Electromagnet pull and field energy

How can the pulling force of an electromagnet be calculated? To help answer this question we'll consider a concrete example of an ordinary electromechanical relay; but keep in mind that the procedure outlined here is applicable generally to other problems involving linear motion caused by a current carrying coil.

Our objective is to find the mechanical force, f, exerted on the armature. Two principles form the basis of our approach. The first is the principle of virtual work. That is, we assume that the armature undergoes an infinitesimally small displacement to reduce the air gap. We then consider how the electrical energy supplied to the coil is distributed between the magnetic field and the relay spring using the second principle: energy balance. We represent this balance by an equation -

We = Wm + Wf
Equation LMA

This is saying that (during movement of the armature) the electrical power supplied at the coil terminals, We, is converted into mechanical work, Wm, and an increase in the magnetic field energy by an amount, Wf.   We derive each term separately.

Assume that -

Electrical energy

Our assumption of zero ohmic losses in the coil will not invalidate the result because we can pretend that the armature displacement is sufficiently large or sufficiently quick that the inductive 'kick' from the coil will represent an indefinitely greater power. Faraday's Law says that the change in flux will give a constant voltage across the coil, V,

V = N dΦ/dt
Equation LMB

The electrical power input is V × I or

P = N I dΦ/dt
Equation LMC

and since this power is applied for a time dt the total electrical energy input is then

We = P dt = (N I dΦ/dt) dt = N I dΦ
Equation LMD

From the definition of inductance

Φ = L I / N
Equation LME

In our case I and N are constant, so

dΦ = (I / N) dL
Equation LMF

Substituting into eqn. LMD -

We = N I (I / N) dL = I2 dL
Equation LMG

Field energy

As the gap is reduced the reluctance decreases. If the coil current is held constant then the flux will increase and so will the field energy. This is true even though there is a smaller volume for the field because the energy is proportional to B2. The standard relationship between inductance and stored energy is

W = ½ L I2
Equation LMH

The coil current has not changed so the increase in field energy must represent an increase in coil inductance, dL. So -

Wf = ½ I2 dL
Equation LMI

Where Wf is the increase in field energy (not its total).

Mechanical energy

The standard equation for mechanical work is simply

Wm = f dx
Equation LMJ

The form that this mechanical energy takes is immaterial. It could be compression of a spring, movement against friction or an increase in momentum. Since there isn't actually going to be any motion we don't care; it's only virtual work that we consider. If it helps, you can pretend that the iron has no mass or momentum; it won't alter the validity of the result.

Applying the balance equation

We can now substitute the results of equations LMG, LMI and LMJ into equation LMA

I2 dL = f dx + ½ I2 dL
Equation LMK

Which is easily solved for the force -

f = ½ I2 dL/dx
Equation LML

From this we draw the conclusions -

Now, although the resulting equation LML is very neat, it says nothing about how the inductance of any particular device may be calculated. Inductance (as well as field energy) is determined by the geometry of the coil and all the iron structures within the field. Finding an approximate expression for the inductance is sometimes possible in simple cases such as the relay but next to impossible for others such as lifting magnets.

One way round this problem might be to use field analysis software to integrate the field energy density throughout the whole volume and thus derive the inductance indirectly. There should be more on this topic within a year.

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E-mail: R.Clarke@surrey.ac.uk
Last revised: 2006 January 21st.