Chapter 30
Inductance
Physics 231
Lecture 10-1
Fall 2008
Magnetic Effects
As we have seen previously, changes in the
magnetic flux due to one circuit can effect
what goes on in other circuits
The changing magnetic flux induces an emf
in the second circuit
Physics 231
Lecture 10-2
Fall 2008
Mutual Inductance
Suppose that we have two coils,
Coil 1 with N1 turns and
Coil 2 with N2 turns
Coil 1 has a current i1 which produces
a magnetic flux, B2 , going through
one turn of Coil 2
If i1 changes, then the flux changes and
an emf is induced in Coil 2 which is
given by
 2  N2
Physics 231
d B2
dt
Lecture 10-3
Fall 2008
Mutual Inductance
The flux through the second coil
is proportional to the current in
the first coil
N2 B 2  M 21i1
where M21 is called the
mutual inductance
Taking the time derivative of this we get
d B 2
di1
N2
 M 21
dt
dt
Physics 231
or
Lecture 10-4
di1
 2   M 21
dt
Fall 2008
Mutual Inductance
If we were to start with the second coil having a varying
current, we would end up with a similar equation with an
M12
We would find that
M 21  M12  M
The two mutual inductances are the same because the mutual
inductance is a geometrical property of the arrangement of
the two coils
To measure the value of the mutual inductance you can use
either
dI1
 2  M
dt
Physics 231
or
Lecture 10-5
dI 2
1   M
dt
Fall 2008
Units of Inductance
V  sec
J
1 Henry  1
1
Amp
Amp 2
Physics 231
Lecture 10-6
Fall 2008
Self Inductance
Suppose that we have a coil having N turns carrying a
current I
That means that there is a magnetic flux through the
coil
This flux can also be written as being proportional to the
current
N B  L I
with L being the self inductance having the same units as the
mutual inductance
Physics 231
Lecture 10-7
Fall 2008
Self Inductance
If the current changes, then the magnetic flux through the
coil will also change, giving rise to an induced emf in the coil
This induced emf will be such as to oppose the change in the
current with its value given by
dI
  L
dt
If the current I is increasing, then
If the current I is decreasing, then
Physics 231
Lecture 10-8
Fall 2008
Self Inductance
There are circuit elements that behave in this manner and they
are called inductors and they are used to oppose any change in
the current in the circuit
As to how they actually affect a circuit’s behavior will be
discussed shortly
Physics 231
Lecture 10-9
Fall 2008
What Haven’t We Talked About
There is one topic that we have not mentioned with
respect to magnetic fields
Just as with the electric field, the magnetic field has
energy stored in it
We will derive the general relation from a special case
Physics 231
Lecture 10-10
Fall 2008
Magnetic Field Energy
When a current is being established in a circuit, work has to be
done
If the current is i at a given instant and its rate of change is
given by di/dt then the power being supplied by the
external source is given by
di
P  VL i  L i
dt
The energy supplied is given by dU  Pdt
The total energy stored in the inductor is then
I
1 2
U  L i di  L I
2
0
Physics 231
Lecture 10-11
Fall 2008
Magnetic Field Energy
This energy that is stored in the magnetic field is available
to act as source of emf in case the current starts to decrease
We will just present the result for the energy density of the
magnetic field
2
1B
uB 
2 0
This can then be compared to the energy density of an
electric field
1
uE   0 E 2
2
Physics 231
Lecture 10-12
Fall 2008
R-L Circuit
We are given the
following circuit
and we then close S1 and
leave S2 open
It will take some finite amount of time for the circuit to reach
its maximum current which is given by I   R
Kirchoff’s Law for potential drops still holds
Physics 231
Lecture 10-13
Fall 2008
R-L Circuit
Suppose that at some time t the
current is i
The voltage drop across the resistor is
given by
Vab  i R
The magnitude of the voltage drop
across the inductor is given by
Vbc  L di
dt
The sense of this voltage drop is that point b is at a higher
potential than point c so that it adds in as a negative quantity
di
 iRL 0
dt
Physics 231
Lecture 10-14
Fall 2008
R-L Circuit
We take this last equation and
solve for di/dt
Notice that at t = 0 when I = 0 we
have that
Also that when the current is no
longer changing, di/dt = 0, that the
current is given by
di  R
  i
dt L L

 di 

 
 dt  initial L
I

R
as expected
But what about the behavior between t = 0 and t = 
Physics 231
Lecture 10-15
Fall 2008
R-L Circuit
We rearrange the original equation and then integrate
di
R
  dt
i   / R 
L
The solution for this is
i

i
t
di '
R
 i ' / R   L  dt '
0
0

1  e  R / L t 
R

Which looks like
Physics 231
Lecture 10-16
Fall 2008
R-L Circuit
As we had with the R-C Circuit, there is a time constant
associated with R-L Circuits
L

R
Initially the power supplied by the emf goes into dissipative
heating in the resistor and energy stored in the magnetic
field
di
 i  i2R  Li
dt
After a long time has elapsed, the energy supplied by the
emf goes strictly into dissipative heating in the resistor
Physics 231
Lecture 10-17
Fall 2008
R-L Circuit
We now quickly open S1
and close S2
The current does not immediately
go to zero
The inductor will try to keep the current, in the same
direction, at its initial value to maintain the magnetic flux
through it
Physics 231
Lecture 10-18
Fall 2008
R-L Circuit
Applying Kirchoff’s Law to the
bottom loop we get
di
 iR  L  0
dt
di
R
  dt
i
L
Rearranging this we have
and then integrating this
0
t
di '
R
 i '   L  dt '
I
0

I  I 0 e  R / L t
0
where I0 is the current at t = 0
Physics 231
Lecture 10-19
Fall 2008
R-L Circuit
This is a decaying exponential
which looks like
The energy that was stored in the inductor will be dissipated
in the resistor
Physics 231
Lecture 10-20
Fall 2008
L-C Circuit
Suppose that we are now given
a fully charged capacitor and
an inductor that are hooked
together in a circuit
Since the capacitor is fully charged
there is a potential difference
across it given by Vc = Q / C
The capacitor will begin to discharge as soon as the switch is
closed
Physics 231
Lecture 10-21
Fall 2008
L-C Circuit
We apply Kirchoff’s Law to this circuit
di q
L  0
dt C
Remembering that
We then have that
di d 2q
 2
dt dt
The circuit equation then becomes
Physics 231
Lecture 10-22
i  dq
dt
d 2q 1

q0
2
LC
dt
Fall 2008
L-C Circuit
This equation is the same as that for the Simple Harmonic
Oscillator and the solution will be similar
q  Q0 cos(t   )
The system oscillates with angular frequency

1
LC
 is a phase angle determined from initial conditions
dq
The current is given by i 
  Q0 sin( t   )
dt
Physics 231
Lecture 10-23
Fall 2008
L-C Circuit
Both the charge on the
capacitor and the current in
the circuit are oscillatory
The maximum charge and
the maximum current occur
p / 2 seconds apart
For an ideal situation, this circuit will oscillate forever
Physics 231
Lecture 10-24
Fall 2008
L-C Circuit
Physics 231
Lecture 10-25
Fall 2008
L-C Circuit
Just as both the charge on the capacitor and the current
through the inductor oscillate with time, so does the energy
that is contained in the electric field of the capacitor and the
magnetic field of the inductor
Even though the energy content of the electric and magnetic
fields are varying with time, the sum of the two at any given
time is a constant
UTotal  U E  U B
Physics 231
Lecture 10-26
Fall 2008
L-R-C Circuit
Instead of just having an L-C
circuit with no resistance, what
happens when there is a
resistance R in the circuit
Again let us start with the capacitor fully charged with a
charge Q0 on it
The switch is now closed
Physics 231
Lecture 10-27
Fall 2008
L-R-C Circuit
The circuit now looks like
The capacitor will start to
discharge and a current will
start to flow
We apply Kirchoff’s Law to this circuit and get
di q
 iR  L   0
dt C
And remembering that
we get
i  dq
dt
d 2q R dq 1


q0
2
L dt LC
dt
Physics 231
Lecture 10-28
Fall 2008
L-R-C Circuit
The solution to this second order differential equation is
similar to that of the damped harmonic oscillator
The are three different solutions
Underdamped
Critically Damped
Overdamped
Which solution we have is dependent upon the relative values
of R2 and 4L/C
Physics 231
Lecture 10-29
Fall 2008
L-R-C Circuit
Underdamped:
4L
R 
C
2
The solution to the second differential equation is then
q  Q0
 R 

t
e  2L 
This solution looks like
2 
 1

R
cos 
 2 t   
  LC 4 L 




The system still oscillates but with
decreasing amplitude, which is
represented by the decaying
exponential
This decaying amplitude is often
referred to as the envelope
Physics 231
Lecture 10-30
Fall 2008
L-R-C Circuit
Critically Damped: R 2 
4L
C
Here the solution is given by
 R 
  2 L  t
t e
R

q  Q0 1 

 2L 
This solution looks like
This is the situation when the
system most quickly reaches
q=0
Physics 231
Lecture 10-31
Fall 2008
L-R-C Circuit
Overdamped:
4L
R 
C
2
Here the solution has the form
q
with
Physics 231
 R  

Q0  2 L  t 
 1
e
2


 R 
2 L e 't  1  2 L e  't 

' 
' 





R


R2
1
'

2
4 L LC
This solution looks like
Lecture 10-32
Fall 2008
L-R-C Circuit
The solutions that have been developed for this L-R-C circuit
are only good for the initial conditions at t = 0 that q = Q0 and
that i = 0
Physics 231
Lecture 10-33
Fall 2008