Download Chapter 30 – Inductance

Chapter 30 – Inductance
I.
Mutual Inductance
A.
Look at two sets of loops with a common axis - loop 1 has N1 turns of wire and a current I1, and loop
2 has N2 turns of wire and a current I2.
loop 1
loop 2
I1
Faraday's Law says that if the current changes in one of
the loops, then an induced emf will be produced in the
other loop.
I2
N1
B.
N2
According to Faraday's Law, an emf (2) will be induced in loop 2 when the flux through loop 2
changes. That is,
2
= -
d( N 2  B2 )
,
dt
where N2B2 is the flux through loop 2 produced by loop 1. The flux N2B2 is then dependent on the
magnetic field (B1) produced by loop 1, which in turn is dependent on the current (I1) in loop 1, that
is,
N2B2  B1  I1 ,
and also then
dN 2  B2 
dt

dB1
dt

dI 1
dt
.
dI1
dI
, or when a constant of proportionality is introduced, 2 = - M21 1 . The constant
dt
dt
M21 is called the “mutual inductance” between loops 1 and 2 due to a change in current in loop 1. In
a like fashion, there is an induced emf in loop 1 that is caused by the change in current in loop 2.
dI
This emf is 1 = - M12 2 , where M12 is called the “mutual inductance” between loops 1 and 2 due
dt
to a change in current in loop 2.
So,
2

Note that the units of mutual inductance are: [M] = volt sec/amp = henry (H).
30-1
C.
As an example find the mutual inductance between two concentric, ideal solenoids: Given two
concentric, ideal solenoids. The inner solenoid has N1 turns, current I1 , length L, and radius R1 . The
outer solenoid has N2 turns, current I2 , length L, and radius R2 .
N2
R1
I1
N1
I2
R2
L
1)
Find the emf induced in solenoid 2 due to a current change
dI1
in solenoid 1, and from this
dt
expression find the mutual inductance, M21 .
2)
Find the emf induced in loop 1 due to a current change
dI2
in loop 2, and from this expression
dt
find the mutual inductance, M12 .
3)
Note the relationship between M21 and M12 and the fact that they do not depend on current or
voltage, only the geometry of the system.
4)
This is just a model of a transformer.
30-2
II.
Self inductance
A.
If current changes in a loop with N turns of wire, then the magnetic field inside the loops must be
changing. If the magnetic field is changing, then according to Faraday's law there must be an
induced emf in the loop. That is,
 =
-
d
(N) .
dt
d
dI
dB
(N) 
, and therefore

dt
dt
dt
proportionality constant, we get
But N  B  I , and

= -L


dI
. Introducing a
dt
dI
,
dt
where L is called the "self inductance" or simply the “inductance” of the loop. This is the voltage
developed across the inductance when the current is changing. The units of self inductance are the
same as the mutual inductance - henries.
B.
As an example, find the self inductance of an ideal solenoid: What is the induced emf in the solenoid
when the current is changing at the rate of dI/dt? The solenoid has a radius a, length b, and number
of turns per length n.
a
b
C.
This emf  is sometimes called the "back emf" because it produces an emf that opposes the change in
the current. That is, if the current in the loops is increasing, then the emf produces an induced
current that opposes this increase (opposite to direction of current flow). If the current in the loops is
decreasing, then the emf produces an induced current that opposes this decrease (same as direction
of current flow).
30-3
D.
The Inductor
1.
A circuit element that is specifically designed to have inductance.
2.
Circuit symbol:
3.
What is the voltage across an inductor?
4.
What does an inductor do?
5.
What happens when current is first delivered to the inductor?
6.
What happens when current flowing in the inductor is cut?
7.
How much power is delivered to or supplied by the inductor?
8.
How much energy is stored in the inductor?
9.
What is the energy density in the magnetic field inside the inductor?
30-4
III. The Inductor in Circuits
A.
The RL Series Circuit. A resistor and inductor are connected
in series.
1.
Switch up at time t = 0.
2.
Switch down at time t = 0. The current = Io at time t = 0.

R
L
30-5
B.
A more complex circuit.
1.
Immediately after the switch is closed, find the current
flowing through the resistors, R and R’.

R
R’
L
2.
After the switch has been closed for a long time, find the current through the resistors.
3.
Assume the switch has been closed for a long time. Now open the switch. Immediately after
the switch is opened, find the current through the resistors.
30-6
C.
LC Series Circuit. Close switch at time t = 0. Let the capacitor have an initial
charge Qo and an initial zero current.
1.
Make a qualitative analysis of charge and current.
2.
Find the charge on the capacitor as a function of time.
3.
Find the current in the loop as a function of time.
4.
Find the energy in the capacitor and the inductor as a function of time.
30-7
+Qo
C
-Qo
L
D.
RLC Series Circuit. At time t = 0, Q = Qo and I = 0. Close the switch
at time t = 0. Find the charge on the capacitor as a function of time.
+Qo
Case I: R = 0, undamped
C
Q = Qo cos t
L
-Qo
R
Case II: R small,
Q  Qo e
Case III:
R

2L
R
<
2L

R
t
2L
1
, underdamped
LC
cos ' t where ' 
1  R
 
LC  2L 
2
1
, critically damped
LC
R
R   2L t

Q  Qo  1 
t e
2L 

Case IV:
R
>
2L
Q  Qo e
1
, overdamped
LC

R 
t
2L 


R 2L
2RL 2  LC1

sinh


2RL 2  LC1 t  cosh 2RL 2  LC1 t

30-8


