# Chapter 15 : Miscellaneous Solved Numerical Problems¶

## Example 15.1 Page No : 392¶

In [1]:
# Variables
R = 0.2     #total resistance of cable in ohms
I = 200.     #current in A
t = 100.     #time in hours
V = 240.     #voltage in volts
c = 0.8     #math.cost of electrical energy in Rs per unit

# Calculations
V1 = I*R     #voltage drop in the cable
#(i)consumer voltage
Vc = V-V1
#(ii)Power loss in the cable
P = I*I*R     #in watts
E = P*t/1000     #energy loss in kWh
C = E*c     #math.cost of energy loss in Rs.

# Results
print 'i)Consumer voltage is %3.1f Volts '%(Vc)
print 'ii)cost of energy loss is Rs %3.2f '%(C)

i)Consumer voltage is 200.0 Volts
ii)cost of energy loss is Rs 640.00


## Example 15.2 Page No : 393¶

In [2]:
# Variables
Vi = 220.     #voltage in volts supplied by dynamo
Vo = 200.     #voltage in volts required for lighting
I = 40.     #current in Amperes

# Calculations
Pi = Vi*I     #power output of dynamo
Po = Vo*I     #power consumed for lighting
L = Pi-Po     #line losses
R = L/(I**2)     #resistance of lines math.since line losses = I**2*R
t = 10     #time in hrs
N = (Po*t)/1000     #no of units of consumed in B.O.T units
Nw = (L*t)/1000     #No of units wasted in B.O.T units

# Results
print 'i)Resistance of lines is  %3.1f Ohms '%(R)
print 'ii)No. of B.O.T units consumed in 10hrs is %3.2f B.O.T units'%(N)
print 'iii)No. of B.O.T units wasted in 10hrs is %3.2f B.O.T units'%(Nw)

i)Resistance of lines is  0.5 Ohms
ii)No. of B.O.T units consumed in 10hrs is 80.00 B.O.T units
iii)No. of B.O.T units wasted in 10hrs is 8.00 B.O.T units


## Example 15.3 Page No : 393¶

In [3]:
# Variables
M = 250000.     #weight of water lifted per hr in kg
h = 50.     #height in metres
g = 9.81     #gravitational const.
WD = M*h*g     #work done by pump per hr in watt-sec
P = WD/3600     #Power output of pump per sec in watts
V = 500.     #supply voltage in volts
Ep = 0.8     #efficiency of pump
Em = 0.9     #efficiency of motor

# Calculations
E = Em*Ep     #overall efficiency
I = P/(V*E)     #current in amperes

# Results
print 'Current drawn by the motor is %3.2f Amperes'%(I)

Current drawn by the motor is 94.62 Amperes


## Example 15.4 Page No : 394¶

In [4]:
import math

# Variables
P = 10.     #Power developed by motor in H.P
N = 600.     #Speed of motor in rpm
#1HP = 735.5Nw-m/sec = 75kgm/sec
a = 75.
b = 735.5

# Calculations
#Torque in kg-m
Tkgm = (P*a*60)/(2*math.pi*N)     #math.since P = 2*pi*NT/60
#Torque in Nw-m
TNwm = (P*b*60)/(2*math.pi*N)     #math.since P = 2*pi*NT/60

# Results
print 'i)Torque in kg.meter is %3.2f kg-m '%(Tkgm)
print 'ii)Torque in Newton.meter is %3.2f Nw-m'%(TNwm)

i)Torque in kg.meter is 11.94 kg-m
ii)Torque in Newton.meter is 117.06 Nw-m


## Example 15.5 Page No : 395¶

In [5]:
# Variables
P = 25.     #Output of diesel engine in kW
s = 12500.     #calorific value of fuel oil in k-cal/kgm
e = 0.35     #overall efficiency of diesel set
P1 = P/e     #input energy required in 1 hour in kWh

# Calculations
P2 = P1*860     #input energy in kcal
m = P2/s     #mass of oil needed per hr in kgm
w = 1000     #weight of 1 ton of oil in kgm
Eg = (P*w)/m     #Energy generated by 1ton of oil in kWh

# Results
print 'i)Mass of oil required per hr is %3.3f kgm '%(m)
print 'ii)Eletrical energy generated per ton of fuel is %4.1f Kwh'%(Eg)

i)Mass of oil required per hr is 4.914 kgm
ii)Eletrical energy generated per ton of fuel is 5087.2 Kwh


## Example 15.10 Page No : 398¶

In [7]:
import math

# Variables
rho = 1.7*(10**-6)     #resistivity of copper in ohm-cm
l = 5     #length in metres
t = 0.005     #thickness in m
D = 0.08     #external diameter in m

# Calculations
d = D-(2*t)     #internal diameter in m
a = math.pi*(D**2-d**2)/4     #cross section area in cm**2
R = rho*l/a     #resistance of copper tube in ohm
R1 = R/(10**-4)     #resistance in micro-ohm

# Results
print 'Thus the resistance of copper tube is %3.2f micro-ohm'%(R1)

Thus the resistance of copper tube is 72.15 micro-ohm


## Example 15.11 Page No : 399¶

In [8]:
# Variables
rho = 1.7*(10**-8)     #resistivity in ohm-m
K = 1/rho     #conductivity in mho/m

# Calculations
a = 0.125*(10**-4)     #cross sectional area of cable in m**2
l = 2000.     #length of cable in meters
G = K*a/l     #conducmath.tance

# Results
print 'Thus conductivity of cable is %e mho/metres '%(K)
print 'and conducmath.tance of cable is %3f mho'%(G)

Thus conductivity of cable is 5.882353e+07 mho/metres
and conducmath.tance of cable is 0.367647 mho


## Example 15.12 Page No : 399¶

In [9]:
# Variables
V = 0.05     #volume in m**3
l = 300     #length in m
R = 0.0306     #resistance of conductor in ohm

# Calculations
rho = R*V/(l**2)     #resistivity of conducting material

# Results
print 'Thus resistivity of conducting material is %e ohm-m'%(rho)

Thus resistivity of conducting material is 1.700000e-08 ohm-m


## Example 15.13 Page No : 399¶

In [10]:
# Variables
rho = 0.67*(10**-6)     #resistivity in ohm-inch
m = 39.4     #1meter  =  39.4inch
m2 = 1525.     #1 meter2 = 1525 square inch

# Calculations
rhoc = rho*m/m2     #resistivity of copper in ohm/m**3
rho1 = rhoc/(10**-6)

# Results
print 'Thus resistivity of copper is %e ohm/m**3'%(rhoc)
print 'which is equal to %2.4f micro-ohm/m**3'%(rho1)

Thus resistivity of copper is 1.731016e-08 ohm/m**3
which is equal to 0.0173 micro-ohm/m**3


## Example 15.14 Page No : 400¶

In [11]:
import math

# Variables
R1 = 0.12     #old conductor resistance in ohm
d1 = 15.     #diameter of old conductor in cm
d2 = 0.4*d1     #diameter of new conductor in cm

# Calculations
a1 = math.pi*(d1**2)/4     #area of cross section of old conductor
a2 = math.pi*(d2**2)/4     #area of cross section of new conductor
#R = rho*l/a = rho*V/a**2
#Henec R is proportional to 1/a**2
R2 = R1*((a1/a2)**2)     #resistance of new conductor

# Results
print 'Thus resistance of new conductor is %2.4f ohm'%(R2)

Thus resistance of new conductor is 4.6875 ohm


## Example 15.15 Page No : 401¶

In [12]:
# Variables
lab = 10.     #la = 10*lb ratio of length of A to length of B.
Aab = 1./2     #Aa = 1/2*Ab ratio of area of A to area of B
RHOab = 1./2     #RHOa = 2*RHOb ratio of resistivity of A to resistivity of B
Ra = 2.     #resistance of A in ohm

# Calculations
Rb = (Ra*Aab)/(lab*RHOab)     #resistance of B in ohm
#Since Ra = RHOa*la/Aa and Rb = RHOb*lb/Ab so from ratio of two we get Rb

# Results
print 'Thus resistance of resistor B is %2.2f ohm'%(Rb)

Thus resistance of resistor B is 0.20 ohm


## Example 15.16 Page No : 402¶

In [13]:
import math

# Variables
RHOo = 10.3*(10**-6)     #resistivity of platinum wire at 0 degree in ohm-cm
d = 0.0074     #diameter of platinum wire
a = math.pi*(d**2)/4     #area of cross section of platinum wire in sq cm
Ro = 4.     #resistance of wire in ohm

# Calculations
l = Ro*a/RHOo     #length of wire in cm
alphao = 0.0038
t = 100     #temp in degree C
R100 = Ro*(1+(alphao*t))

# Results
print 'Thus length of wire required is %3.2f cms'%(l)
print 'and Resistance of wire at 100 degreeC is %2.2f ohms'%(R100)

Thus length of wire required is 16.70 cms
and Resistance of wire at 100 degreeC is 5.52 ohms


## Example 15.17 Page No : 403¶

In [14]:
# Variables
Ra = 1.     #resistance of A in ohm
lab = 20.     #ratio of length of A to length of B
Aab = 1./3     #ratio of area of A to area of B

# Calculations
#resistivity is same for both wires
Rb = Ra*(Aab/lab)     #resistance of wire B in ohm
#math.since Ra = rho*la/Aa and Rb = rho*lb/Ab so from ratio of both we get Rb

# Results
print 'Thus resistance of wire B is %2.4f omhs'%(Rb)

Thus resistance of wire B is 0.0167 omhs


## Example 15.19 Page No : 405¶

In [15]:
# Variables
I1 = 2./(2+3)     #current across 2V battery in circuit EBD in A
Vbe = 3*I1     #voltage dropp across BE in V
I2 = 4./(5+3)     #current across 4V battery in circuit AFC in A

# Calculations
Vaf = 3*I2     #voltage dropp across AF in V
V = Vbe+4-Vaf     #sum of potential drops starting from E and ending at F

# Results
#V is the P.D. between E and F
print 'Thus the P.D. between E and F is %2.1f Volts'%(V)

Thus the P.D. between E and F is 3.7 Volts


## Example 15.20 Page No : 405¶

In [16]:
# Calculations
#Let current in XA = I, in XY = I1, in AY = I-40, in YB = I-40+I1-60, in BX = I+I1-150.
#By Kirchhoff's second law, in circuit XAYA I-I1 = 20
# and in circuit XAYBX 25I+15I1 = 1950
I1 = (1950-500.)/(15.+25)     #in Amperes

# Results
print 'Thus the current in branch XY is I1 = %2.2f Amps'%(I1)

Thus the current in branch XY is I1 = 36.25 Amps


## Example 15.21 Page No : 407¶

In [17]:
# Variables
A = 30.     #area of hysteresis material in cm**2
s1 = 0.4     #scale is 1cm = 0.4Wb/m**2
s2 = 400.     # and 1cm = 400AT/m

# Calculations
V = 1.2*(10**-3)
f = 50     #frequency in Hz
H = A*s1*s2     #hysteresis loss/m**3/cycle in joules
Hp = H*V*f     #hysteresis power loss in Watts

# Results
print 'Thus hysteresis power loss is %3.2f Watts'%(Hp)

Thus hysteresis power loss is 288.00 Watts


## Example 15.22 Page No : 407¶

In [18]:
# Variables
d = 7500.     #density of iron in kg/m**3
w = 12.     #weight of iron in kgm

# Calculations
V = w/d     #volume of iron in m**3
f = 25.     #frequency in Hz
N = 3600.*f     #number of cycle per hour
A = 300     #area in joules/m**3
E = A*V*N     #Total energy loss per hour in joules

# Results
print 'Thus total energy loss per hour is %5.2f Joules'%(E)

Thus total energy loss per hour is 43200.00 Joules


## Example 15.23 Page No : 407¶

In [19]:
# Variables
l = 0.5     #length of coil in meters
d = 0.1     #diameter of coil
N = 1500.     #no of turns of coil

# Calculations
a = math.pi*(d**2)/4     #cross sectional area of coil in m**2
Ur = 1     #relative permeability
Uo = 4*math.pi*(10**-7)     #permeability
I = 8     #current in A
L = ((N**2)*a*Uo*Ur)/l     #self inducmath.tance of coil in H
E = (1./2)*L*(I**2)     #Energy stored in Joules

# Results
print 'Thus Self Inducmath.tance of coil is %2.3f H'%(L)
print 'and Energy stored is %1.2f Joules'%(E)

Thus Self Inducmath.tance of coil is 0.044 H
and Energy stored is 1.42 Joules


## Example 15.24 Page No : 408¶

In [20]:
# Variables
N = 600.     #number of turns on the coil
I = 2.     #current pasmath.sing through solenoid in A
l = 0.6     #length of solenoid in meter
H = N*I/l     #magnetic field at the centre in AT/m
Ur = 1.     #relative permeability

# Calculations
Uo = 4*math.pi*(10**-7)     #permeability
d = 0.025     #diameter in meters
a = math.pi*(d**2)/4     #cross sectional area of coil in m**2
phi = Uo*Ur*H*a     #flux in Wb

# Results
print 'Thus Magenetic field at centre is %3.2f AT/m'%(H)
print ' and Flux is %e Wb'%(phi)

Thus Magenetic field at centre is 2000.00 AT/m
and Flux is 1.233701e-06 Wb


## Example 15.25 Page No : 408¶

In [21]:
# Variables
Ur = 1.     #relative permeability
B = 1.257     #flux density in Wb/m**2

# Calculations
Uo = 4*math.pi*(10**-7)     #permeability
H = B/(Uo*Ur)     #magnetimath.sing force in AT/m
l = 0.004     #length of air gap in meter
AT = H*l     #AT required for the air gap

# Results
print 'Thus AT required for the air gap is %3.1f '%(AT)

Thus AT required for the air gap is 4001.2


## Example 15.26 Page No : 409¶

In [22]:
# Variables
D = 0.3     #diameter of anchor ring in m
l = math.pi*D     #length of iron ring in m
N = 400.     #number of turns on the iron ring
a = 0.0012     #area of cross section of iron path in m**2
Ur = 1000.     #relative permeability

# Calculations
Uo = 4*math.pi*(10**-7)     #permeability
I = 2     #current in A
phi = (N*I)/(l/(Uo*Ur*a))     #flux through iron path in WB
phi1 = phi/(10**-3)     #flux in mWb

# Results
print 'Thus flux through iron path is %2.2f mWb'%(phi1)

Thus flux through iron path is 1.28 mWb


## Example 15.27 Page No : 409¶

In [23]:
# Variables
a = 0.01     #crosssectional area of ring in m**2
Uo = 4*(math.pi)*(10**-7)     #absolute permeability
lf = 1.25     #leakage factor
Ur = 400.     #permeability
N = 175.     #no of turns

# Calculations
phig = 0.8*(10**-3)     #flux through air gap in Wb
Bg = phig/a     #Flux density in air gap in Wb/m**2
Hg = Bg/Uo     #magnetimath.sing force in air gap in AT/m
Lg = 0.004     #length of air gap in m
ATg = Hg*Lg     #AT required for air gap in AT
phii = phig*lf     #flux through iron path in Wb
Bi = phii/a     #Flux density in iron path in Wb/m**2
Hi = Bi/(Uo*Ur)     #magnetimath.sing force in iron path in AT/m
Li = 1.5     #length of iron path in m
ATi = Hi*Li     #At required for iron path in AT
AT = ATi+ATg     #total AT required
I = ATg/N     #Magnetimath.sing current required in A

# Results
print 'Thus the magnetimath.sing current required is %2.2f Amps'%(I)

Thus the magnetimath.sing current required is 1.46 Amps


## Example 15.28 Page No : 411¶

In [24]:
# Variables
SI = 0.2     #steady current in A
t = 0.2     #time in sec
Q = SI*t     #charge given to condenser in Coulomb
V = 220.     #PD across condenser in Volts

# Calculations
C = Q/V     #Capacitance of condenser in F
C1 = C*(10**6)     #Capacitance in mircoF

# Results
print 'Thus the Charge of condenser is %2.2f Coulomb'%(Q)
print 'And the Capacitance of condenser is %3.2f microF'%(C1)

Thus the Charge of condenser is 0.04 Coulomb
And the Capacitance of condenser is 181.82 microF


## Example 15.29 Page No : 411¶

In [25]:
# Variables
C = 2*(10**-6)     #capacitance of condenser in F
V = 10000     #PD across condenser in Volts

# Calculations
E = (1./2)*C*(V**2)     #energy stored in condenser in Joules
H = E/4.2     #heat produced in the wire in calories

# Results
print 'Thus heat produced in the wire is %2.2f calories'%(H)

Thus heat produced in the wire is 23.81 calories


## Example 15.30 Page No : 411¶

In [26]:
# Variables
V = 15*(10**3)     #potential difference applied in V
A = 0.02     #surface area of plate in m**2
d = 0.001     #dismath.tance between plates in m
C = 4.5*(10**-10)     #Capacitance of capacitor in F
Ko = 8.854*(10**-12)     #consmath.tant

# Calculations
K = (C*d)/(Ko*A)     #dielectric consmath.tant
q = C*V     #charge on condenser in C
D = q/A     #Electric flux density in C/m**2

# Results
print 'Thus the Charge of condenser is %e Coulomb'%(q)
print 'And the electric flux density of condenser is %e microF'%(D)

Thus the Charge of condenser is 6.750000e-06 Coulomb
And the electric flux density of condenser is 3.375000e-04 microF


## Example 15.31 Page No : 412¶

In [27]:
# Variables
m = 0.6     #mass of water in kgm
S = 4200.     #specific heat of water
T1 = 100.     #temperature in degreeC
T2 = 10.     #temperature in degreeC
t = 5*60.     #time in sec
V = 230.     #Supply voltage in Volts

# Calculations
H = m*S*(T1-T2)     #Heat required to raise the temp of water from 0 to 100 degree. in J
e = 0.78     #efficiency of kettle
Ei = H/e     #Energy input in Joules
Ei1 = Ei/(100*3600)     #Energy input in kWh
W = Ei/t     #Rating of kettle in watts
R = (V*V)/W     #Resistance of heating element in ohms

# Results
print 'Thus Resistance of heating element is %2.1f ohms'%(R)

Thus Resistance of heating element is 54.6 ohms


## Example 15.32 Page No : 413¶

In [28]:
# Variables
m1 = 120.     #mass of water to be heated in kg
m2 = 20.     #mass of copper math.tank in kg
S1 = 1.     #specific heat of water
S2 = 0.095     #specific heat of copper
T1 = 10.     #temp in degreeC
T2 = 60.     #temp in degreeC

# Calculations
H = (m1*S1*(T2-T1))+(m2*S2*(T2-T1))     #heat required to raise the temp of water  and math.tank in kcal
H1 = H*4200     #heat required in Joules
e = 0.8     #thermal efficiency
E = H1/e     #Energy input in joules
E1 = E/(1000*3600)     #energy input in kWh
r = 3     #rating of heater in kW
t = E1/r     #time taken in hours

# Results
print 'Thus the time taken to raise the temp is %2.3f hours'%(t)

Thus the time taken to raise the temp is 2.963 hours


## Example 15.33 Page No : 414¶

In [29]:
import math

# Variables
rho = 5*(10**-5)     #specific resistance for steel in ohm-cm
U = 1.     #relative permeability
d = 0.15     #depth of penetration in cm

# Calculations
f = (rho*(10**9))/(U*d*d*4*(math.pi**2))     #frequency required in cycles per sec
f1 = f/1000     #frquency in k.cycles/sec

# Results
print 'Thus the frequency required is %3.3f k.cycles/sec'%(f1)

Thus the frequency required is 56.290 k.cycles/sec


## Example 15.34 Page No : 414¶

In [30]:
import math

# Variables
v = 50.*20*2     #Volume of board to be heated in cm**3
Mw = 0.56     #weight of wood in gm/cm**3
m = Mw*v/1000     #mass of wood in kgm
S = 0.35     #specific heat of wood
t = 15./60     #time in hrs
f = 30*(10**6)     #frequency in cycles/sec
t2 = 150.;t1 = 30.     #temp in degreeC
H = m*S*(t2-t1)     #heat required to raise the temp in kcal
Hw = H*1000/860     #heat required in kW

# Calculations
P = Hw/t     #power required in Watts
e = 0.5     #efficiency of dielectric heating process
Pi = P/e     #power input required in Watts
Ko = 8.854*(10**-12)     #absolute permittivity
K = 5     #relative permittivity
A = 0.5*0.2     #area in m
i = 0.02
C = Ko*K*A/i     #capacitance of parallel plate capacitor in F
Xc = 1/(2*math.pi*f*C)     #capacitive reacmath.tance in ohms
cosx = 0.05
tanx = 19.97
R = Xc*tanx     #resistance
V = math.sqrt(Pi*R)     #voltage in volts
Ic = V/Xc     #current through the board in Amps

# Results
print 'Thus the power required is %2.1f Watts'%(Pi)
print 'And Voltage across the board is %3.2f volts'%(V)
print 'And the current through the board is %2.3f Amps'%(Ic)

Thus the power required is 437.6 Watts
And Voltage across the board is 457.64 volts
And the current through the board is 19.095 Amps


## Example 15.35 Page No : 416¶

In [31]:
# Variables
m = 2.     #quantity of aluminium to be melted in kg
t1 = 15.
t2 = 660.     #temp in degreeC
S = 0.212     #specific heat of aluminium
L = 78.8     #latent heat of aluminium in kcal/kg

# Calculations
H = (m*S*(t2-t1))+(m*L)     #total heat required to melt Al in kcal
i = 5     #input to furnace in kW
E = i*(1000*10*60)     #Energy input to furnace in watt-sec
E1 = E/4180     #energy input in kcal
e = H*100/E1     #efficiency of furnace

# Results
print 'Thus the efficiency of furnace is %2.3f percent'%(e)

Thus the efficiency of furnace is 60.123 percent


## Example 15.36 Page No : 417¶

In [32]:
# Variables
O = 5*735.5     #output of motor in W
e = 0.85     #efficiency of motor
c = 2.     #math.cost of energy per unit in Rs

# Calculations
I = O/e     #input of motor in Watts
t = 4     #time in hrs
E = I*t/1000     #energy consumed in kWh
C = c*E     #math.cost of umath.sing the motor in Rs

# Results
print 'Thus the math.cost of umath.sing the motor is %2.3f Rs'%(C)

Thus the math.cost of umath.sing the motor is 34.612 Rs


## Example 15.37 Page No : 417¶

In [33]:
# Variables
I = 2.5*(10**-3)     #current in Amp
t = 30*(10**-3)     #time in sec
Q = I*t     #charge pasmath.sing through the person in Coulumbs

# Calculations
e = 1.602*(10**-19)     #charge of 1 electron in C
N = Q/e     #no of electrons pasmath.sing through the person

# Results
print 'Thus the no of electrons pasmath.sing through the person is %e electrons'%(N)

Thus the no of electrons pasmath.sing through the person is 4.681648e+14 electrons


## Example 15.38 Page No : 417¶

In [34]:
# Variables
#(a)Finding resistance between 2 ends
l = 1     #length in m
a = 2.5*(10**-2)*0.05*(10**-2)     #area of cross section in m**2
rho = 1.724*(10**-8)     #specific resistance of copper in ohm-m

# Calculations
R = rho*l/a     #resistance of the strip in ohm
#(b) Finding resistance between 2 faces
l1 = 0.05*(10**-2)     #length in m
a1 = 2.5*(10**-2)*1     #area of cross section in m**2
R1 = rho*l1/a1     #resistance in ohm

# Results
print 'Thus the resistance of the strip is %e ohms '%(R)
print 'And the resistance between the faces is %e ohms'%(R1)

Thus the resistance of the strip is 1.379200e-03 ohms
And the resistance between the faces is 3.448000e-10 ohms


## Example 15.39 Page No : 418¶

In [35]:
# Variables
m = 2.     #weight of water to be heated in kg
t2 = 98.
t1 = 15.     #temp in degreeC
s = 1.     #specific heat of water
V = 200.     #voltage in volts

# Calculations
H = m*s*(t2-t1)     #energy required to raise the temp of water in kcal
H1 = H*4200     #energy in Watt-sec or Joules
e = 0.85     #efficiency of kettle
E = H1/e     #energy input required in watt-sec
E1 = E/(1000*3600)     #energy input in kWh
c = 35.     #math.cost per unit in paise
C = c*E1     #ocst of energy used in paise
t = 10./60     #time in hrs
W = E1*1000/t     #wattage of kettle in watts
R = V*V/W     #resistance of heating element in ohms

# Results
print 'Thus the resistance of heating element is %2.0f ohms'%(R)
print 'And the math.cost of energy used is %2.0f paisa'%(C)

Thus the resistance of heating element is 29 ohms
And the math.cost of energy used is  8 paisa


## Example 15.40 Page No : 418¶

In [1]:
import math

# Variables
phi = 70000./(10**8)     #flux to be set up in Wb math.since 10**8lines  = 1Wb
d = 0.03     #diameter in m
a = math.pi*d*d/4     #area of cross section in m**2
B = phi/a     #flux density in Wb/m**2
Lg = 0.002     #length of air gap in m
Ls = (math.pi*0.2)-Lg     #length of steel path
Uo = 4*math.pi*(10**-7)     #absolute permitivity
Ur = 800.     #relative permitivity of steel

# Calculations
Hg = B/Uo
Hs = B/(Uo*Ur)
AT = (Hg*Lg)+(Hs*Ls)     #total ampere turns required
N = 500.     # no of turns
I = AT/N     #exciting current in amps

# Results
print 'Thus the value of exciting current is %2.3f A'%(I)

Thus the value of exciting current is 4.386 A