What Is an Example of an Electromagnet? Real-World Specs & Practice Problem

When a student asks, "what is an example of an electromagnet?", the textbook answer is usually a steel nail wrapped in copper wire connected to a D-cell battery. But on a jobsite, in a manufacturing plant, or on an electrical engineering exam, an electromagnet is a precision electromechanical actuator. Real-world examples include the coils inside AC contactors that start 50-horsepower motors, solenoid valves that control hydraulic fluid, and massive DC lifting magnets used in scrap yards to move tons of ferrous metal.

At their core, all these devices rely on the same principle: passing current through a coil generates a magnetomotive force (MMF) that drives magnetic flux through a core and an air gap. To master this concept, we need to move past the nail-and-wire analogy and look at actual industrial specifications, followed by a rigorous circuit calculation.

Real-World Examples of Electromagnets in Industry

Before solving magnetic circuit problems, you need to understand the physical parameters of real devices. The table below details four common industrial electromagnets, highlighting the vast differences in coil resistance, ampere-turns (NI), and core materials. Notice that higher voltage does not necessarily mean higher ampere-turns; the magnetic requirement is dictated by the mechanical work the device must perform.

Device / Model	Application	Nominal Voltage	Coil Resistance	Typical Ampere-Turns (NI)	Core Material
Schneider Electric LC1D09 Contactor	Motor Starting	24V DC	14.4 Ω	~1,200 AT	Silicon Steel Laminations
ASCO 8210G15 Solenoid Valve	Fluid Control	120V AC	21 Ω (impedance)	~800 AT	430 Stainless Steel
MW5-130 Lifting Magnet	Scrap Handling	220V DC	4.8 Ω	~15,000 AT	Low Carbon Cast Steel
MagneMotion Linear Motor	Factory Automation	48V DC	2.0 Ω	~3,500 AT	Neodymium / Iron

Source data compiled from manufacturer datasheets and standard magnetic circuit theory.

Practice Problem: Calculating Flux Density in a Lifting Magnet

Method and Theorem Applied

We will use Hopkinson’s Law (also known as Magnetic Ohm’s Law), which states that Magnetomotive Force (𝒡) equals Magnetic Flux (Φ) multiplied by Reluctance (ℛ): 𝒡 = Φ ℛ. This is the direct magnetic analog to V = IR in electrical circuits. A comprehensive breakdown of this analogy can be found in Georgia State University’s HyperPhysics magnetic circuit module.

Step-by-Step Algebraic Solution

Step 1: Identify Constants and Calculate MMF

• Permeability of free space (μ₀) = 4π × 10^-7 T·m/A ≈ 1.2566 × 10^-6 T·m/A
• MMF (𝒡) = N × I = 500 turns × 5 A = 2,500 A·t

Step 2: Calculate Reluctance of the Steel Core (ℛ_c)

Formula: ℛ_c = l_c / (μ₀ × μ_rc × A)

• ℛ_c = 1.2 / (1.2566 × 10^-6 × 2000 × 0.05)
• ℛ_c = 1.2 / 0.00012566
• ℛ_c = 9,549 A·t/Wb

Step 3: Calculate Reluctance of the Air Gap (ℛ_g)

Formula: ℛ_g = l_g / (μ₀ × μ_rg × A). Note that for air, μ_rg = 1.

• ℛ_g = 0.002 / (1.2566 × 10^-6 × 1 × 0.05)
• ℛ_g = 0.002 / (6.283 × 10^-8)
• ℛ_g = 31,831 A·t/Wb

Step 4: Calculate Total Reluctance (ℛ_tot)

Since the core and gap are in series:

• ℛ_tot = ℛ_c + ℛ_g = 9,549 + 31,831 = 41,380 A·t/Wb

Step 5: Calculate Magnetic Flux (Φ)

• Φ = 𝒡 / ℛ_tot = 2,500 / 41,380 = 0.0604 Wb

Step 6: Calculate Flux Density (B)

• B = Φ / A = 0.0604 / 0.05 = 1.208 T (Tesla)

Answer Sanity Check

Independent Verification and Common Exam Traps

The Trap: Physical Size vs. Magnetic Resistance

The most common mistake students make on this type of exam problem is the "Physical Size Trap." The steel core is 1.2 meters long, while the air gap is only 0.002 meters long. Because the core is 600 times longer physically, a beginner might intuitively assume the core dominates the circuit’s reluctance.

However, look at the math: the air gap provides 77% of the total reluctance (31,831 out of 41,380 A·t/Wb). Even though the gap is microscopic compared to the core, air is 2,000 times less permeable than the steel. In magnetic circuits, material permeability almost always overrides physical length when dealing with small air gaps.

How to Verify the Answer Independently

You can quickly verify your answer using an engineering approximation. In many magnetic circuits, if the core permeability is high enough, we assume the core has infinite permeability (μ_rc = ∞), meaning its reluctance drops to zero. Let’s test this assumption:

1. Assume ℛ_tot ≈ ℛ_g = 31,831 A·t/Wb.
2. Calculate approximate flux: Φ_approx = 2,500 / 31,831 = 0.0785 Wb.
3. Calculate approximate B-field: B_approx = 0.0785 / 0.05 = 1.57 T.

Our exact calculation yielded 1.208 T, while the idealized gap-only calculation yielded 1.57 T. This independent check proves two things: first, our exact math is logically sound (it must be lower than the idealized version because the core does consume some MMF); second, it proves that for this specific device, you cannot safely ignore the core reluctance, as doing so would overestimate the lifting force by roughly 30%.

Frequently Asked Questions

Does the shape of the electromagnet core change the calculation?
Yes. The formula ℛ = l / (μA) assumes a uniform cross-sectional area. If the core flares out at the poles (a common design in lifting magnets to reduce fringing flux), you must calculate the reluctance in segments, using the specific area and length for each geometric section, then sum them in series.

Why do AC electromagnets (like contactors) have shading coils?
AC current passes through zero 120 times a second (on a 60Hz grid). Without a shading coil—a shorted copper ring embedded in the pole face—the magnetic flux would drop to zero every half-cycle, causing the contactor to violently chatter and destroy its mechanical contacts. The shading coil acts as a shorted transformer secondary, generating an out-of-phase magnetic field that holds the armature closed during the zero-crossings.

How do I convert Flux Density (Tesla) to actual Lifting Force (Newtons)?
Use Maxwell’s pulling force formula: F = (B² × A) / (2 × μ₀). For our problem, F = (1.208² × 0.05) / (2 × 1.2566 × 10^-6) ≈ 29,000 Newtons, or roughly 2.9 metric tons of lifting force.