# Chapter 5: Electromagnetism¶

## Example 5.1, Page 145¶

In [1]:
#Variable declaration
N = 100;                                 # Number of turns
delta_phi = 10e-03;                      # Flux linked with coil, Wb
delta_t = 2e-03;                         # Time during which flux changes, s

#Calculations
e =((-N)*delta_phi)/delta_t;            # Average induced emf, V

#Result
print "The average emf induced in the coil = %3d V"%e

The average emf induced in the coil = -500 V


## Example 5.2, Page 146¶

In [2]:
#Variable declaration
N = 250;                                     # Number of turns
delta_phi1 = 20e-03;                         # Flux linked with coil, Wb
delta_phi2 = -16e-03;                        # Flux linked with coil, Wb
delta_t1 = 0.05;                             # Time, s
delta_t2 = 0.01;                             # Time, s

#Calculations
e_1 =((-N)*delta_phi1)/delta_t1;             # Average induced emf, V
e_2 =((-N)*delta_phi2)/delta_t2;             # Average induced emf, V

#Results
print "Change in flux in first case = %4.2f weber"%delta_phi1
print "Emf induced in first case = %3d volts"%e_1
print "Change in flux in second case = %4.2f weber"%delta_phi2
print "Emf induced in second case = %3d volts"%e_2

Change in flux in first case = 0.02 weber
Emf induced in first case = -100 volts
Change in flux in second case = -0.02 weber
Emf induced in second case = 400 volts


## Example 5.3, Page 147¶

In [3]:
#Variable declaration
e = 100;                                          # Induced emf, V

#Calculations
# For simplification let (delta_phi)/(delta_t) = k
k = 0.1;                                         # Rate of chage of flux linked with coil, Wb/s
# Since e =((-N)*delta_phi)/delta_t, soling for N
N = (e)/k;                                      # Number of turns

#Result
print "The number of turns on the coil = %4d"%N

The number of turns on the coil = 1000


## Example 5.4, Page 149¶

In [36]:
import math

#Variable declaration
v = 5;                                # Velocity, m^2
theta =(math.pi/3);                       # Angle, degrees
phi = 1.6e-03;                        # Flux, Wb
l = 0.1;                              # Length of pole face, m
d = 0.4;                              # Breadth of pole face, m

#Calculations
A = l*d;                              # Cross-sectional area of pole face, m^2
B = phi/ A;                           # Flux density, T
e =( B*l*v)*math.sin(theta);               # Induced emf, V

#Result
print "The emf induced = %5.4f V"%e

The emf induced = 0.0173 V


## Example 5.5, Page 149¶

In [37]:
import math

#Variable declaration
l = 0.15;                                 # Effective length of conductor, m
v = 8;                                    # Velocity, m^2
theta = 55;                     # Angle, degrees
e = 2.5;                                   # Induced emf, V

#Calculations
# Since e = B*l*v*sin(theta), solving for B
B = e/(l*v*math.sin(theta*math.pi/180));                    # Flux density, T

#Result
print "The density of the field = %5.3f Tesla"%B

The density of the field = 2.543 Tesla


## Example 5.6, Page 149¶

In [38]:
import math

#Variable declaration
l = 2.2;                                 # Effective length of conductor, m
B =38e-06;                               # Flux density, T
theta = (math.pi/2);                         # Angle, degrees
v = 800/36;                              # Velocity, m^2

#Calculations
e = B*l*v*math.sin(theta);                    # Induced emf, V

#Result
print "The emf induced in the axle = %4.2f mV"%(e/1e-03)

The emf induced in the axle = 1.84 mV


## Example 5.7, Page 152¶

In [11]:
import math

#Variable declaration
l = 0.22;                                 # Effective length of conductor, m
B = 0.35;                               # Flux density, T
I = 3;                                  # Current, A
theta = (math.pi/2);                        # Angle, degrees

#Calculations
# Since the force exerted on the conductor placed in magnetic field is directly proportional to the flux density ,
#the value of current flowing through the conductor, and the length of conductor lying inside the field, therefore
F = B*I*l*math.sin(theta);                   # Force, N

#Result
print "The force exerted on the conductor = %5.3f N"%F

The force exerted on the conductor = 0.231 N


## Example 5.8, Page 153¶

In [39]:
import math

#Variable declaration
phi = 2.5e-03;                             # Flux, Wb
l = 0.05;                                  # Effective length of pole, m
d = 0.03;                                  #  Effective width of pole, m
F = 1.25;                                  # Force exerted on conductor, N
A = l*d;                                   # Cross-sectional area of pole face, m^2
B = phi/A;                                # Flux density, T
theta = (math.pi/2);                          # Angle, degrees

#Calculations
# Since F = B*I*l*sin(theta), solving for I
I = F/(B*l*math.sin(theta));                   # Current in conductor, A
theta_2 = (math.pi/4);                        # New angle, degrees
F_2 = B*I*l*math.sin(theta_2);                  # Force exerted on conductor, N

#Results
print "The value of the current = %2d A"%I
print "The force exerted on conductor when placed at 45 degrees to the field = %5.3f newton"%F_2

The value of the current = 14 A
The force exerted on conductor when placed at 45 degrees to the field = 0.884 newton


## Example 5.9, Page 154¶

In [13]:
#Variable declaration
l = 0.015;                                    # Length of coil, m
d = 0.006;                                     # Width of  coil, m
B = 1.2;                                      # Flux density, T
I = 1e-02;                                    # Current, a
r = d/2;                                      # Radius of rotation, m

#Calculations
# Since torque is given by the product of force and distance, therefore, we have
T = 2*B*I*l*r;                               # Torque, Nm

#Result
print "The torque exerted on the coil = %4.2f micro-Nm"%(T/1e-06)


The torque exerted on the coil = 1.08 micro-Nm


## Example 5.10, Page 155¶

In [14]:
#Variable declaration
N = 80;                                 # Number of turns
l = 0.02;                               # Length of coil, m
r = 0.012;                              # Radius of coil, m
I = 45e-06;                              # Current in coil, A
T = 1.4e-06;                              # Torque exerted on coil, Nm
A = l*r;                                  # Cross-sectional area of coil, m^2

#Calculations
# Since T = 2*B*I*l*r, solving for B
B = T/(2*A*N*I);                             # Flux density, T

#Result
print "The flux density produced by the pole pieces = %4.2f T"%B

The flux density produced by the pole pieces = 0.81 T


## Example 5.11, Page 158¶

In [15]:
#Variable declaration
d = 0.035;                      # Distance between two parallel conductors, m
I_1 = 50;                       # Electric current in first coil, A
I_2 = 40;                      # Electric current in second coil, A

#Calculations
F = ((2e-07)*I_1*I_2)/d;         # Force exerted by conductors, N

#Result
print "The force exerted between the conductors = %4.1f mN"%(F/1e-03);

The force exerted between the conductors = 11.4 mN


## Example 5.12, Page 158¶

In [40]:
import math

#Variable declaration
d = 2;                           # Distance between two parallel conductors, m
I_1 = 1000;                     # Electric current in first coil, A
I_2 = 300;                      # Electric current in second coil, A

#Calculations
B = (mew_o*I_1)/d;              # Flux density due to first coil, T
F = ((2e-07)*I_1*I_2)/d;         # Force exerted by conductors, N

#Result
print "The flux density at a distance of %1d m from the centre of a conductor carrying a current of %4d A = %5.3f mT"%(d, I_1, B/1e-03);
print "Force exerted by conductors = %2d mN"%(F/1e-03);

The flux density at a distance of 2 m from the centre of a conductor carrying a current of 1000 A = 0.628 mT
Force exerted by conductors = 30 mN


## Example 5.13, Page 163¶

In [17]:
#Variable declaration
R_c = 40;                           # Resistance of coil, ohm
I_fsd = 5e-04;                      # Full-scale deflection current, A
I = 3;                              # Current reading, A

#Calculations
V_c = I_fsd*R_c;                    # Potential difference, V
# Since I = I_s + I_fsd, solving for I_s
I_s = I-I_fsd;                       # Shunt current, A
# From Ohm's law, V_c = I_s*R_s, solving for R_s
R_s = V_c/I_s;                       # Shunt resistance, ohm

#Result
print "The value of required shunt resistance = %4.2f milli-ohm"%(R_s/1e-03);

The value of required shunt resistance = 6.67 milli-ohm


## Example 5.14, Page 163¶

In [18]:
#Variable declaration
R_c = 40;                           # Resistance of coil, ohm
I_fsd = 5e-04;                      # Full-scale deflection current, A
I_fsd = 5e-04;                      # Full-scale deflection current, A
V = 10;                             # Voltage reading range, V
V_c = 0.02;                          # Potential difference across coil resistance, V

#Calculations
# From Ohm's law, V = I_fsd*R, solving for R
R = V/I_fsd;                         # Total resistance, ohm
# Since R = R_m + R_c, solving R_m
R_m = R - R_c;                       # Multiplier resistance, ohm

#Result
print "The required value of multiplier resistance = %5.2f kilo-ohms"%(R_m*1e-03);

The required value of multiplier resistance = 19.96 kilo-ohms


## Example 5.15, Page 164¶

In [19]:
#Variable declaration
R_c = 1500;                             # Coil resistance, ohm
I_fsd = 75e-06;                         # Full-scale deflection current, A
I = 5;                                  # Current range, A
V = 10;                                  # Voltage range, V

#Calculations
# Part (a)
# Using Ohm's law,
V_c = I_fsd*R_c;                         # Potential difference across coil resistance, V
# Since I = I_s + I_fsd, solving for I_s
I_s = I-I_fsd;                       # Shunt current, A
# From Ohm's law, V_c = I_s*R_s, solving for R_s
R_s = V_c/I_s;                       # Shunt resistance, ohm

# Part (b)
# Since = V = V_m + V_c, solving for V_m
V_m = V - V_c;                        # Potential difference across multiplier resistance, V
# From Ohm's law, V_m = I_fsd*R_m, solving for R_m
R_m = V_m/I_fsd                      # Multiplier resistance, ohm

#Results
print "The required value of shunt resistance = %4.1f mega-ohm"%(R_s/1e-03);
print "The required value of multiplier resistance = %4.1f mega-ohm"%(R_m*1e-03);

The required value of shunt resistance = 22.5 mega-ohm
The required value of multiplier resistance = 131.8 mega-ohm


## Example 5.16, Page 166¶

In [20]:
#Variable declaration
R_1 = 30.;                           # Resistance, ohm
R_2 = 70.;                           # Resistance, ohm
R_in = 200.;                          # Internal resistance of meter, ohm
V = 12.;                              # Supply voltage, V

#Calculations
# Using voltage divider rule, we have
V_2t = (R_2 /(R_1 + R_2))*V               #  True value of p.d across resistance R_2, V
# Since the resistances R_2 and R-in are parallel, so their equivalent resistance is given their parallel combination
R_BC = (R_2 * R_in)/(R_2 + R_in);                       # Resistance, ohms
# Using the potential divider technique,
V_2i = (R_BC / ( R_BC + R_1 ))*V                       # Indicated value of p.d across by voltmetre, volts
err = (( V_2i-V_2t ) / V_2t)*100                      # Percentage error in the reading

#Results
print "The p.d. indicated by the meter = %3.1f V"%V_2i
print "The percentage error in the reading = %4.2f percent"%err

The p.d. indicated by the meter = 7.6 V
The percentage error in the reading = -9.50 percent


## Example 5.17, Page 168¶

In [21]:
#Variable declaration
R_in = 200.;                                   # Internal resistance of meter, kilo-ohms
V = 10.;                                       # Supply voltage, volts
R_1 = 10.;                                     # Resistance, kilo-ohms
R_2 = 47.;                                     # Resistance, kilo-ohms

#Calculations
V_1 = R_1/(R_1+R_2)*V                         # P.d across resistance R_1, V
V_2 = R_2/(R_1+R_2)*V                         # P.d across resistance R_2, V

# Part (a)
R_AB = (R_1 * R_in)/(R_1 + R_in);                      # Resistance, kilo-ohms
V_AB = (R_AB / ( R_AB + R_2 ))*V                       # True value of p.d across by voltmetre, V
R_BC = (R_2 * R_in)/(R_2 + R_in);                      # Resistance, kilo-ohms
V_BC = (R_BC / ( R_BC + R_1 ))*V                        # Indicated value of p.d across by voltmetre, V

# Part (b)
# Error for V_1 measurement
error_AB = (V_AB - V_1)/V_1*100                        # Percentage error in the reading
#Error for V_2 measurement
error_BC = (V_BC-V_2)/V_2*100                           # Percentage error in the reading

#Results
print "The p.d. indicated by the meter across  first resistor = %4.2f V"%V_AB
print "The p.d. indicated by the meter across second resistor = %4.2f V"%V_BC
print "Percentage error for V_1 measurement = %4.2f percent"%error_AB
print "Percentage error for V_2 measurement = %4.2f percent"%error_BC

The p.d. indicated by the meter across  first resistor = 1.68 V
The p.d. indicated by the meter across second resistor = 7.92 V
Percentage error for V_1 measurement = -3.96 percent
Percentage error for V_2 measurement = -3.96 percent


## Example 5.18, Page 176¶

In [23]:
#Variable declaration
L = 0.25;                                           # Self-inductance, H
delta_I = 250e-03;                                  # Change in current, A
delta_t = 25e-03;                                  # Time, s

#Calculations
e = ((L)*delta_I)/(delta_t);                      # Induced emf, V

#Result
print "The value of emf induced = %3.1f V"%e

The value of emf induced = 2.5 V


## Example 5.19, Page 176¶

In [24]:
#Variable declaration
e = 30.;                   # Induced emf, V

#Calculations
# For simplicity, let rate of change of current i.e delta_I/delta_t = k
k = 200;                               # Rate of change of current, ampere-second
# Since e = ((-L)*delta_I)/(delta_t), solving for L
L = e/k;                               # Self-inductance, H

#Result
print "The inductance of the circuit = %4.2f H"%L

The inductance of the circuit = 0.15 H


## Example 5.20, Page 176¶

In [25]:
#Variable declaration
L = 50e-03;                               # Self-inductance, H
e = 8;                                   # Induced emf, V

#Calculations
# Since e = ((-L)*delta_I)/(delta_t), solving for delta_I/delta_t,and for simplicity letting the  rate of change of
#current i.e delta_I/delta_t = k
k = e/L;                    # Rate of change of current, As

#Result
print "The rate of change of current = %3d A/s"%k

The rate of change of current = 160 A/s


## Example 5.21, Page 178¶

In [26]:
#Variable declaration
N = 150;                             # Number of turns in a coil
I = 10;                              # Electric current flowing through coil, A
phi = 0.10;                          # Flux, Wb
delta_t = 0.1;                       # Time, s

#Calculations
# Part (a)
L  = (N * phi)/I                    # Self-inductance, H
delta_I = 20;                        # Change in current, A
# Part (b)
e = abs((-L*delta_I)/(delta_t));        # Induced emf, V

#Results
print "The inductance of the coi = %3.1f H"%L
print "The emf induced in the coil = %2d V"%e

The inductance of the coi = 1.5 H
The emf induced in the coil = 300 V


## Example 5.22, Page 178¶

In [27]:
#Variable declaration
I_1 = 8;                                   # Electric current, A
I_2 = 2;                                   # Electric current, A
N = 3000;                                 # Number of turns in a coil
phi_1 = 4e-03;                            # Flux, Wb
delta_t = 0.1;  # Reversal time of current, s

#Calculations
L  = (N * phi_1)/I_1;                    # Self-inductance, H
delta_I = I_1 - I_2;                        # Change in current, A
e = ((L)*delta_I)/(delta_t);        # Induced emf, V

#Result
print "The emf induced in the coil = %2d volts"%e

The emf induced in the coil = 90 volts


## Example 5.23, Page 179¶

In [41]:
import math

#Variable declaration
N_1 = 600;                                 # Number of turns in a coil in first case
N_2 = 900;                                 # Number of turns in a coil in second case
N_3 = 900;                                 # Number of turns in a coil in third case
l = 45e-03;                               # Effective length of coil, m
A = 4e-04;                                # Cross-sectional area of coil, m^2
mew_r1 = 1;                                # Relative permeability in first case
mew_r2 = 1;                                # Relative permeability in second case

#Calculations
# Part (a)
mew_r3 = 75;                                # Relative permeability in third case
L_1 = (mew_o*mew_r1*(N_1**2)*A)/l;          # Self-inductance of coil in first case, H

# Part (b)
# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore,
#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2
L_2 = (L_1*(N_2**2))/(N_1**2);          # Self-inductance of coil in second case, H

# Part (c)
# Since mew_r3 = 75*mew_r2, keeping all other quantities same we have
L_3 = mew_r3*L_2;          # Self-inductance of coil in third case, H

#Results
print "Self-inductance of coil in first case = %4.2f mH"%(L_1/1e-03);
print "Self-inductance of coil in second case = %5.3f mH"%(L_2/1e-03);
print "Self-inductance of coil in third case = %5.3f H"%L_3

Self-inductance of coil in first case = 4.02 mH
Self-inductance of coil in second case = 9.048 mH
Self-inductance of coil in third case = 0.679 H


## Example 5.24, Page 182¶

In [29]:
#Variable declaration
N_A = 2000;                                 # Number of turns in a coil A
N_B = 1500;                                 # Number of turns in a coil B
I_A = 0.5;                                  # Electric current in coil A, A
phi_A = 60e-06;                             # Flux linked with coil A, Wb

#Calculations
# Part (a)
L_A = (N_A*phi_A)/I_A;                      # Self-inductance of coil A
phi_B = 0.83*(60e-06);                     # Flux linked with coil B, Wb

# Part (b)
M = (N_B*phi_B)/I_A;                        # Mutual inductance of the two coils, H

#Results
print "Self-inductance of coil A = %4.2f H"%L_A
print "Mutual inductance of the two coils = %5.3f H"%M

Self-inductance of coil A = 0.24 H
Mutual inductance of the two coils = 0.149 H


## Example 5.25, Page 183¶

In [42]:
import math

#Variable declaration
N = 400;                                 # Number of turns in a coil
l = 0.25;                                # Effective length of coil, m
A = 4.5e-04;                             # Cross-sectional area, m^2
mew_r = 180;                             # Relative permeability

#Calculations
L = (mew_o*mew_r*(N**2)*A)/l          # Self-inductance of coil, H

#Result
print "The self inductance of the coil = %2d milli-henry"%(L/1e-03);

The self inductance of the coil = 65 milli-henry


## Example 5.26, Page 183¶

In [43]:
import math

#Variable declaration
L_1 = 65e-03;                         # Self-inductance of first coil, H
delta_I = 1.5;                        # Change in current, A
delta_t = 3e-03;                      # Time, s
k = 0.95;                             # 95 percent of flux produced
N_1 = 400;                                 # Number of turns in a coil A
N_2 = 650;                                 # Number of turns in a coil B

#Calculations
# Part (a)
# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore,
#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2
L_2 = (L_1*(N_2**2))/(N_1**2)          # Self-inductance of second coil , H

# Part (b)
M = k*math.sqrt(L_1*L_2);                 # Mutual inductance of two coils, H

# Part (c)
e_1 = ((L_1)*delta_I)/(delta_t);              # Induced emf in  first coil, V

# Part (d)
e_2 = (M*delta_I)/delta_t;                     # Induced emf in  second coil, V

#Results
print "The self-inductance of coil 2 = %3d mH"%(L_2/1e-03)
print "The value of mutual inductance = %3d mH"%(M/1e-03)
print "The self-induced emf in coil 1 = %4.1f V"%e_1
print "The mutually induced emf in coil 2 = %2d V"%e_2

The self-inductance of coil 2 = 171 mH
The value of mutual inductance = 100 mH
The self-induced emf in coil 1 = 32.5 V
The mutually induced emf in coil 2 = 50 V


## Example 5.27, Page 185¶

In [32]:
#Variable declaration
L = 50e-03;                                 # Self-inductance of coil, H
I = 0.75;                                  # Electric current in coil, A

#Calculations
W = (L*(I**2))/2                            # Energy stored, J

#Result
print "Energy stored in the inductor = %4.1f mJ"%(W/1e-03)

Energy stored in the inductor = 14.1 mJ


## Example 5.28, Page 185¶

In [44]:
import math

#Variable declaration
L_1 = 25e-03;                             # Self-inductance of first coil, H
L_2 = 40e-03;                             # Self-inductance of  second coil, H
I = 0.25;                                  # Electric current in coils, A
k =0.8;                                    # Coupling coefficient

#Calculations
# Part (a)
W_1 = (L_1*(I**2))/2;                       # Energy stored in first coil, J
W_2 = (L_2*(I**2))/2;                       # Energy stored in second coil, J
M = k*math.sqrt(L_1*L_2);                       # Mutual inductance of coils

# Part (b)
W_M = M*(I)*(I);                           # Energy stored due to mutual inductance of coils, J
W_sa = W_1 + W_2 + W_M;                        # Energy stored by two inductors when connected in series aiding, J
W_so = W_1 + W_2 - W_M;                        # Energy stored by two inductors when connected in series opposition, J

#Results
print "Energy stored in first coil = %4.2f mJ"%(W_1/1e-03)
print "Energy stored in second coil = %4.2f mJ"%(W_2/1e-03)
print "Energy stored by two inductors when connected in series aiding = %3.1f mJ"%(W_sa/1e-03)
print "Energy stored by two inductors when connected in series opposition = %4.2f mJ"%(W_so/1e-03)

Energy stored in first coil = 0.78 mJ
Energy stored in second coil = 1.25 mJ
Energy stored by two inductors when connected in series aiding = 3.6 mJ
Energy stored by two inductors when connected in series opposition = 0.45 mJ


## Example 5.29, Page 189¶

In [34]:
#Variable declaration
V_2 = 60;                         # Output voltage, V
V_1 = 240;                        # Input voltage, V
N_2 = 500;                        # Secondary turns

#Calculations
# Part (a)
# For simplicity let V_1/V_2 = N_1/N_2 = k
k = V_1/V_2                      # Turns ratio

# Part (b)
# Since V_1/V_2 = N_1/N_2, solving for N_1
N_1 = k*N_2;                       # Primary turns

#Results
print "The required turns ratio = %1d:1"%k
print "The number of primary turns = %4d"%N_1

The required turns ratio = 4:1
The number of primary turns = 2000


## Example 5.30, Page 190¶

In [35]:
#Variable declaration
R_L = 15;                               # Load resistor, ohms
V_2 = 240.;                              # Terminal p.d at secondary, V
V_1 = 600;                              # Supply voltage, V

#Calculations
# Part (a)
# Since V_1/V_2 = N_1/N_2 = k
k = V_1/V_2;                            # Turns ratio

# Part (b)
I_2 = V_2/R_L;                          # Current drawn by the load, A
P_2 = V_2*I_2;                          # Power drawn by the load, W

# Part (c)
I_1 = P_2/V_1                           # Current drawn from the supply, A

#Results
print "The transformer turns ratio = %3.1f:1"%k
print "The current drawn by the load = %2d A"%I_2
print "The power drawn by the load = %4.2f W"%(P_2*1e-03);
print "The current drawn from the supply = %3.1f A"%I_1

The transformer turns ratio = 2.5:1
The current drawn by the load = 16 A
The power drawn by the load = 3.84 W
The current drawn from the supply = 6.4 A