**Q.** For an electromagnetic system, show that the energy stored in a magnetic field is equal to the area enclosed b/w magnetization curves for open and closed position of the armature and flux linkage - current locus during the armature moment.

**Sol. **Consider a simple magnetic relay. Initially the armature is in the open position. When switch S is closed, current I is established in the N turn - coil. The flux set up depends upon mmf N_{i} and the reluctance of the magnetic path. The magnetic field thus produced, creates North and South poles and as a result of it., there is established a magnetic force tending to shortern the air - gap.

If the armature is not allowed to move, the mechanical work done, dW_{mech.} Is zero.

Therefore,

dW_{elec} = 0 + dW_{fld}

(dW_{elec} = ** **dW_{mech.} + dW_{fld})

This shows that when the movable part of any physical system is kept fixed, the entire electrical energy input is stored in the magnetic field.

dW_{fld} = dW_{elec}

and dW_{fld} = dW_{elec} = i.dψ = F.dΦ

If the initial flux is zero, then the magnetic field energy stored W_{fld}, in establishing a flux Φ_{1} or flux linkage ψ_{1}, is given by

W_{fld} = ^{ψ}_{1} _{0}(i.d ψ_{1} = ^{Φ}_{1} _{0}(F.d Φ

I and F must be expressed in terms of ψ and Φ.

When the armature is held in open position then the most of the m.m.f. is consumed in the air - gap and it is likely that magnetic saturation may not occur.

W_{fld} = ^{Φ}_{1} _{0}( dW_{fld} = ^{Φ}_{1} _{0}(F.dQ = area OABO

W_{fld} = ^{ψ}_{1} _{0}(d. W_{fld} = ^{ψ}_{1} _{0}(i.d ψ_{1} = area OABO

Area OACO = (d. W_{fld} = ^{F}_{1 0}( Φ.dF = ^{i}_{1} _{0}( ψ.di

This area OACO is called the co - energy W_{fld}

** **

W^{'}_{fld} = ^{F}_{1 0}( Φ.dF = ^{i}_{1} _{0}( ψ.di

Above, ψ and Φ must be expressed in terms of F and I respectively. Co - energy has no physical significance, it is however useful in calculating the magnetic forces.

With no magnetic saturation,

Area OABO = Area OACO

Or W_{fld} = W^{'}_{fld}

And W_{fld} + W^{'}_{fld} = Area OCABO = Φ_{1}F_{1} = ψ_{1}i_{1}

In general, for magnetic circuit,

W_{fld} = W^{'}_{fld} = 1/2 ψi = 1/2FΦ