Download Chapter 13 INDUCTANCE

Chapter 13 INDUCTANCE
• Introduction
• Self inductance
• Mutual inductance
• Transformer
• RLC circuits
• AC circuits
Figure 1 Self inductance in a circuit. The light bulbs
serve as voltmeters. When the current is switched on
the light across the inductance is bright because of the
• Summary
large emf across this coil while the coil across the resistor
is low because the current is low. In the steady case
only the bulb across the resistor is bright because of
the voltage drop. When the circuit is broken the energy
INTRODUCTION
stored in the self inductance is dissipated in the bulb
Faraday’s important contribution was his discovery that across the inductance.
a changing magnetic flux induces an emf in a circuit.
His relation is given as:
fact that the circulation of the electric field can be
non-zero for changing magnetic flux in a closed cirΦ
=−
cuit. Faraday’s law underlies much of the technology

that is used in modern life. This chapter probes techwhere the electromotive force  is given by:
nical aspects of induction.
I
→
→ → −
−
→ −
v × B) · l
 =  ( E + −
• Magnetic energy


and the magnetic flux Φ is given by:
Z
→
→ −
−
Φ =   B · S



The negative sign in Faraday’s law is a statement of
Lenz’s Law. Faraday’s Law encompasses two phenomena, the induced electric field in a fixed circuit due
to a changing magnetic flux, and motional emf due to
motion of the circuit in a magnetic field. Einstein’s
Theory of Relativity shows that these two phenomena
are manifestations of the same physics that result from
changing frames of reference. It was shown that independent of whether the circuit moves or not, Faraday’s
law is equivalent to the statement that:
I
→
− −
→
E · l = −



Z
 



−→
→
B −
· S

This relation is derived easily from Faraday’s Law for
→
−
the special case of a fixed circuit since then only B is
time dependent.
Faraday’s Law provides a direct linkage of electric
and magnetic fields that occurs for dynamical situations, that is changing magnetic fields. It leads to the
SELF INDUCTANCE
According to Faraday’s Law, a changing magnetic flux
in a circuit induces an emf that resists such a change.
Consider an isolated circuit. If the current in this circuit changes then the magnetic field produced by the
current in this circuit will induce an emf in the same
circuit. This is called the back emf because it opposes the change in the magnetic flux enclosed by the
circuit, that is, the circuit exhibits ”inertia”. This is
called self inductance and is illustrated by the demonstration where turning off the magnet current causes
a current to flow in the light bulb in the circuit shown
in figure 1.
Consider that this circuit is designated by the letter
 Then the flux in circuit  due to the current in the
same circuit will be written as Φ . Using Faraday’s
law we have:
 = −
Φ

This can be written as:
 = −

Φ 
= −
 

where the self inductance  is defined as:
97
Figure 3 Mutual inductance.
Figure 2 Self inductance of a solenoid.
MUTUAL INDUCTANCE
 ≡
Φ

Note the negative sign in the equation for emf which
results from Faraday’s law.
Example: Self inductance of a solenoid.
Consider that the solenoid has  turns, length 
radius , and carries a current  as shown in figure 2.
Ampère’s Law can be used to show that the magnetic
field in a solenoid is axial with a magnitude:
 = 0



The flux linkage Φ , taking into account that the
 field is uniform across the solenoid, and that it is
linked  times since the coil has  turns.
Φ =
Z
→
− −
→
2
B  · S =  2  = 0
2  


Thus the self inductance  is given as:
Φ
2
= 0
 =
2


The self inductance  is just a simple geometric
number for any coil.
NB, Many books use =turns per unit length in
his formula, whereas these notes use  = the total
number of turns; be careful not to get confused.
The SI unit of inductance is the Henry after the US
scientist. The SI unit of magnetic flux Φ is the weber.


Thus, since  = Φ
 =  =  The Henry
is a large unit. For example, for the above case let
 = 103 ,  = 002,  = 01 and 0 = 410−7 ,
3

then  = 16. For this coil, if 
 = 10 
per second, then the induced back emf will be 16 volts.
98
Mutual inductance is the induced emf in one circuit 
due to a changing  field produced by a second circuit
. Faraday’s Law can be written as:
 = −
Φ
Φ 

=−
= −

 

where the mutual inductance  is defined as:
Φ

Calculation of the mutual inductance for any pair
of circuits can be a complicated integral. One uses the
Biot Savart law to compute the B field at circuit 
due to circuit 
→
−
I
−−→ 0
 dl × rc

B =
2

4 

 ≡

Knowing the magnetic field, then one can compute the
flux linkage in circuit  due to the  field produced by
circuit .
Z
→
−−→ −
B · S
Φ =
 

Φ =
0 
4
I



Z

−
→
−
→
dl × rc
 · S
2

Thus the mutual inductance is given by

≡
=
Φ

−
−
→
Z →
I
0
dl × rc
 · S
2

4 



This complicated double integral can be simplified mathematically using Stokes Theorem, to give that:
 =
0
4
→
I I −→ −
l · l

 
This non-trivial step gives that the double integral is
a symmetric geometric factor. This relation is called
Neumann’s Formula.
 
Φ
= 0 2


You can easily compute the mutual inductance due to
the magnetic flux due to  in  and you will obtain
the same mutual inductance relation. Typical values
might be  =  = 103   = 01  = 001 then
one obtains  = 40.
=
TRANSFORMER
Figure 4 Concentric solenoids
The above proof can be repeated for the emf in 
due to a changing current in circuit . That is;
 = −
Φ

Φ 
=−
= −

 

The transformer is a nice example of use of inductance.
Consider two tightly-coupled circuits such as two concentric solenoids with an alternating emf   applied
to the primary coil  and a resistor dissipating energy
connected to the secondary coil . Consider that all
of the magnetic flux Φ goes through both coils. Then
the flux linkage for the primary circuit  is
leading to mutual inductance
 =
0
4
Φ =  Φ
→ −→
I I −
l · l
=  = 

 
That is; the mutual inductance is symmetrical whether
one is considering the flux in  due to  or vice versa.
Thus we can write for the two coils that:
while the flux linkage for the secondary is
Φ =  Φ
If the magnetic flux is time dependent then we have
  = −
Φ

Similarly for the secondary

  = −

  = −


where the mutual inductance  is a geometrical factor
expressing the degree of coupling of the magnetic flux
between two separate circuits. Note the negative sign
remaining from Faraday’s and Lenz’s laws.
Mutual inductance between two concentric solenoids
In general the computation of mutual inductance
is non trivial. However, one can easily calculate the
mutual inductance between concentric solenoids. Consider the system shown in figure 4 where the radii of
the coils are such that    . The magnetic field in
 due to circuit  is
  = −
 = 0
The  field from  only extends over an area 2 .
Thus the flux linkage in circuit  due to the magnetic
flux from  is:
Thus the mutual inductance is:
Φ

gives


=


That is the voltage ratio equals the turns ratio. The
perfect transformer does not dissipate energy in the
transformer, thus we must have power conserved, that
is:
   =   
Thus:


=


The non perfect transformer can be solved using
Kirchhoff’s loop rule that the sum of emfs around the
primary circuit equals zero:



Φ =  2  = 0  2
Eliminating
Φ






−
=0


where we have to include the induced emf due to self
inductance as well as the mutual inductance term.
For the secondary the voltage across the resistor
  =   Then Kirchhoff’s loop rule gives:
  − 
99
  − 


−
=0


Multiply the first equation by  and the second
equation by  and then take the difference of these
equations gives
(   −    ) = (  −  2 )


Figure 5 The transformer.
Obviously the closest coupling of magnetic flux occurs for self inductance where a coil is coupled perfectly
to the magnetic flux it generates. Thus we must have
that:
 ≤ 
Figure 6 RLC circuit.
 ≤ 
Thus :
 2 ≤  
For perfect coupling of magnetic flux between the
coupled circuits, then:
 2 =  
In the case of perfect coupling then the right-hand
side of the previous equation relating the emfs is zero,
therefore:
   =   
For perfect coupling this equation gives the same equations as given above;




=
=
=




Note that the transformer only works for oscillating
currents and emf’s, otherwise 
 = 0 However, the
ratio of voltages is independent of frequency in this
elementary theory. For perfect coupling and a resistive
load, then the primary and secondary waveforms are
in phase and the solution is simple.
The ability of the transformer to easily and efficiently transform voltages for AC power is the reason
that AC is used for power distribution.
100
RLC CIRCUITS
It is useful to consider the response of simple circuits
involving resistance , capacitance , and inductance
. The response of general LRC circuits to AC input
signals is important because of many applications to
technology. However, the discussion of such response
requires a detailed discussion of both the amplitude
and phase of the output relative to the input waveforms. The following is a brief summary of some concepts of AC circuits.
Consider the series combination of ,  and 
shown in figure 6. Assume that initially the capacitor
is charged with charge 0 when the switch is closed at
a time  = 0. Using Kirchhoff’s loop rule, and knowing
that voltage across the capacitor  = 
  then:


−
= 


From charge conservation, Kirchhoff’s node rule,
we have:
+

=0

Using these two equations gives a second order differential equation.
2   

+
+
=0
2

 

Figure 7 Damped RLC circuit response compared with
undamped solution when R=0.
Since  =  this also can be written as:
2
2 
 

+
+
=0
2
 

If you have studied second order homogeneous differential equations you will know that the solution for
1
2
 4
light damping, that is, 
2 is:
1

Figure 8 When 
 4
2 the motion is overdamped
1
2
leading to an exponetial decay. When 
= 4
2 the
system is critically damped leading to the most rapid
damping. Critical damping is used for meter systems to
ensure that the needle reaches the correct value in the
shortest time.

 () = 0 − 2  [ sin  +  cos ]
where:
2 =
1
2
−

42
The time dependence is that of a damped harmonic
oscillation with angular frequency  and damping time
constant  = 2
 as shown in figure 7. Note that for
 = 0 there is no damping and one has a constant har1
monic oscillation with angular frequency  = √
For
the damped case the frequency  is slightly reduced.
1
2
 4
On the other hand, when 
2  the relation
for  2 is negative leading to an imaginary value for 
producing a non-oscillatory over-damped motion that
decays exponentially as shown in figure 8.
If one applies an sinusoidal voltage from a power
supply then one will have the phenomena of resonance
when the applied frequency approaches the resonant
frequency of the circuit as will be dicussed next lecture.
Figure 9 The Tesla coil.
Tesla Coil
The Tesla coil provides a nice example of RLC circuits
coupled to transformers. The first transformer raises
the 110 60 Hz primary voltage to 15 , 60 Hz. The
small spark gap breaks down at 15 kV stimulating the
101
Figure 11 The series resonant    circuit and the
corresponding phasor diagram.
Capacitor C
Since
 = 
and from charge conservation

Figure 10 Phase relations between current (solid line)
and voltage (shaded) for a resistor, capacitor, and
inductor. The phasor diagram is shown on the right.
LC circuit to oscillate at about 500 kHz. The rate of
change of current in the primary of the second transformer is 10,000 times what it would be at 60 Hz. The
turns ratio for the second coil then produces 300 kV
across the final spark gap.
AC CIRCUITS
This discussion leads naturally to the topic of AC circuits which is of considerable technical importance.
This relates to the response of R,L,C circuits to an
applied sinusoidal voltage.
This topic is not included in this course and the
examinations because of the mathematical complexity.
However, for your education it is useful to recognize
the basic elements of AC circuits.
Consider an applied voltage that is a cosine function of time
 = 0 cos 
It is useful to define an impedance  by

=

Resistor 
Ohm’s law gives that

0
=
cos 


Thus the current and voltage are in phase as shown in
figure 10a and the impedance is
=
 = 
102


=


= −0 sin 
=
= 0 cos( +

)
2
Thus as shown in figure 10b for a capacitor the current leads the voltage by 90◦ and the impedance
is
1
 = −

and 90◦ out of phase. This is obvious in that you can
only change the voltage across a capacitor by having
current flow into the capacitor to change the stored
charge.
Inductor L
Since by Kirchhoff’s rules for circuit figure 10c,
 −

=0

Thus
0 cos  = 


By integration this gives
0
sin  = 

This can be rewritten as
=
0

cos( − )

2
Thus for an inductor the the current lags the voltage by 90◦ and the impedance is
 = 
and 90◦ out of phase. This is obvious in the the back
emf opposes change of the  field, that is the current
when a voltage is applied.
Thus using Kirchhoff’s rules for the RLC circuit in
figure 11 it can be seen that the magnitude of the effective impedance  can be calculate using Pythagorus
Theorem to have a magnitude given by
"
µ
1
|| = 2 +  −

¶2 # 12
and the voltage leads the current by a phase angle
µ
1 ¶
 − 
−1
 = tan

Figure 12 Build up of magnetic energy as the current
increases in a RL circuit connected to a battery after the
switch is closed at t=0.
This series    is an example of the fact that
any combination of passive impedances can be represented as a net load having a resultant complex impedance Z such that
V = IZ
where  has an inphase resistive component
 = || cos 
Figure 13 LC circuit.
and a reactive, or out of phase, component
 = || sin 
It is interesting that the resistive load dissipates
power where
1
 =   =  2
2
where the factor of 12 comes from the fact that the
average of (cos )2 over one complete cycle is 12.
However a reactive load does not dissipate power since
the voltage and current are out of phase then the product of   gives a (cos  sin ) term which averages
to zero over one complete cycle of oscillation.
Because of the mathematical complexity of this
topic this discussion will not be pursued further. It
is suggested that you skim over chapter 31 of Giancoli
to get an broader impression of this topic.
MAGNETIC ENERGY



This is equivalent to the statement that the energy
provided by the battery equals the energy dissipated
in the resistor plus the energy stored in the self inductance. Thus we have that the energy stored in the
inductance is:
 0  =  2  + 
 = 
Integrating the energy from  = 0 to the final value
gives the magnetic energy stored in the self inductance
as:
=
Z

0
 =
1 2

2
Consider a simple LC circuit, shown in figure 13,
with oscillating current and charge . The total energy
is distributed between the capacitor and the inductor
as:
Energy stored in Inductor
Since forces occur between magnetic circuits, energy
must be stored in the magnetic field. Consider the
system shown in figure 12.
Using Kirchhoff’s loop rule we have:


Consider that in a time dt, a charge  = 
flows. The work done by the battery is given by  0 
This equals:
 0 =  + 
  =  + 
  =
1 2 1 2
+ 
2 
2
Note that the energy oscillates between the capacitor, when Q is maximum and  = 0 to the inductor
when  = 0 and I is a maximum. This is analogous to
103
Ampère’s law gives the magnetic field inside the
toroid is
 
() = 0
2 
Integrating over the rectangular cross section inside
the toroid windings, gives the magnetic flux inside the
windings to be
Z 
Z 


() = 0  
Φ=
2

 

0
  ln( )
2

Thus the flux linkage for the N turns wrapped around
the toroid is


Φ  =  Φ = 0  2  ln( )
2

This gives that the self inductance
Φ=
Figure 14 N turn toriod with inner radius , outer
radius , and thickness .
harmonic oscillations of a pendulum where the energy
oscillates between kinetic energy and potential energy.
The inertia in the inductance is analogous to moment
of inertia in the kinetic energy term for angular motion
of the pendulum. The energy stored in the capacitor is
analogous to the gravitational potential energy stored
at the extreme positions of the pendulum oscillation.
Energy Density in a Magnetic Field
It is more useful to express the stored magnetic energy
density in terms of the magnetic field B just as the
electric energy density was expressed in terms of the
electric field E In the case of the electric field, the
stored electric energy for a capacitor, of  = 12  2
was used to show that the electric energy can be expressed as the integral of the electric energy density
 in vacuum
1
  = 0  2
2
Thus the total stored energy in the electric field in
vacuum
Z
1
 =
0  2 
 2

where the integral is taken over all space
For the magnetic field it will be shown later that
the magnetic energy  = 12  2 can be expressed
in terms of the magnetic energy density   in vacuum
 =
2
20
Thus the total stored energy in the magnetic field in
vacuum is
Z
1 2

 =
 2 0

The equivalence of this expression and  = 12  2
can be illustrated by considering the toriod shown in
figure 14.
104
Φ


= 0  2 ln( )

2

Therefore the stored magnetic energy
=
1 2


 = 0 ( )2 ln( )
2
4

Consider the integral of
Z
1 2

 =
 2 0
 =

Knowing that
0  
2 
and that the volume element of a ring inside the torus
is d = 2 gives
¶2
Z µ
1
0  
2
 =
20
2 
Z 


 = 0 ( )2
4
 
That is
1


 =  2 = 0 ( )2 ln( )
2
4

which is the same relation obtained using  = 12  2 
That is, the two expressions for magnetic energy give
the same answer for this case. In fact it can be proven,
using vector differential calculus, that this is always
true.
As a result, the most general expression for the
total electromagnetic energy can be written in terms of
the electric and magnetic fields as given by the integral
over all space of the energy density  
() =
1
1 2
 = ( 0  2 +
)
2
2 0
Z
1
1 2
( 0  2 +
)
  =

2
2 0

This is especially useful for discussions of electromagnetic waves.
SUMMARY
The concepts of self inductance and mutual inductance
and some applications have been discussed. It was
shown that the induced emf in an isolated circuit can
be written as:


where the self inductance L is defined as:
  = −
 ≡
Φ

Similarly, the coupling between two circuits is given
by


where the mutual inductance  is defined as:
  = −
 ≡
Φ

Elements of the response of circuits with L, R, and
C were discussed as well as applications to the transformer and the induction coil.
It was shown that the energy stored in an inductor
is given by
1
 =  2
2
Thus the total energy stored in a circuit having both
inductors and capacitors is:
  =  + 
  =
1 2 1 2
+ 
2 
2
It is especially useful to express the total energy
stored in an electromagnetic field in terms the energy
density of the E and B fields.
Z
1
1 2
( 0  2 +
)
  =

2
2 0

This form will be used in discussing electromagnetic
radiation.
Reading assignment: Giancoli, Chapter 30 plus
skim through Chapter 31.
105