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