### Demagnetising and Cross Magnetizing Conductors

The conductors which are responsible for producing demagnetizing and distortion effects are shown in the Fig.1. Fig. 1

The brushes are lying along the new position of MNA which is at angle θ  from GNA. The conductors in the region AOC = BOD = 2θ  at the top and bottom of the armature are carrying current in such a direction as to send the flux in armature from right to left. Thus these conductors are in direct opposition to main field and called demagnetizing armature conductors.
The remaining armature conductors which are lying in the region AOD and BOC carry current in such a direction as to send the flux pointing vertically downwards i.e. at right angles to the main field flux. Hence these conductors are called cross magnetizing armature conductors which will cause distortion in main field flux.
These conductors are shown in the Fig. 2 Fig. 2

1.1 Calculation of Demagnetizing and Cross Magnetizing Amp-Turns
Let us the number of demagnetizing and cross magnetizing amp-turns.
Let       Z    = Total number of armature conductors
P = Number of poles
I = Armature conductor current in Amperes
= Ia/2 for simplex wave winding
= Ia/P for simplex lap winding
θm = Forward lead of brush in mechanical degrees.
The conductors which are responsible for demagnetizing ampere-turns are lying in the region spanning 4 θm degrees. The region is between angles AOC and BOD, as shown in the Fig. 2.
...      Total number of armature conductors lying in angles AOC and BOD.
Since two conductors from one turn,

The conductor which are responsible fro cross magnetizing ampere turns are lying between the angles AOD and BOC, as shown in the Fig.2.
Total armature-conductors / pole = Z/P
From above we have found an expression for demagnetizing conductors per pole.

Since two conductors from one turn,

If the brush shift angle is given in electrical degrees then it should be converted into mechanical degrees by using the relation,

Example :
A wave wound 4 pole d.c. generator with 480 armature conductors supplies a current of 144 A. The brushes are given an actual lead of 10o. Calculate the demagnetizing and cross magnetizing amp turns per pole.
Solution :
P = 4,            Z = 480,      Ia = 144 A
For wave wound,
I =  Ia/2
= 144/2 = 72 A
θm = 10o

1. 2. 