Solenoid Coil Electromagnetic Force Calculator
Estimate the pulling force of a solenoid coil from the number of turns, current, core cross-sectional area, and air gap length. Enter your coil parameters to get the electromagnetic force in newtons.
Solenoid Magnetic Force Estimator
Solenoid Electromagnetic Force Formula
Where:
- \(F\) — electromagnetic force (N)
- \(N\) — number of turns in the coil
- \(I\) — current through the coil (A)
- \(A\) — cross-sectional area of the core (m²)
- \(g\) — length of the air gap between core and armature (m)
- \(B\) — magnetic flux density in the gap (T)
- \(\mu_0\) — permeability of free space, \(4\pi \times 10^{-7}\) H/m
The first equation is the standard force law for a magnetic field acting across an air gap. Because the air gap dominates the magnetic reluctance of the circuit, the flux density can be approximated as \(B \approx \mu_0 N I / g\), which combines into the working formula used by the calculator.
Solenoid Coil Electromagnetic Force – Calculation Example
Example: a solenoid actuator has 100 turns, carries a current of 2 A, the core cross-section is 2 cm², and the armature sits 1 mm away from the core.
Convert the units: \(A = 2~\text{cm}^2 = 2 \times 10^{-4}~\text{m}^2\), \(g = 1~\text{mm} = 0.001~\text{m}\).
\[ F = \frac{4\pi \times 10^{-7} \times (100 \times 2)^2 \times 2 \times 10^{-4}}{2 \times (0.001)^2} \]
\[ F = \frac{4\pi \times 10^{-7} \times 40000 \times 2 \times 10^{-4}}{2 \times 10^{-6}} \approx 5.03~\text{N} \]
Result: the solenoid produces an electromagnetic force of approximately 5.03 N, or about 0.51 kgf of pull.
How to use the calculator
- Enter the Number of turns in the coil (default: 100).
- Enter the Current flowing through the coil in amperes.
- Enter the Core area — the cross-sectional area of the magnetic core — in cm².
- Enter the Air gap, the distance between the core and the armature (the moving part), in mm.
- Read the Electromagnetic Force in newtons below the inputs.
What is solenoid electromagnetic force?
A solenoid is a coil of wire that generates a magnetic field when current flows through it. When the coil is wrapped around a ferromagnetic core and placed near a movable armature, the magnetic field pulls the armature toward the core, converting electrical energy into mechanical motion. This pulling force is what the calculator estimates.
The force arises from the energy stored in the air gap. As the gap closes, the magnetic circuit's reluctance drops and the field does mechanical work on the armature. The classic result, valid for a uniform field across a small gap, is \(F = B^2 A / (2\mu_0)\) — the force grows with the square of the flux density and with the pole face area.
Two practical consequences follow from the formula. First, the force scales with \((NI)^2\), so doubling either the current or the number of turns quadruples the force. Second, the force scales with \(1/g^2\), meaning the pull rises very steeply as the armature approaches the core. A solenoid that can only manage a few newtons at a 5 mm gap may deliver hundreds of newtons once seated.
Solenoids are found in relays, valves, door locks, starters, printers, and robotic grippers. This calculator is most useful during the early design stage, when you need to check whether a given coil geometry and drive current can produce enough force to move a load at the required stroke.
Assumptions and limitations
- Linear, unsaturated core. The formula assumes the iron's permeability is so high that the air gap dominates the total reluctance. If the core saturates, the actual force will be lower than predicted.
- Fringing is neglected. Real flux bulges out at the edges of the gap, effectively increasing the area. The effect is usually small for gap lengths much shorter than the pole dimensions.
- Uniform field. The gap should be small and the pole faces parallel for the \(1/g^2\) relationship to hold.
- DC operation. For AC solenoids, use the RMS current; inductance and eddy-current effects are not modeled.
- Copper losses and heating are not included. Check the coil's resistance and duty cycle separately; a coil that produces enough force but overheats is still a failed design. For winding-related parameters, see the Physical Properties of Coil and Wire Gauge calculators.
Practical tips
To increase force without raising current, add turns or reduce the gap — both are more effective than increasing the core area, because force depends on \((NI)^2\) and \(1/g^2\) but only linearly on \(A\). Keep in mind that more turns mean higher resistance and a longer, weaker magnetic path if the coil grows.
If you are designing the driver circuit or analyzing the coil as a circuit element, the Straight Wire Inductance and Series Inductance calculators can help you estimate the inductive load the solenoid presents.
Frequently Asked Questions
How do I calculate solenoid force by hand without a calculator?
Multiply the number of turns by the current to get the magnetomotive force, square it, multiply by the core area in m² and by \(4\pi \times 10^{-7}\), then divide by twice the air gap squared (in meters). Example: 500 ampere-turns, a 3 cm² core, and a 2 mm gap give \(F = 4\pi \times 10^{-7} \times 250000 \times 3 \times 10^{-4} / (2 \times 4 \times 10^{-6}) \approx 11.8\) N.
Why is the air gap so important to solenoid force?
Force is inversely proportional to the square of the gap, so halving the gap quadruples the pull. Going from 4 mm to 1 mm gives 16 times more force. This is also why a solenoid's force varies enormously across its stroke: the specified force usually applies at the maximum gap, and once the armature seats, the pull is far higher than the datasheet's nominal rating.
What unit conversions do I need for the solenoid force formula?
The formula only works in SI units, so convert area from cm² to m² by dividing by 10,000, and gap length from mm to m by dividing by 1,000. Current stays in amperes. To interpret the result, note that 1 newton ≈ 0.102 kgf, so a 10 N solenoid lifts roughly a 1 kg mass in the vertical orientation.
What is a typical force for small solenoids?
Miniature solenoids usually deliver 1–10 N at their rated stroke, general-purpose industrial types range from 20–100 N, and large actuators like starter solenoids or heavy door locks can exceed several hundred newtons. If your calculated force is far below the load you need, reduce the air gap or increase the ampere-turns rather than the core area, since area only affects force linearly.
Why does my real solenoid produce less force than calculated?
The most common cause is core saturation: once the iron's flux density approaches about 1.5–2 T, the permeability drops and the actual force falls below the prediction. Other contributors include flux fringing at the pole edges, leakage flux that never crosses the gap, joints in the magnetic path, and non-parallel pole faces. Expect real devices to deliver 50–80% of the ideal value at longer strokes.
Can I use this calculator for AC solenoids?
Only approximately. Enter the RMS current, but be aware that AC coils also have inductive reactance that limits current and eddy currents that reduce the effective flux. A true AC solenoid force also pulsates at twice the line frequency and averages about half the peak value. For accurate AC design, you need the coil's inductance and impedance, not just the turns and current.
How can I increase solenoid force without raising the current?
Add more turns or shorten the air gap — both are far more effective than enlarging the core. Since force scales with \((NI)^2\), doubling the turns doubles the ampere-turns and quadruples the force at the same current. The trade-offs are higher coil resistance (which reduces current for a voltage-driven coil), longer winding time, and more copper cost, so verify the power supply can still deliver the required current.