B. Sc. Physics (H.R.

K) Chapter 38: Inductance

19
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

INDUCTANCE

Inductance
Consider a coil wound on a cylindrical core. Assume that the current in the coil either increases or
decreases with time. When the current flow through the coil, a magnetic field will be set up inside the coil as
shown in figure below.

As the current changes with time, the magnetic flux through the coil also changes and induces an emf
in the coil. This induced emf e
L
is directly proportional to the rate of change of current
dI
dt
.
e
L

JI
Jt

e
L
= -I
JI
Jt

The negative sign is due to the fact that induced emf oppose the cause which produce it. The symbol I
is the inductance or self inductance of the coil which is described as:
The ratio between the induced emf and the rate of change of current
I = -
e
L
[
JI
Jt

The unit of inductance is Henry. The inductance of a coil is one Henry if an emf of one volt is induced
in it, when the current changes at the rate of one ampere per second.
1 Ecnry =
1 :olt
[
1 ompcrc
scconJ

= 1 IA
-1
s
Inductance of a Solenoid
Consider a solenoid of length l, having N number of turns. If the current I flows through solenoid, then
magnetic field B produced is given by:
B =

0
N I
I
= p
0
n I
Where n =
N
I
is the number of turns per unit length.
If A is the area of each circular turn of the solenoid, then the flux
B
passing through one turn is:

B
= (B)(A)

B
= p
0
n IA
The magnetic flux passing through all the N turns is given by:

B
= Np
0
n IA
If the current changes through the solenoid, then the magnetic flux also changes with respect to the time, i.e.,
J
B
Jt
= Np
0
n A
JI
Jt

B. Sc. Physics (H.R.K) Chapter 38: Inductance

20
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

According to the Faradays law of electromagnetism, the induced
emf e is described as:
e = -
dq
B
dt
= -Np
0
n A
dI
dt
---------------- (1)
If L is the self inductance of the solenoid, then
e

= -I
dI
dt
---------------- (2)
Comparing (1) and (2), we get:
I = Np
0
n A
As n =
N
I
N = nl
I = (nl)p
0
n A
I = p
0
n
2
l A
This expression gives the self inductance of a solenoid.
Inductance of a toroid
Let N is the number of turns in a toroid through current I is flowing and r is mean radius of the toroid.
Let a and b are the inner and outer radii of toroid, respectively, and h is its height.

The magnetic field produced at any point inside toroid can be find out using expression:
B =
p
0
NI
2nr

Let b Jr = JA is the small area element, then the magnetic flux passing through the small area element
will be:
J
B
= B JA
J
B
= B b Jr
Integrating we get:

B
= _ B b Jr
=b
=u

The magnetic flux passing through all N turns is:

B
= _ N B b Jr
=b
=u

B
= _ N _
p
0
NI
2nr
] b Jr
=b
=u
=
p
0
N
2
I b
2n
_
Jr
r
=b
=u

B
=
p
0
N
2
I b
2n
ln|r|
u
b
=
p
0
N
2
I b
2n
(lnb - lno)

B
=

0
N
2
I h
2n
ln
b
u

q
B
I
=

0
N
2
h
2n
ln
b
u
------------------ (1)
The self inductance of a toroid is:
I =
q
B
I
------------------ (2)

B
I

B
= I I
By comparing (1) and (2), we get:
B. Sc. Physics (H.R.K) Chapter 38: Inductance

21
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

I =
p
0
N
2
b
2n
ln
b
o

So, the inductance of a toroid depends only on geometrical factors.
Inductors with Magnetic Materials
When a magnetic field B
0
acts on a ferromagnetic material, then total magnetic field B is increased
which can be expressed as:
B = p

B
0

Where p

is the relative permeability of permeability constant of the magnetic material.
Similarly, the inductance of a coil is increased, if some magnetic material is placed inside the coil. If I
0

and I is the inductance with and without the magnetic material, respectively, then:
I = p

I
0
------------------ (1)
Now the inductance of solenoid without magnetic material is:
I
0
= p
0
n
2
l A
Putting values of I
0
in equation (1), we have:
I = p

p
0
n
2
l A
The effective values of permeability constant p

for a ferromagnetic material have the values in range
of 1u
3
to 1u
4
. Therefore the inductance of a coil can be increased by a factor of 1u
3
to 1u
4
with the core of a
ferromagnetic substance.

Growth of Current in LR Circuit
Consider a series circuit of a resistance R and inductance
connected in series with a battery of emf e as shown in the figure.
When the switch is open, no current will flow through the
circuit. So at time t = u, I = u.
When the switch is closed, the current starts flowing
through the circuit. But due to inductance, this current takes some
time to get its maximum value. (During this period, the current is called transient current)
Initially, when the current is changing, the back emf I
dI
dt
appears across the inductance coil which dies away
with the passage of time. And its value falls to zero, when the current is maximum.
Let I
R
and I
L
are the values of potential drop across resistance R and inductance I, respectively. Then,
by applying the Kirchhoffs rule:
e -I
R
- I
L
= u
I
L
= e - I
R

I
dI
dt
= e - IR
dI
s-IR
=
dt
L

Multiplying both sides by R, we get:
RdI
s-IR
= R
dt
L

Integrating between the limits,
At t = u, I = u;
B. Sc. Physics (H.R.K) Chapter 38: Inductance

22
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

At t = t, I = I
]
RdI
s-IR
I
0
= ] R
dt
L
t
0

|ln(e -IR)|
0
I
= -
R
I
|t|
0
t

ln(e - IR) - ln(e) = -
R
I
t
ln_
e -IR
e
] = -
R
I
t
e -IR
e
= c
[-
R
L
t

IR = e - e c
[-
R
L
t

IR = e _1 - c
[-
R
L
t
]
I =
e
R
_1 - c
_-
t
L
R
,
_
_
Let
L
=
L
R
= Inductive time constant and
s
R
= I
0
= Maximum Current
I = I
0
_1 - c
[-
t
:
L

]
This is the equation of growth of current
Special Cases:
At time t =
L

I = I
0
_1 - c
[-
:
L
:
L

] = I
0
(1 - c
-1
) = I
0
(1 - u.S7)
I = u.6S I
0

At t =
I = I
0
(1 - c
(- )
) = I
0
_1 -
1

] = I
0
(1 - u)
I = I
0

Inductive Time Constant
It is the interval of time during which the current grows to 6S % of its maximum value.
The graph shows that current through the circuit increases exponentially.
As
I = I
0
_1 - c
[-
R
L
t
] ------------------ (1)
I
R
= IR = I
0
R _1 - c
[-
R
L
t
] = e _1 - c
[-
R
L
t
] ------------------ (2)
i.e., I
R
increases exponentially upto e.
Differentiating equation (1) with respect to time, we get:
JI
Jt
= -
R
I
I
0
_u - c
[-
R
L
t
]
I
L
= I
dI
dt
= I
0
R c
[-
R
L
t
= e c
[-
R
L
t
------------------ (3)
i.e., I
L
decreases exponentially upto u.


B. Sc. Physics (H.R.K) Chapter 38: Inductance

23
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

Decay of Current in LR Circuit
When the S is connected to the b, as shown in the figure, the current start decaying exponentially with
time and becomes zero after some time.
If I
R
and I
L
are the values of potential difference across the resistance R and inductance L,
respectively. Then by applying the Kirchhoffs Rule:
I
R
+ I
L
= u
IR + I
dI
dt
= u
I
dI
dt
= -IR

dI
I
= -
R
L
Jt
Integrating between the limits,
At t = u, I = I
0
;
At t = t, I = I
]
dI
I
I
I
0
= ] R
dt
L
t
0

|ln(I)|
I
0
I
= -
R
I
|t|
0
t

ln(I) - ln(I
0
) = -
R
I
t
ln[
I
I
0
= -
R
L
t
I
I
0
= c
[-
R
L
t

I = I
0
c
[-
R
L
t

I = I
0
c
_-
t
L
R
,
_

Let
L
=
L
R
= Inductive time constant
I = I
0
c
[-
t
:
L


This is the equation of decay of current in an LR circuit.

Special Cases:
At time t =
L

I = I
0
c
[-
t
:
L

= I
0
( c
-1
) = I
0
( u.S7)
I = u.S7 I
0

At t =
I = I
0
( c
(- )
) = I
0
_
1

] = I
0
( u)
I = u
Inductive Time Constant
It is the interval of time during which the current decays to 6S % of its maximum value.
As I = I
0
c
_-
t

L
]
------------------ (1)
The potential difference across R is
I
R
= IR = I
0
R c
[-
R
L
t

I
R
= e c
[-
R
L
t
------------------ (2)
Differentiating equation (1) with respect to time, we get:
JI
Jt
= -
R
I
I
0
c
[-
R
L
t

B. Sc. Physics (H.R.K) Chapter 38: Inductance

24
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

I
L
= I
JI
Jt
= - I
0
R c
[-
R
L
t

I
L
= - e c
[-
R
L
t
------------------ (3)

Energy Stored in the Magnetic Field
Consider a resistor and an inductor is connected in series with a battery of emf e. According to the
Kirchhoffs 2
nd
rule;
e -I
R
- I
L
= u
e = I
R
+I
L

e = IR + I
dI
dt

Multiplying both sides by I, we have:
eI = I
2
R +II
dI
dt

Here eI is the power supplied by the source, I
2
R is the power dissipated in resistance and II
dI
dt
is the
energy supplied per time in the inductance coil, where the magnetic field also exists.
If u
B
is the total energy stored in the magnetic field of inductance coil, then the energy stored per unit
time is:
Ju
B
Jt
= II
JI
Jt

Ju
B
= II JI
Integrating both sides, we get:
_ Ju
B
0
B
0
= _ II JI
I
0

u
B
=
1
2
II
2

This is the expression of energy stored in the magnetic
field of a current carrying inductor. This equation is similar to the
expression of energy stored in the electric field of a capacitor.
u
L
=
1
2

q
2
C

The energy stored in the inductor can be dissipated through
joule heating in the resistor, in the similar way the energy stored in
the capacitor is dissipated through the joule heating in the resistor during the discharging of a capacitor.
Energy Density
The energy density u
B
in case of an inductor can be find out by using expression:
Encrgy cnsity =
Lncg Stocd
voIumc

If u
B
is the energy stored in the magnetic field of an inductor of length l having cross-sectional area A, then the
energy density in the expression will be:
u
B
=
u
B
A l

As u
B
=
1
2
II
2
, therefore
B. Sc. Physics (H.R.K) Chapter 38: Inductance

25
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

u
B
=
[
1
2
II
2

A l
=
II
2
2A l

In case of a current carrying coil, the induce I can be find out by
using expression: I = p
0
n
2
l A
u
B
=
(p
0
n
2
l A)I
2
2A l
=
p
0
n
2
l AI
2
2A l

u
B
=
p
0
n
2
I
2
2
=
p
0
2
n
2
I
2
2p
0

The magnetic field strength
inside a current carrying coil is
B = p
0
n I
u
B
=
B
2
2p
0

This expression is similar to the energy density inside the electric field of a capacitor, which is expressed as:
u
L
=
e
0
E
2
2

where e
0
is the permittivity of free space and E is the electric
field strength inside the capacitor.
Electromagnetic Oscillations (Qualitative discussion)
Consider an oscillating circuit which consists of a capacitor
and an inductor. If the capacitor is initially charged and the
switch is then closed, we find that both the current in the
circuit and the charge on the capacitor oscillate between
maximum positive and negative values. If the resistance of
the circuit is zero, no energy is transformed to internal energy.
In the following analysis, we neglect the resistance in the
circuit. We also assume an idealized situation in which energy
is not radiated away from the circuit.
When the capacitor is fully charged, the energy u
L
in
the circuit is stored in the electric field of the capacitor and is
equal to
u
L
=
1
2

(q
mux
)
2
C

At this time, the current in the circuit is zero, and
therefore no energy is stored in the inductor.
u
B
= u
As the switch S is closed, the capacitor begins to discharge and
the energy stored in its electric field decreases. The discharge
of the capacitor represents a current in the circuit, and hence some energy is now stored in the magnetic field of
the inductor. Thus, energy is transferred from the electric field of the capacitor to the magnetic field of the
B. Sc. Physics (H.R.K) Chapter 38: Inductance

26
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

inductor. When the capacitor is fully discharged, it stores no energy. At this time, the current reaches its
maximum value, and all of the energy is stored in the inductor.
(u
L
)
mn
= u, (u
B
)
mux
=
1
2
I(I
mux
)
2


The current continues in the same direction, decreasing in magnitude, with the capacitor eventually
becoming fully charged again but with the polarity of its plates now opposite the initial polarity. This is
followed by another discharge until the circuit returns to its
original state of maximum charge q
mux
. The energy continues to
oscillate between inductor and capacitor. When the energy stored
in the capacitor is maximum, the energy stored in inductor is
minimum and vice versa. But the total energy u = u
L
+ u
B

remains constant.
As the energy in such circuit is used to oscillate between
capacitor and inductor, therefore, it is also called Oscillating
Circuit.

LC Circuit: Electromagnetic Oscillations (Quantitative
analysis)
Let u
L
and u
B
are the values of energy stored in the electric field of capacitor and magnetic field of an
inductor, respectively.
So the total energy u stored in the circuit is:
u = u
L
+ u
B

u =
1
2

q
2
C
+
1
2
II
2

Differentiating the above equation, we get
Ju
Jt
=
J
Jt
_
1
2

q
2
C
+
1
2
II
2
_
u = cosntont
u =
q

C

Jq
Jt
+II

JI
Jt


q

C

Jq
Jt
+ II

JI
Jt
= u
I =
dq
dt

q

C

Jq
Jt
+ I.
Jq
Jt

.
J
Jt
_
Jq
Jt
] = u

Jq
Jt
_
q

C
+ I.
J
Jt
_
Jq
Jt
]_ = u

q

C
+ I
J
2
q
Jt
2
= u

d
2
q
dt
2
+
q

LC
= u -------------- (1)
For a block spring system, the differential equation of a mass spring system is:
B. Sc. Physics (H.R.K) Chapter 38: Inductance

27
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

d
2
x
dt
2
+

m
= u
The solution of this equation is:
=
m
cos(t + )
Where
m
is the amplitude of oscillating mass spring system and is its initial phase. Similarly, the solution of
equation (1) will be:
q = q
m
cos(t + )
And
dq
dt
= q
m
sin(t + )
d
2
q
dt
2
=
2
q
m
cos(t + )
Putting the values of q and
d
2
q
dt
2
in equation (1), we get:
q
m
cos(t + )
2

1
LC
= u

2

1
IC
= u
=
1
IC

=
1
2n

1
IC

This is the expression of frequency of the LC circuit of electromagnetic oscillations.
The energy stored in the electric field of a capacitor is:
u
L
=
1
2

q
2
C
=
1
2
q
m
cos(t + )
2

u
L
=
1
2C
q
m
2
cos
2
(t + )
The energy stored in the magnetic field of an inductor will be:
u
B
=
1
2
II
2

u
B
=
1
2
I _
Jq
Jt
_
2
=
1
2
I _
J
Jt
q
m
cos(t + )_
2

u
B
=
1
2
I
2
q
m
2
sin
2
(t + )
u
B
=
1
2
I _
1
IC
q
m
2
sin
2
(t + )_
u
B
=
1
2C
q
m
2
sin
2
(t + )
The total energy stored in the LC circuit is:
u = u
L
+ u
B
=
1
2
q
m
2
cos
2
(t + ) +
1
2C
q
m
2
sin
2
(t + )
u =
1
2
q
m
2
cos
2
(t + ) + sin
2
(t + )
u =
1
2
q
m
2

As the q
m
is the maximum value of charged stored in the capacitor, so the total energy stored in LC circuit is
constant.
B. Sc. Physics (H.R.K) Chapter 38: Inductance

28
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

Damped and Forced Oscillations
A resistance is always present in every real LC circuit. When
we take this resistance into account, we find that the total
electromagnetic energy is not constant but decreases with time as it is
dissipated as internal energy in the resistor. For a LC circuit, the total
energy u of the circuit is described as:
u = u
L
+ u
B

Where u
L
is the energy stored in the electric field of a charged
capacitor and u
B
is the energy stored in the magnetic field of inductor.
u =
1
2

q
2
C
+
1
2
II
2

Differentiating the above equation, we get
d0
dt
=
d
dt

1
2

q
2
C
+
1
2
II
2
------------------- (1)
The total energy of LC circuit is no longer constant but rather

d0
dt
= I
2
R
Thus the equation (1) will become:
d
dt

1
2

q
2
C
+
1
2
II
2
= I
2
R

q

C

dq
dt
+II

dI
dt
= -I
2
R
I =
dq
dt

q

C

dq
dt
+I.
dq
dt

.
d
dt
[
dq
dt
= -[
dq
dt

2
R

dq
dt

q

C
+ I.
d
dt
[
dq
dt
= -[
dq
dt

2
R

q

C
+ I.
d
dt
[
dq
dt
= -
dq
dt
R
I
d
2
q
dt
2
+
dq
dt
R +
q

C
= u
The general solution of the above equation will be:
q = q
m
c
[-
Rt
2L

cos(

t + )
Here

2
- [
R
2L

2

The figure shows the charged stored on the capacitor in a damped LC circuit as the function of time. The current
decreases exponentially with time. Moreover the frequency

is strictly less than the frequency of

undammed oscillations.

Forced Oscillations and Resonance
The RLC series circuit is similar to a mechanical system consisting of a simple harmonic oscillator in a
damping medium. The oscillating system consist of mass m attached to a spring of spring constant .
B. Sc. Physics (H.R.K) Chapter 38: Inductance

29
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera


If the frequency of the oscillator is equal to the natural frequency of system

m
, the there will be
resonance and the amplitude of the oscillation will be maximum.
In the similar way, if

is the frequency of sinusoidal voltage source, then the sinusoidal voltage is

expressed as:
e = e
m
cos

t
Where e
m
is the peak value of sinusoidal voltage. The current flowing through
the circuit will be:
I = I
m
cos(

t - )
Where I
m
is the maximum value of sinusoidal current.
If the angular frequency of external voltage source is equal to the
natural frequency =
1
LC
of the circuit, then there will be resonance and the
maximum current will flow through the circuit.
The resistance R provides the damping in the circuit. Greater the
resistance, smaller will the current flow through the circuit. Such oscillations
which are produced in oscillating system are called forced oscillations.

Growth of Current in RC Series Circuit
Consider a resistance R and a capacitance C is connected in series with a battery of emf e as shown in
the figure. The S
1
and S
2
are the two switches.

When the switch S
1
is closed, keeping the switch S
2
, the capacitor begins to charge and the voltage
across the capacitor v

begins to increase. Let v

is the potential drop across the resistor, then by applying the

Kirchhoffs law:
e - I
R
- I
C
= u
e - IR -
q
C
= u
e -
q
C
= IR
B. Sc. Physics (H.R.K) Chapter 38: Inductance

30
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

I =
Jq
Jt

e
q
C
= R
Jq
Jt

1
R
Jt =
Jq
[e
q
C

Multiplying equation by
1
C
on both sides:

1
C
Jq
[e
q
C

=
1
RC
Jt
Integrating the above equation, we get:
_

1
C
Jq
[e
q
C

q
0
=
1
RC
_ Jt
t
0

ln[e
q
C

0
q
=
1
RC
|t|
0
t

ln[e
q
C
ln(e) =
1
RC
(t u)
ln_
e
q
C
e
_ =
t
RC

ln[1
q
eC
=
t
RC

1
q
eC
= c
[-
t
RC

q
eC
= 1 c
[-
t
RC

q = eC _1 c
[-
t
RC

]
Where

= RC is the capacitive time constant, therefore:

q = eC _1 c
[-
t

] ------------------- (1)
When t = , the capacitor will fully charged to its saturation value q
0

q
0
= eC _1 c
[-

] = eC(1 c
(-)
) = eC _1
1
c

] = eC _1
1

] = eC
The equation (1) will become:
q = q
0
_1 c
[-
t

] ------------------- (2)
Growth of Current
Differentiating equation (2) with respect to time, we get:
Jq
Jt
=
J
Jt
_q
0
_1 c
[-
t
:

]_ =
J
Jt
_q
0
q
0
c
[-
t
RC

_
I = u q
0
_
1
RC
] c
[-
t
RC

I =
1
R
q
0
C
c
[-
t
RC

B. Sc. Physics (H.R.K) Chapter 38: Inductance

31
Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

q
0
C
= e
I =
e
R
c
[-
t
RC

e
R
= I
0

I = I
0
c
[-
t
RC

Special Cases
At t = u
q = eC _1 - c
[-
0
:

] = eC(1 - c
0
) = eC(1 - 1 )
q = u
At t =

q = eC _1 - c
[-
:

] = eC(1 - c
(-1)
) = eC _1 -
1
c
]