Solenoid Magnetic Field
B = µ0 · N · I / l
Quick answer
This calculator solves B = µ₀ × N × I / l (= µ₀ × n × I) for Flux Density. With Number of Turns 0.5, Current 2 A, Solenoid Length 10 cm, the result is 0.1257 G. It solves in 4 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
Inside a long air-core solenoid the magnetic field is nearly uniform and equals the permeability times the number of turns per unit length times the current. Adding a ferromagnetic core multiplies the field by the relative permeability. This is the design equation for electromagnets, relays, inductors, and solenoid actuators.
Variables
- B — Axial flux density inside the solenoid (T)
- N — Total number of turns
- I — Current (A)
- l — Length of the solenoid (m)
- µ₀ = 4π×10⁻⁷ H/m —
Practical Notes
With a high-permeability core, replace µ₀ by µ₀·µr—the field can rise by thousands of times until the core saturates. The field outside an ideal long solenoid is essentially zero.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| B | Flux Density | T | G, mT, T |
| N | Number of Turns | — | — |
| I | Current | A | pA, nA, µA, mA, A, kA |
| l | Solenoid Length | cm | mm, cm, m, km |
Step by step
- 1. What you know: Number of Turns 0.5, Current 2 A, Solenoid Length 10 cm
- 2. The relation: B = µ₀ × N × I / l (= µ₀ × n × I)
- 3. Substitute: N = 0.5, I = 2 A, l = 10 cm
- 4. Result: Flux Density = 0.1257 G
Worked examples
Minimum range
Given: Number of Turns 0.025, Current 2 mA, Solenoid Length 1 mm
Result: Flux Density = 6.28 × 10⁻⁴ G
Low range
Given: Number of Turns 0.25, Current 20 mA, Solenoid Length 10 mm
Result: Flux Density = 0.006283 G
Typical
Given: Number of Turns 0.5, Current 2 A, Solenoid Length 10 cm
Result: Flux Density = 0.1257 G
High range
Given: Number of Turns 0.8, Current 100 A, Solenoid Length 1 m
Result: Flux Density = 1.005 G
Maximum range
Given: Number of Turns 8, Current 1 kA, Solenoid Length 10 m
Result: Flux Density = 10.05 G
Solving for each variable
Solve for Flux Density
Number of Turns 0.025, Current 2 mA, Solenoid Length 1 mm → 6.28 × 10⁻⁴ G
Solve for Number of Turns
Flux Density 10 G, Current 2 mA, Solenoid Length 1 mm → 397.9
Solve for Current
Flux Density 10 G, Number of Turns 0.025, Solenoid Length 1 mm → 31.83 A
Solve for Solenoid Length
Flux Density 10 G, Number of Turns 0.025, Current 2 mA → 6.28 × 10⁻⁵ mm
How Flux Density changes with Number of Turns
| Number of Turns (input) | Flux Density (result) |
|---|---|
| 0.05 | 0.01257 G |
| 0.125 | 0.03142 G |
| 0.25 | 0.06283 G |
| 0.375 | 0.09425 G |
| 0.5 | 0.1257 G |
| 0.75 | 0.1885 G |
| 1 | 0.2513 G |
| 2 | 0.5027 G |
| 3.5 | 0.8796 G |
| 5 | 1.257 G |
The same result in other units
| Unit | Value |
|---|---|
| G | 0.1257 |
| mT | 0.01257 |
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<p><a href="https://vbatpower.com/formulas/solenoid-field/">Solenoid Magnetic Field</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Solenoid Magnetic Field?
Use B = µ₀ × N × I / l (= µ₀ × n × I). With Number of Turns 0.5, Current 2 A, Solenoid Length 10 cm, Flux Density works out to 0.1257 G.
What is the formula for Solenoid Magnetic Field?
The formula is B = µ₀ × N × I / l (= µ₀ × n × I). Inside a long air-core solenoid the magnetic field is nearly uniform and equals the permeability times the number of turns per unit length times the current.
How do you find Number of Turns when you know the other values?
Rearrange for Number of Turns: with Flux Density 10 G, Current 2 mA, Solenoid Length 1 mm, Number of Turns is 397.9.
How do you find Current when you know the other values?
Rearrange for Current: with Flux Density 10 G, Number of Turns 0.025, Solenoid Length 1 mm, Current is 31.83 A.
What units does Solenoid Magnetic Field use?
Flux Density is shown in G, mT. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
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