# Chapter 4: Magnetic Fields and Circuits¶

## Example 4.1, Page 116¶

In [1]:
#Variable declaration
A = 6e-04;                  # Cross-sectional area of pole face, metre-square
phi = 30e-06;              # Flux, Wb

#Calculations
B = phi/A;                  # Flux density, T

#Result
print "The flux density at the pole face = %2d mT"%(B/1e-03)

The flux density at the pole face = 50 mT


## Example 4.2, Page 116¶

In [2]:
#Variable declaration
A = 45e-06;                    # Cross sectional area of pole face, metre-square
B = 0.6;                      # Flux density, T

#Calculations
# Using formula B = phi/A, solving for phi
phi = B*A;                   # Flux, Wb

#Result
print "The flux produced by pole face = %2d micro-wWb"%(phi/1e-06);

The flux produced by pole face = 27 micro-wWb


## Example 4.3, Page 117¶

In [3]:
#Variable declaration
N = 1500;                      # Number of turns in a coil
A = 5e-04;                     # Cross- sectional area of of coil, metre-square
phi = 0.2e-03;                 # Flux, Wb
I = 0.75;                      # Coil-current, A

#Calculations
# Since m.m.f is the product of the current and the number of turns, therefore, we have
F = N*I;                      # Magnetomotive force, At
B = phi/A;                    # Flux density, T

#Result
print "The m.m.f and flux density produced are %4d At and %3.1f T respectively"%(F, B);

The m.m.f and flux density produced are 1125 At and 0.4 T respectively


## Example 4.4, Page 117¶

In [4]:
#Variable declaration
N = 600;                      # Number of turns in a coil
F = 1500.;                      # Magnetomotive force, At

#Calculations
# Since magnetomotive force,F = N*I, solving for I
I = F/N;                      # Excitation-current, A

#Result
print "The excitation current required = %3.1f A"%I;

The excitation current required = 2.5 A


## Example 4.5, Page 118¶

In [5]:
import math

#Variable declaration
I = 0.4;                  # Current, A
N = 550;                  # Number of turns in a coil
d = 8e-02;                # Diameter, m
l = (math.pi*d);              # Average length of the magnetic circuit, m

#Calculations
# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have
H = (N*I)/l;                # Magnetic field strength, At/m

#Result
print "The magnetic field strength inside the toroid = %6.2f At/m"%H

The magnetic field strength inside the toroid = 875.35 At/m


## Example 4.6, Page 120¶

In [6]:
#Variable declaration
A = 15e-04;                   # Cross-sectional area of core, metre-square
mew_r1 = 65;                  # Relative permeability of core
phi_1 = 2e-04;                # Flux, Wb
mew_r2 = 800.;                 # Changed relative permeability of core

#Calculations
B_1 = phi_1/A;                # Flux density, T
mew_r = mew_r2/mew_r1;        # Relative permeability of core
# Since cross-sectional area of core A remains constant, therefore, we have mew_r = B_1/B_2 , solving for B_2
B_2 = mew_r*B_1;              #  New flux density, T
# Since B_2 = phi_2/A, solving for phi_2
phi_2 = B_2*A;                # New flux, Wb

#Result
print "The new flux and flux density are %5.3f mWb and %5.3f T respectively"%(phi_2/1e-03, B_2);

The new flux and flux density are 2.462 mWb and 1.641 T respectively


## Example 4.7, Page 120¶

In [8]:
import math

#Variable declaration
r = 0.04;                           # Mean radius of torod, m
A = 3e-04;                          # Csa of toroid, m^2
mew_o = 4*(math.pi)*1e-07;              # Permeability of free space
mew_r = 150;                        # Relative permeability of toroid
N = 900;                            # Number of turns on coil
I = 1.5;                            # Coil current, A
l = 2*(math.pi)*r;                      # Effective length of toroid, m

#Calculations&Results
# Part (a)
# Since m.m.f is the product of the current and the number of turns, therefore, we have
F = N*I;                            # Magnetomotive force, At
print "The m.m.f of toroid = %4d At"%F

# Part (b)
# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have
H = F/l;                               # Magnetic field strength, At/m
print "The magnetic field strength = %6.1f At/m"%H;

# Part (c)
B = (mew_r*mew_o*H);                   # Flux density, T
phi = B*A;                             # Flux, Wb
print "The flux and flux density are %6.2f micro-weber and %6.4f T respectively"%(phi/1e-06,B)

The m.m.f of toroid = 1350 At
The magnetic field strength = 5371.5 At/m
The flux and flux density are 303.75 micro-weber and 1.0125 T respectively


## Example 4.8, Page 120¶

In [9]:
import math

#Variable declaration
r = 3e-02;                            # Radius of toroid, m
A = 4.5e-04;                          # Cross-sectional area of toroid, metre-square
N = 500;                              # Number of turns
phi = 250e-06;                        # Flux, Wb
mew_o = 4*(math.pi)*(1e-07);              # Permeability of free space
mew_r = 300;                          # Relative permeability

#Calculations
l = 2*(math.pi)*r;                        # Effective length, m
B = phi/A;                            # Flux density, T
# Since B = (mew_r)*(mew_o)*H, solving for H
H = B /((mew_r)*(mew_o));              # Magnetic field strength, At/m
# Since H = F/l, solving for F
F = H*l;                               # Magnetomotive force, At
# Since mmf,F = N*I, solving for I
I = F/N;                               # Electric current, A

#Result
print "The value of current needs to be passed through the coil = %4.2f A"%I

The value of current needs to be passed through the coil = 0.56 A


## Example 4.9, Page 121¶

In [10]:
import math

#Variable declaration
# Part (a)
I = 0.2;                    # Electric current, A
l = 5e-02;                  # Effective length, m
A = 7e-04;                  # Cross-sectional area, metre-square
d = 0.5e-03;                # Diameter, m
mew_r = 1;                  #Relative permeability for wood

#Calculations
N = l/d;                    # Number of turns
# Since mmf is the product of the current and the number of turns, therefore, we have
F = N*I;                     # Magnetomotive force, At
# Part (b)
# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have
H = F/l;                               # Magnetic field strength, At/m
B = ( mew_r * mew_o * H );             #  Flux density, T
# Part (c)
phi = B * A;                           # Flux, Wb

#Result
print "The mmf produced = %2d At"%F
print "The flux density produced = %3d micro-tesla"%(B/1e-06);
print "The flux produced = %5.3f micro-weber"%(phi/1e-06);

The mmf produced = 20 At
The flux density produced = 502 micro-tesla
The flux produced = 0.352 micro-weber


## Example 4.10, Page 125¶

In [11]:
import math

#Variable declaration
N = 1000;                                     # Number of turns on coil
r = 0.1;                                      # Mean radius of toroid, m
phi = 0.1775e-03;                             # Flux density(value from graph), Wb
A = math.pi*1e-04;                               # Csa of toroid, m^2
H = 88;                                       # Magnetic field strength(value from graph), At/m
B = phi/A;                                    # Flux density, T

#Calculations&Results
# Part (a)
l = 2*math.pi*r;                                  # Effective length of toroid, m
# Since H = (N*I)/l, solving for I
I = (H*l)/N ;                                 # Electric current in coil, A
print "Coil current = %4.1f mA"%(I/1e-03);

# Part (b)
# Since B = mew_o * mew_r * H, solving for mew_r
mew_r = B/(mew_o*H);                         #Relative permeability of toroid
print "The relative permeability of toroid = %4d"%mew_r;

Coil current = 55.3 mA
The relative permeability of toroid = 5109


## Example 4.11, Page 125¶

In [12]:
import math

#Variable declaration
l = 0.15;                       # Mean length, m
N = 2500;                      # Number of turns
I = 0.3;                       # Electric current, A

#Calculations
# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have
H = (N*I)/l;                   # Magnetic field strength, At/m
B = 0.75;                      # Flux density( value taken from graph ), T
# Since B = ( mew_r * mew_o * H ), solving for mew_r
mew_r = B/(mew_o * H);               # Relative permeability

#Results
print "The flux density of given toroid = %3.2f T "%B
print "The relative permeability of given toroid = %5.1f"%mew_r

The flux density of given toroid = 0.75 T
The relative permeability of given toroid = 119.4


## Example 4.12, Page 126¶

In [13]:
import math

#Variable declaration
l = 0.1875;                         # Mean length, m
A = 8e-05;                          # Cross- sectional area of of coil, metre-square
N = 750;                            # Number of turns
phi = 112e-06;                      # Flux, Wb
l_gap = 0.5e-03;                    # Average length of the magnetic circuit,m
B = phi/A;                          # Flux density, Wb
H = 2000;                           # Magnetic field strength( value taken from graph ), At/m

#Calculations
F_Fe = H*l;                         # The m.m.f in the iron part of the circuit, At
# Since F = I*N, solving for I
I = F_Fe/N;                         # Coil current under normal conditions, A
# Since B = mew_o * H_gap, solving for H_gap
H_gap = B/mew_o;                    # Magnetic field strength, At/m
# Since H_gap = F_gap/l_gap, solving for F_gap
F_gap = H_gap * l_gap;              # The mmf in the air part of the circuit, At
F = F_Fe + F_gap;                  # Total circuit mmf, At
I_new = F/N;                       # Current required to maintain the flux at its original value, A

#Results
print "The coil current required to produce a flux of %3d micro-weber in the toroid = %3.1f A "%(phi/1e-06, I);
print "Current required to maintain the flux at its original value = %5.3f A"%(I_new);

The coil current required to produce a flux of 112 micro-weber in the toroid = 0.5 A
Current required to maintain the flux at its original value = 1.243 A


## Example 4.13, Page 127¶

In [14]:
#Variable declaration
l_A = 0.25;                  # Mean length of circuit A, m
l_B = 0.15;                  # Mean length of circuit A, m
A_A = 11.5e-04;              # Cross-sectional area of circuit A, metre-square
A_B = 12e-04;                # Cross-sectional area of circuit B, metre-square
phi = 1.5e-03;               # Flux, Wb
N = 1000;                    # Number of turns

#Calculations
B_A = phi/A_A;              # Flux density linked with circuit A, T
B_B = phi/A_B;              # Flux density linked with circuit B, T
H_A = 1470;                 # Magnetic field strength of circuit A( value taken from graph ), At/m
H_B = 845;                  # Magnetic field strength of circuit B( value taken from graph ), At/m
# Since H = F/l, solving for F
F_A = H_A * l_A;                # Magnetic field strength of circuit A, At/m
F_B = H_B * l_B;                # Magnetic field strength of circuit B, At/m
F = F_A + F_B;                   # Total circuit m.m.f, At/m
I = F/N;                          # Coil current, A

#Result
print "Coil current in the magnetic circuit = %5.3f A"%I

Coil current in the magnetic circuit = 0.494 A


## Example 4.14, Page 130¶

In [15]:
import math

#Variable declaration
A = 8e-04;                                # Cross-sectional area, metre-square
d = 24e-02;                               # Mean diameter of iron ring, m
phi = 1.2e-03;                            # Flux, Wb
mew_r = 1200;                             # Relative permeability
mew_air = 1;                             # Permeability for air
l_gap = 3e-03;                           # Mean length, m

#Calculations
l_Fe = (math.pi) * d;                        # Mean length of iron circuit, m
S_Fe = l_Fe/(mew_r * mew_o *A);          # Reluctance of iron circuit, At/Wb
S_gap = l_gap/(mew_air * mew_o *A);        # Reluctance of gap, At/Wb
S = S_Fe + S_gap;                        # Total circuit reluctance, At/Wb
# Since phi = F/S, solving for F
F = phi*S;                               # Magnetomotive force, At

#Result
print "The required mmf = %5.1f At"%F

The required mmf = 4331.0 At


## Example 4.15, Page 130¶

In [16]:
import math

#Variable declaration
N = 500;                                       # Number of turns on first section's coil
phi = 2e-03;                                   # Flux produced by first section, Wb
l_1 = 85e-02;                                  # Length of first section, m
l_2 = 65e-02;                                  # Length of second section, m
l_3 = 0.1e-02;                                 # Length of third section, m
A_1 = 10e-04;                                  # Csa of first section, m^2
A_2 = 15e-04;                                  # Csa of second section, m^2
A_3 = 12.5e-04;                                # Csa of second section, m^2
mew_r1 = 600;                                  # Relative permeability of first section
mew_r2 = 950;                                  # Relative permeability of second section
mew_r3 = 1;                                    # Relative permeability of third section

#Calculations&Results
# Part (a)
S_1 = l_1/(mew_r1 * mew_o * A_1);              # Reluctance of first section, At/Wb
S_2 = l_2/(mew_r2 * mew_o * A_2);              # Reluctance of first section, At/Wb
S_3 = l_3/(mew_r3 * mew_o * A_3);              # Reluctance of first section, At/Wb
S = S_1 + S_2 + S_3;                           # Total reluctance of the circuit, At/Wb
print "Total reluctance of the circuit = %4.2fe+06 At/Wb"%(S*1e-06);

# Part (b)
# Since phi = F/S, solving for F
F = phi*S;                                    # Magnetomotive force, At
# Since F = N*I, solving for I
I = F/N;                                      # Electric current in first section, A
print "Electric current in first section = %4.2f A"%I

Total reluctance of the circuit = 2.13e+06 At/Wb
Electric current in first section = 8.51 A