Solenoid Force Calculator:
Enter the values of number of coil turns, n and current, I(A), permeability constant, c(N.m2/A2), cross-sectional area, A(m2) and distance, d(m) to determine the value of Force, F(N).
Solenoid Force Formula:
A solenoid is an electromagnet formed by winding a wire coil around a cylindrical core, generating a magnetic field when electric current flows through it. The solenoid’s force is a magnetic pulling force produced at its core or on an attached armature, depending on the solenoid’s configuration. This force is used in various mechanical and electromagnetic systems such as actuators, relays, and valves.
The strength of the force depends on the number of coil turns (n), the electric current (I), the area of the solenoid cross-section (A), the permeability constant (c), and the distance between the solenoid and the object being attracted (d). The force increases with higher current or more turns of the coil and decreases with larger distance d.
Force, F(N) in Newtons equals the square of the product of the number of coil turns, n and current, I(A) in amperes multiplied by the permeability constant, c(N.m2/A2) in Newton metre square per ampere square and the solenoid cross-sectional area, A(m2) in metre square, divided by twice the square of the distance, d(m) in metres.
Force, F(N) = (n * I(A))2 * c(N.m2/A2) * A(m2) / 2 * d2(m)
F(N) = force in Newtons, N.
n = number of coil turns.
I(A) = current in amperes, A.
c(N.m2/A2) = permeability constant in Newton metre square per ampere square, N.m2/A2.
A(m2) = cross-sectional area in metre square, m2.
d(m) = distance in metres, m.
Solenoid Force Calculation:
1.A solenoid with 500 turns carries a current of 2A. The cross-sectional area is 0.002m2, the permeability constant is 4π * 10−7N.m2/A2 and the distance to the object is 0.05m. Calculate the force.
Given: n = 500turns, I(A) = 2A, c(N.m2/A2) = 4π * 10−7N.m2/A2, A(m2) = 0.002m2, d(m) = 0.05m.
Force, F(N) = (n * I(A))2 * c(N.m2/A2) * A(m2) / 2 * d2(m)
F(N) = (500 * 2)2 * 12.56 * 10‑7 * 0.002 / 2 * 0.052
F(N) = 1000000 * 12.56 * 10‑7 * 0.002 / 2 * 0.0025
F(N) = 25.12 * 10-4 / 0.005
F(N) = 0.502N.
2.Given: n = 400turns, I(A) = 3A, c(N.m2/A2) = 4π * 10−7N.m2/A2, F(N) = 0.5N, d(m) = 0.05m.
Force, F(N) = (n * I(A))2 * c(N.m2/A2) * A(m2) / 2 * d2(m)
A(m2) = F(N) * 2 * d2(m) / (n * I(A))2 * c(N.m2/A2)
A(m2) = 0.5 * 2 * 0.052 / (400 * 3)2 * 12.56 * 10‑7
A(m2) = 1 * 0.0025 / 1440000 * 12.56 * 10-7
A(m2) = 0.0025 / 1.806
A(m2) = 0.001385m2.