Chapter 1 Magnetic circuits

Example 1.1,Page number 8

In [1]:
from math import pi
import math   
# given
Bc=0.8              #flux-density in tesla
Hc=510.0            #magnetic intensity
Bg=0.8              #flux density in the core
A=0.00000125663     #vacuum permeability
lg=0.0015           #air-gap lengths
lc=0.36             #mean-core path
N=500.0             #no.of turns in the coil
# calculations
Fg=Bg/A*(2*lg)
Fc=Hc*lc
F=Fc+Fg              #total mmf required
i=F/N                #current in the circuit
Pre=Bc/Hc*1000       #permeability of core
RelPre=Pre/A*10**-3  #relative permeability
F=Hc*lc
i=F /N                 #current required
Pre=round(Pre,3)       #permeability of core
RelPre=round(RelPre,2)  
i=round(i,3)
print 'permeability of core is ',Pre
print 'relative permeability of core is',RelPre
print 'The current in circuit is ',i,'A'

permeability of core is  1.569
relative permeability of core is 1248.28
The current in circuit is  0.367 A

Example 1.2,Page number 9

In [2]:
from math import pi
A=0.00000125663             #vacuum permeability
lc=36.0*10.0**-2.0         #core-length
N=500.0          
i=4.0                #current in the circuit
lg=2.0*10.0**-3.0    #air-gap length
m=-A*(lc/lg)
c=(N*i*A)/(lg)
Hc=(N*i)/(lc)  #flux density
Hc=round(Hc,0)
print 'The flux density is',Hc,'At/m'

The flux density is 5556.0 At/m

Example 1.3,Page number 11

In [3]:
import math
#given
N1=N2=500.0       #no.of turns
I1=I2=10.0        #current
Ibafe=3*52*10**-2
A=4.0*pi*10**-7   #vacuum permeability
b=1200           #relative permeability of theferro-magnetic material
Ag=0.0004
Ac=4*10^-4
lg=5*10^-3      #air-gap length
Ibecore=0.515
c=0.0002067
d=0.0004134     #airgap flux
#calculations
F1=N1*I1
F2=N2*I2
Pre=1200*A     #permeability
Rbafe=(Ibafe)/(Pre*Ac)
Rg=lg/(A*Ag)
Rbecore=Ibecore/(Pre*Ac)
Bg=d/(Ag)       # airgap fluxdensity
Hg=Bg/A*10**-6  #magnetic intensity
Bg=round(Bg,3)
Hg=round(Hg,3)
print 'the airgap flux density is',Bg,'T'
print 'The magnetic intensity in the air gap is ',Hg,'*10**6 At/m'

the airgap flux density is 1.034 T
The magnetic intensity in the air gap is  0.822 *10**6 At/m

Example 1.5,Page number 15

In [4]:
from math import pi
# given 
Irad=20         #inner-radius of the coil
Orad=25         #outer-radius of the coil
Dia=22.5        #diameter of the coil                        
N=250           #no.of turns of the coil
i=2.5           #coil current
B=1.225
# calculations
l=2*pi*Dia*10**-2
radius=1/2*(Irad+Orad)
H=(N*i)/l
A=pi*((Orad -Irad)/2)**2*10**-4        # Cross-sectional area
z=(1.225)*(pi*6.25*10**-4)
y=(N*z)
L=(y/i)
core=(B/H)          
l=(2*pi*22.5*10**-2)
Rcore=2599.64*10**2                  # formula for Rcore=(l)/(core*A)
L=(N**2)/(Rcore)
L=round(L,2)*1000
print 'Inductance in the coil is',L,'mH'

Inductance in the coil is 240.0 mH

Example 1.7,Page number 24

In [5]:
import math
# given
n=500.0     #no.of turns
E=100.0     #amplitude
A=0.001     #cross-sectional area of the core
b=1/120
f=1.2       #given max.flux density value
#calculations
max1=0.833             #Formula for max1=(E/1000)*(b)
max2=(f*A)
E=(120*n*max2*2) # result
max1=round(max1,2)
E=round(E,2)
print 'Maximum flux is',max1,'mWb'
print 'Maximum value of E is',E,'V'

Maximum flux is 0.83 mWb
Maximum value of E is 144.0 V

Example 1.8,Page number 31

In [6]:
from math import pi
#given
lg=0.4             #air-gap length
Bg=0.8             #flux density
Hm=42.0*10.0**3.0
A=4.0*pi*10.0**-7.0  #vacuum permeability
Ag=2.5*10.0**-4.0
Bm=0.95
#calculations
Hg=Bg/A
lm=(lg/Hm)*Hg
Am=(Bg*Ag)/(Bm)*10000
lm=round(lm,2)
Am=round(Am,3)
print 'The dimension Im is',lm,'cm'
print 'The dimension Am is',Am,'cm**2'

The dimension Im is 6.06 cm
The dimension Am is 2.105 cm**2