1870.67 \text{ A}\cdot\text{t}$$

$$I = \frac{1870.67}{500} \approx 3.74 \text{ A}$$

Sanity Checks and Independent Verification

Before moving on to the next exam question or firing up your bench power supply, run these checks to ensure your answer holds up to physical reality.

1. Order of Magnitude and Physical Dominance
Notice that the MMF drop across the tiny 2 mm air gap ($1818.93 \text{ A}\cdot\text{t}$) is roughly 35 times larger than the drop across nearly 40 cm of solid steel ($51.74 \text{ A}\cdot\text{t}$). This perfectly aligns with magnetic theory: air has a relative permeability of 1, while the steel is operating at $\mu_r \approx 7345$. The air gap dominates the reluctance of the circuit. If your calculation showed the steel requiring more MMF than the gap, you would immediately know you either saturated the core past 2.0 T or forgot to convert millimeters to meters.

2. Unit Verification
$H$ is in Amperes/meter. Length is in meters. Multiplying them yields Ampere-turns ($\text{A}\cdot\text{t}$). Dividing total $\text{A}\cdot\text{t}$ by the unitless turn count ($N$) leaves Amperes. The dimensional analysis is flawless.

3. Independent Verification via Reluctance
You can verify this answer by calculating the total reluctance ($\mathcal{R}$) of the circuit instead of using field intensities.

• Core Reluctance: $\mathcal{R}_c = \frac{l_c}{\mu_c A_c} = \frac{0.398}{(0.00923)(10 \times 10^{-4})} \approx 43,120 \text{ A}\cdot\text{t/Wb}$
• Gap Reluctance: $\mathcal{R}_g = \frac{l_g}{\mu_0 A_g} = \frac{0.002}{(4\pi \times 10^{-7})(1.05 \times 10^{-3})} \approx 1,515,833 \text{ A}\cdot\text{t/Wb}$
• Total Reluctance: $\mathcal{R}_{total} = 43,120 + 1,515,833 = 1,558,953 \text{ A}\cdot\text{t/Wb}$
• Total MMF: $\mathcal{F} = \Phi \times \mathcal{R}_{total} = (1.2 \times 10^{-3}) \times 1,558,953 \approx 1870.74 \text{ A}\cdot\text{t}$

Dividing $1870.74 \text{ A}\cdot\text{t}$ by 500 turns yields 3.74 A, confirming our previous step-by-step derivation. The minor decimal variance is strictly due to rounding $\mu_c$ in the table lookup versus the direct $B/H$ division.

Frequently Asked Questions

Why do we use Ampere's Law instead of just $B = \mu H$ for the whole circuit?
Because $\mu$ is not constant in ferromagnetic materials. The B-H curve is highly non-linear. You can only use $B = \mu H$ for the air gap (where $\mu = \mu_0$) or for the steel if you are operating deep in the linear, unsaturated region below 0.5 T. At 1.2 T, you must rely on empirical B-H table lookups.

What happens if I ignore the 5% fringing factor in a real build?
If you ignore fringing, you will calculate a higher $B_g$, leading to a higher required $H_g$, and ultimately overestimate the required current by about 2.5%. In a precision relay design, this could mean specifying a 4.0 A power supply when a 3.75 A supply would suffice, wasting budget and increasing thermal dissipation.