The Core Question: What Is an Example of Magnetism in a Circuit?
When students or hobbyists ask, "what is an example of magnetism" in a practical electrical context, the most quantifiable and ubiquitous answer is the solenoid (the electromagnetic coil found in relays, contactors, and inductors). Unlike permanent magnets, electromagnets allow us to control magnetic flux density ($B$) precisely by adjusting current and coil geometry. This makes them the perfect subject for circuit theory exams and bench validation.
Below is a classic exam-style practice problem. We will calculate the magnetic field inside an air-core solenoid and then make a concrete hardware decision to build it.
You are designing an air-core RF choke (solenoid) for a filter circuit. The coil has 400 turns of wire, a total winding length of 0.1 meters, and carries a continuous DC current of 2.0 Amps.
1. Calculate the magnetic flux density ($B$) inside the center of the coil in Tesla.
2. Select the correct AWG wire size to wind this coil safely without thermal failure.
Method Selection: Ampère’s Law vs. Biot-Savart
To solve for the magnetic field, we must choose the right theorem. The two primary methods for calculating magnetic fields from currents are the Biot-Savart Law and Ampère’s Law.
- Biot-Savart Law: Requires integrating the vector contributions of every infinitesimal wire segment. It is universally applicable but mathematically brutal for anything other than a single straight wire or a simple loop.
- Ampère’s Law: Exploits symmetry. By drawing a closed Amperian loop where the magnetic field is either constant or zero, we can bypass complex calculus entirely.
The Verdict: We use Ampère’s Law. Because the solenoid's length (0.1m) is significantly greater than its assumed diameter, the magnetic field inside is highly uniform and parallel to the axis, while the field outside is effectively zero. This symmetry perfectly satisfies Ampère's conditions. For a long solenoid, Ampère's Law simplifies to the standard formula:
$B = \mu_0 \cdot \mu_r \cdot n \cdot I$
(Where $n$ is turn density, $N/L$). You can review the derivation of this via Georgia State University's HyperPhysics resource on solenoids.
Step-by-Step Solution: Calculating the Magnetic Flux Density
Let’s break down the algebra without skipping steps. We are working in SI units (meters, amps, Tesla).
Step 1: Identify the known variables.
- Total turns ($N$) = 400
- Length ($L$) = 0.1 m
- Current ($I$) = 2.0 A
- Permeability of free space ($\mu_0$) = $4\pi \times 10^{-7}$ T·m/A
- Relative permeability of air ($\mu_r$) = 1 (Air is non-magnetic)
Step 2: Calculate turn density ($n$).
$n = N / L$
$n = 400 / 0.1$
$n = 4,000$ turns per meter.
Step 3: Substitute all values into the Ampère's Law solenoid equation.
$B = (4\pi \times 10^{-7}) \times (1) \times (4,000) \times (2.0)$
Step 4: Group the powers of 10 and simplify.
$B = 4\pi \times 10^{-7} \times 8,000$
$B = 4\pi \times 10^{-7} \times (8 \times 10^3)$
$B = 32\pi \times 10^{-4}$
Step 5: Calculate the final decimal value.
$B = 32 \times 3.14159 \times 10^{-4}$
$B \approx 100.53 \times 10^{-4}$
$B \approx 0.01005$ Tesla (or 10.05 mT)
The Trap & Sanity Check: Verifying the Answer
Sanity Check (Order of Magnitude):
The Earth’s magnetic field is approximately 50 µT (0.00005 T). Our calculated field is 10.05 mT (0.01005 T). Dividing 0.01005 by 0.00005 gives roughly 200. It is entirely reasonable that a 2-Amp coil with 4,000 turns/meter generates a field about 200 times stronger than the Earth's ambient field. The units (Tesla) are correct, and the magnitude passes the real-world sniff test.
Independent Verification:
On the bench, you don't trust the math until you measure it. To verify this independently, wind the physical coil and place a linear Hall-effect sensor (like the SS49E) directly in the bore. Power the coil with a bench supply limited to 2.0A. The SS49E outputs 1.4 mV/Gauss. Since 10.05 mT equals 100.5 Gauss, you should see a voltage offset of approximately 140.7 mV above the sensor's 2.5V quiescent baseline. (For more on magnetic measurement, see the All About Circuits DC textbook chapter on magnetism).
Decision Path: Selecting the Wire Gauge for the Coil
Knowing the magnetic field is only half the engineering task. We must select a wire gauge that can handle 2.0A continuously inside a tightly wound coil, where heat dissipation is poor. We will use standard copper magnet wire (polyurethane or polyester-imide insulated).
| AWG Size | Max Ampacity (Chassis) | Max Ampacity (Tight Coil Derated) | Decision Path (If-Then) |
|---|---|---|---|
| 26 AWG | 1.3 A | ~0.8 A | IF current < 1A, pick 26 AWG. ELSE reject (will melt at 2A). |
| 24 AWG | 2.1 A | ~1.4 A | IF current < 1.5A, pick 24 AWG. ELSE reject (thermal runaway risk in a dense coil). |
| 22 AWG | 3.5 A | ~2.4 A | IF current is 2.0A, PICK 22 AWG. Provides a 20% thermal safety margin. |
| 20 AWG | 5.0 A | ~3.5 A | IF space is unconstrained, pick 20 AWG. ELSE reject (too bulky for 400 turns in 0.1m). |
Concrete Pick: Based on the decision matrix, you must purchase 22 AWG Copper Magnet Wire (e.g., MWS Wire Industries PN: 22AWG-PE or equivalent). At 0.64mm diameter, 400 turns will occupy roughly 256mm of linear space if wound single-layer, meaning you will need to wind it in approximately 3 layers to fit your 0.1m (100mm) bobbin length. This confirms 22 AWG is physically viable and electrically safe.
Frequently Asked Questions
What happens if I use a ferromagnetic core instead of air?
If you insert an iron core, the relative permeability ($\mu_r$) jumps from 1 to anywhere between 200 and 5,000. In our formula, $B$ would multiply by that factor. However, the trap is that ferromagnetic materials saturate. Once the core reaches roughly 1.5 to 2.0 Tesla, $\mu_r$ effectively drops back toward 1, and adding more current yields almost zero increase in magnetic field. Air cores do not saturate, which is why they are preferred in high-current RF applications.
Does the diameter of the solenoid affect the magnetic field?
In the idealized Ampère's Law equation for a "long" solenoid, diameter does not appear. The field strength depends only on turn density ($n$) and current ($I$). However, if the diameter approaches the length of the coil, the "long solenoid" assumption breaks down, edge effects dominate, and the actual field at the center will be lower than the formula predicts.
Can I use stranded wire instead of solid magnet wire?
Technically yes, but practically no. Stranded wire has a larger overall diameter due to the insulation and air gaps between strands. This reduces your turn density ($n$) for a given volume, lowering your magnetic field. Furthermore, standard PVC insulation on stranded wire will melt under the heat of a tightly packed 2A coil. Always use enamel-coated solid magnet wire for solenoids.
