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H
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mH
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μH
2,500
turns/m
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H
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mH
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μH
2,500
turns/m
The Solenoid Inductance Calculator computes the self-inductance of a solenoid coil using $$L = \mu_0 \mu_r n^2 A l = \frac{\mu_0 \mu_r N^2 A}{l}$$ where N is the total turns, A is the cross-sectional area, l is the length, and μᵣ is the core relative permeability.
Inductance measures a coil's ability to store energy in its magnetic field and to resist changes in current. It is a critical parameter in filter design, power supply regulation, motor control, RF circuits, and energy storage applications.
Self-inductance relates the magnetic flux linkage through a coil to the current producing it: $$L = \frac{N\Phi}{I}$$. For a solenoid with uniform internal field $$B = \mu_0 \mu_r n I$$, the flux through each turn is $$\Phi = BA$$, and the total flux linkage is:
$$N\Phi = N \cdot (\mu_0 \mu_r n I) \cdot A = \mu_0 \mu_r \frac{N^2}{l} A \cdot I$$
Dividing by I gives the inductance:
$$L = \mu_0 \mu_r \frac{N^2 A}{l} = \mu_0 \mu_r n^2 A l$$
Key observations:
The energy stored in an inductor is $$W = \frac{1}{2}LI^2$$. The voltage across an inductor is $$V = L\frac{dI}{dt}$$ — it opposes changes in current, which is the basis of inductive filtering and energy storage.
This formula is for an ideal solenoid (length ≫ diameter). For shorter coils, correction factors (Nagaoka coefficient) improve accuracy. Multi-layer coils and toroidal shapes require modified formulas.
The calculated inductance tells you how much the coil resists current changes and how much energy it stores. Higher inductance means stronger filtering of AC signals and more stored energy. For circuit design, the inductance determines the time constant (L/R in RL circuits) and resonant frequency (with a capacitor: $$f = 1/(2\pi\sqrt{LC})$$). Ensure the inductance value matches your circuit requirements.
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A small 20-turn air-core coil (3 cm long, ~2.5 cm diameter) has ~8.4 μH — suitable for MHz-range RF circuits.
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With an iron core (μᵣ = 2000), a 200-turn coil achieves ~0.8 H — useful for mains frequency filtering and chokes.
The self-inductance of a solenoid is $$L = \frac{\mu_0 \mu_r N^2 A}{l}$$, where N is total turns, A is cross-sectional area, l is length, and μᵣ is relative permeability of the core. Equivalently, with turn density n = N/l: $$L = \mu_0 \mu_r n^2 A l$$.
Each of the N turns contributes to the total magnetic flux, and each turn also links with the flux from all other turns. The flux through one turn is proportional to N (since B ∝ nI = NI/l), and there are N turns contributing to the total flux linkage, giving NΦ ∝ N² and thus L ∝ N².
A ferromagnetic core (iron, ferrite) amplifies the magnetic field by its relative permeability μᵣ. Since inductance is proportional to μᵣ, an iron core with μᵣ = 2000 increases inductance 2000-fold compared to air. This is why practical inductors and transformers use magnetic cores — achieving large inductance in compact sizes.
The Nagaoka coefficient (K) corrects the ideal solenoid formula for finite length-to-diameter ratios. For short, wide coils (length ≈ diameter), K can be 0.5–0.8, significantly reducing inductance from the ideal formula. For long thin solenoids (length ≫ diameter), K approaches 1 and the ideal formula is accurate.
An inductor stores energy $$W = \frac{1}{2}LI^2$$ in its magnetic field. For example, a 1 H inductor carrying 10 A stores 50 J. This stored energy is released when the current decreases, which is why inductors produce voltage spikes when switched off abruptly — requiring snubber circuits for protection.
Self-inductance (L) relates a coil's own flux linkage to its own current. Mutual inductance (M) relates the flux linkage in one coil due to current in another coil. Mutual inductance is the basis of transformers: $$M = k\sqrt{L_1 L_2}$$, where k is the coupling coefficient (0 to 1).
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