Chapter 12 Electromagnetic induction

Example 12.1 Page no 665

In [6]:
#Given
a=20                   #mWb
a1=-20                  #mWb
t=2*10**-3              #s
N=100

#Calculation
a2=(a1-a)*10**-3
e=(-N*a2)/t

#Result
print"Average e.m.f induced in the coil is", e,"V"

Average e.m.f induced in the coil is 2000.0 V

Example 12.2 Page no 665

In [12]:
#Given
r=5*10**-2                         #m
N=1
B=0.35
t=0.12                               #S

#Calculation
import math
A=math.pi*r**2
a1=B*A
a2=-B*a1
e=(N*a1)/t

#Result
print round(e,2),"V"

0.02 V

Example 12.3 Page no 665

In [24]:
#Given
A=10**-2                           #m**2
a=45                               #degree
B1=0.1                             #T
R=0.5                              #ohm
t=0.7                              #S

#Calculation
import math
a1=B1*A*math.cos(a*3.14/180.0)
a2=0
a3=a1-a2
e=a3/t
I=e/R

#Result
print"Current during this time interval is", round(I*10**3,1),"*10**-3 A"
print"Magnitude of induced emf is",round(e*10**3,0),"*10**-3 V"

Current during this time interval is 2.0 *10**-3 A
Magnitude of induced emf is 1.0 *10**-3 V

Example 12.4 Page no 666

In [28]:
#Given
I=1.2*10**-3                              #A
N=1.0
R=10                              #ohm

#Calculation
e=I*R
a=e/N

#Result
print"Necessary rate is", a*10**2,"*10**-2 Wb/second"

Necessary rate is 1.2 *10**-2 Wb/second

Example 12.5 Page no 666

In [47]:
#Given
r=10**-1                   #m
B=3.0*10**-5                  #T
t=0.25                           #S
N=500
R=2                                #ohm

#Calculation
import math
a1=B*math.pi*r**2*math.cos(0*3.14/180.0)
a2=B*math.pi*r**2*math.cos(180*3.14/180.0)
a3=a1-a2
e=(N*a3)/t
I=e/R

#Result
print"Magnitude of the emf is", round(e*10**3,1),"*10**-3 V"
print"Current induced in the coil is",round(I*10**3,1),"*1)**-3 A"

Magnitude of the emf is 3.8 *10**-3 V
Current induced in the coil is 1.9 *1)**-3 A

Example 12.6 Page no 666

In [53]:
#Given
e=10**-2                        #V
B=5*10**-5                      #T
r=0.5                          #m
N=1

#Calculation
import math
A=math.pi*r**2
n=(e*N)/(math.pi*r**2*B)

#Result
print"Rate of rotation of the blade is", round(n,1),"revolutions/second"

Rate of rotation of the blade is 254.6 revolutions/second

Example 12.7 Page no 667

In [59]:
#Given
a=12
b=7
t=2

#Calculation
e=((a*t)+b)*10**-3

#Result
print"(i) Magnitude of induced emf is", e*10**3,"mV"
print"(ii) The current induced in the coil will be anticlockwise"

(i) Magnitude of induced emf is 31.0 mV
(ii) The current induced in the coil will be anticlockwise

Example 12.8 Page no 673

In [63]:
#Given
B=1                          #T
l=0.5                        #m
v=40                         #m/s

#Calculation
import math
e=B*l*v*math.sin(60*3.14/180.0)

#Result
print"emf induced in the conductor is", round(e,2)

emf induced in the conductor is 17.32

Example 12.9 Page no 673

In [68]:
#Given
g=9.8
h=10
B=1.7*10**-5
l=1               #m

#Calculation
import math
v=math.sqrt(2*g*h)
e=B*l*v

#Result
print"Potential difference between its end is", e*10**4,"*10**4 V"

Potential difference between its end is 2.38 *10**4 V

Example 12.10 Page no 673

In [77]:
#Given
v=72 *(5/18.0)                         #Km/h
B=40*10**-6                    #T
A=40
l=2                           #m
t=1.0
N=1

#Calculation
A=l*v
a=B*A
e=N*a/t

#Result
print"e.m.f generated in the axle of the car", e*10**3,"mV"

e.m.f generated in the axle of the car 1.6 mV

Example 12.11 Page no 673

In [84]:
#Given
w=1000/60.0
r=0.3
B=0.5                      #T

#Calculation
v=w*r
vav=v/2.0
e=B*r*vav

#Result
print"e.m.f induced is",e,"V"

e.m.f induced is 0.375 V

Example 12.12 Page no

In [91]:
#Given
r=0.5                             #m
n=2                                 #r.p.s
B=0.4*10**-4                        #T

#Calculation
import math
w=2*math.pi*n
e=0.5*B*r**2*w

#Result
print"Magnitude of induced e.m.f between the axle and rim is", round(e*10**5,2)*10**-5,"V"

Magnitude of induced e.m.f between the axle and rim is 6.28e-05 V

Example 12.13 Page no 674

In [96]:
#Given
R=1                    #m
B=1
f=50

#Calculation
import math
e=math.pi*R**2*B*f

#Result
print"e.m.f between the centre and the matallic ring is", round(e,1),"V"

e.m.f between the centre and the matallic ring is 157.1 V

Example 12.14 Page no 679

In [100]:
#Given
N=500
a=1.4*10**-4                   #Wb
l=2.5                         #A

#Calculation
L=(N*a)/l

#Result
print"Inductance of the coil is", L*10**3,"mH"

Inductance of the coil is 28.0 mH

Example 12.15 Page no 679

In [106]:
#Given
L=130*10**-3                   #H
I1=20                           #mA
I2=28                           #mA
t=140.0*10**-3                    #S

#Calculation
l=I2-I1
e=(-L*l)/t

#Result
print"Magnitude of induced e.m.f is", round(e,2),"*10**-3 V"
print"Direction oppose the increase in current"

Magnitude of induced e.m.f is -7.43 *10**-3 V
Direction oppose the increase in current

Example 12.16 Page no 679

In [115]:
#Given
N=4000
l=0.6                              #m
r=16*10**-4                        #m

#Calculation
u=4*math.pi*10**-7
L=(u*N**2*((math.pi*r)/4.0))/l
Liron=N*L

#Result
print"Inductance of the solenoid is", round(Liron,0),"H"

Inductance of the solenoid is 168.0 H

Example 12.17 Page no 679

In [119]:
#Given
L=10.0                       #H
e=300                        #V
t=10**-2                       #S

#Calculation
dl=(e*t)/L
a=e*t

#Result
print"Charge in magnetic flux is", a,"Wb"

Charge in magnetic flux is 3.0 Wb

Example 12.18 Page no 680

In [125]:
#Given
L=10*10**-3
I=4*10**-3
N=200.0

#Calculation
N1=L*I
a=N1/N

#Result
print"Total flux linked with the coil is", N1,"Wb"
print"Magnetic flux through the cross section of the coil is",a,"Wb"

Total flux linked with the coil is 4e-05 Wb
Magnetic flux through the cross section of the coil is 2e-07 Wb

Example 12.19 Page no 680

In [134]:
#Given
L=500*10**-3
I1=20*10**-3                  #A
I2=10*10**-3                  #A

#Calculation
U1=0.5*L*I1**2
U2=0.5*L*I2**2

#Result
print "Magnetic energy stored in the coil is",U1*10**6,"*10**-4 J"
print"New value of energy is",U2,"J"

Magnetic energy stored in the coil is 100.0 *10**-4 J
New value of energy is 2.5e-05 J

Example 12.20 Page no 680

In [144]:
#Given
E=12
R=30.0                    #ohm
L=0.22 

#Calculation
I0=E/R
I=I0/2.0
P=E*I
dl=(E-(I*R))/L
du=L*I*dl

#Result
print"(i) Energy being delivered by the battery is", P,"W"
print"(ii) ENergy being stored in the magnetic field of inductor is",du,"W"

(i) Energy being delivered by the battery is 2.4 W
(ii) ENergy being stored in the magnetic field of inductor is 1.2 W

Example 12.21 Page no 680

In [147]:
#Given
L=2.0                         #H
i=2                              #A

#Calculation
U=0.5*L*i**2

#Result
print"Amount of energy spent during the period is", U,"J"

Amount of energy spent during the period is 4.0 J

Example 12.22 Page no 686

In [150]:
#Given
e=1500                         #V
dl=3                               #A
dt=0.001                             #s

#Calculation
M=(e*dt)/dl

#Result
print"Mumtual induction between the two coils is", M,"H"

Mumtual induction between the two coils is 0.5 H

Example 12.23 Page no 686

In [155]:
#Given
N2=1000
I1=5.0                          #A
a2=0.4*10**-4                  #Wb
dl=-24                             #A
dt=0.02                              #S

#Calculation
M=(N2*a2)/I1
eb=(-M*dl)/dt

#Result
print"(i) Mutual induction between A and B is", M,"H"
print"(ii) e.m.f induced by the coil is", eb

(i) Mutual induction between A and B is 0.008 H
(ii) e.m.f induced by the coil is 9.6

Example 12.24 Page no 687

In [168]:
#Given
N=1200
A=12*10**-4                         #m**2
r=0.15                                     #m
N2=300
a=0.05

#Calculation
import math
u=4*math.pi*10**-7
L=(u*N**2*A)/(2*math.pi*r)
M=(u*N*N2*A)/(2*math.pi*r)
dl=2/a
e=M*dl

#Result
print"(i) Self inductance of the toroid is", round(L*10**3,1),"*10**-3 H"
print"(ii) Induced e.m.f. in the second coil is",round(e,3),"V"

(i) Self inductance of the toroid is 2.3 *10**-3 H
(ii) Induced e.m.f. in the second coil is 0.023 V

Example 12.25 Page no 688

In [181]:
#Given
I=2.0
a1=20*10**-2
x=0.15
A2=0.3*10**-2

#Calculation
import math
u=4*math.pi*10**-7
B1=(u*I*a1**2)/(2.0*(a1**2+x**2)**1.5)
a=B1*math.pi*A2**2
M=a/I

#Result
print"(i) Flux linking the bigger loop is", round(a*10**11,1)
print"(ii) Mutual induction between the two loops is",round(M*10**11,2),"!0**-11 H"

(i) Flux linking the bigger loop is 9.1
(ii) Mutual induction between the two loops is 4.55 !0**-11 H

Example 12.26 Page no 688

In [188]:
#Given
l=0.5                        #m
n=20                         #turns
r=50                         #cm
A1=40*10**-4                  #m**2
n1=25
A2=25*10**-4                   #m**2

#Calculation
u=4*math.pi*10**-7
N=n*r
N2=n1*r
M=(u*N*N2*A2)/l

#Result
print"Mutual induction of the system is",round(M*10**3,2),"*10**-3 H"

Mutual induction of the system is 7.85 *10**-3 H