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Induction
Self-Inductance, RL Circuits
X XX X
X XX XX X
X XX X
a
b
e
b
B
I
e
L
(dI / dt )
I
R
1
L
I
a
L
L/R
e1
VL
f( x ) 0.5
0.0183156
0
0
1
2
3
4
t
Today...
• Mutual Inductance – applications!
• Concept of Self-Inductance
• Definition of Self-Inductance
• Calculation of Self-Inductance for Simple
Cases
• RL Circuits
• Energy in Magnetic Field
Mutual Inductance
• Demo: current is induced in one coil
when the current is changed in a
neighboring coil
b
a
• We can describe this effect
quantitatively in terms of the mutual
inductance, the ratio of flux through the
loop to current in opposite loop.
M
 ab  ba

Ib
Ia
ab = flux through coil “a”
due to current in coil “b”
• Note M has this symmetry (not obvious perhaps, but true)
• Change current in loop a  changes Ba  change flux through
loop b  induces current in b to oppose the change  loop b
produces a field Bb which opposes the change of Ba.
• The same thing happens with just one loop. This is called selfinductance.
Applications of Mutual Inductance
iron
• Transformers (still to come)
– Change one AC voltage into another
• Airport Metal Detectors
– Pulsed current  pulsed magnetic field
 Induces emf in metal
– Ferromagnetic metals “draw in” more B
 larger mutual inductance  larger emf
– Emf  current (how much, how long it
lasts, depends on the resistivity of the
material)
– Decaying current produces decaying
magnetic field
 induces current in receiver coils
– Magnitude & duration of signal depends
on the composition and geometry of the
metal object.
N2
V2 
V1
N1
e
~
V1
V2
N1
N2
(primary) (secondary)
Applications of Mutual Inductance
• Pacemakers
– It’s not easy to change the battery!
– Instead, use an external AC supply.
– Alternating current
 alternating B
 alternating ФB inside “wearer”
 induces AC current to power
pacemaker
~
Applications of Mutual Inductance
Self-Inductance
• Consider the loop at the right, Rloop=0.
• switch closed  current starts to flow in
“inductance”
the loop. (is it infinite??) I = V/ 0 !!! No.
limits dI / dt
X XX X
X XX XX X
X XX X
a
b
• Therefore, magnetic field produced in the area
enclosed by the loop. (B proportional to I)
• Therefore, flux through loop increases as the current increases
• Therefore, emf induced in loop opposing initial direction of
current flow because it opposes increasing flux (Faraday’s
Law)
this emf acts like a battery to oppose
the real battery – reduce current flow]
• Fact: the current turns on at such a rate to give V(a)-V(b) =
battery voltage:
dI d B
dt

dt
 V (a)  V (b)
• Self-Induction: changing current through a loop inducing an
opposing emf in that same loop.
Self-Inductance
• The magnetic field produced by the
current in the loop shown is
proportional to that current: B  I
• The flux, therefore, is also
proportional to the current.
I
 B   B  dS  I
• We define this constant of
proportionality between flux and
current to be the inductance, L.
L
B
I
• Combining with Faraday’s ε   d  B   d ( LI )   L dI
dt
dt
dt
Law gives the emf induced
by a changing current:
dI
ε  L
dt
Preflight 17:
Two simple pieces of wire A and B are shaped
into almost complete loops. The loose ends of
each loop are connected to identical batteries.
Assume the loops have the same total resistance,
and that they do not interfere with each other.
2) Which loop has the greatest flux through it (assume the loops
have the same current in them)?
a) A
b) B
c) same
Recall the B-field of current loop (at the center): B = m0I/(2R)
The area of the loop is A = p R2

 The flux through the loop increases with R.
 The flux through loop A is bigger than the flux through loop B.
3) Which loop has the greatest self inductance?
a) A
b) B
c) same

Flux is  B   B  dS
Self inductance is defined as
so if I is the same, LA > LB.
L
B
I
Self-Inductance
• The inductance of an inductor (a set of coils in some geometry;
e.g., solenoid) then, can be calculated from its geometry alone if
the device is constructed from conductors and air (similar to the
capacitance of a capacitor).
• If extra material (e.g., an iron core) is added, the inductance will
increase (just as adding a dielectric increases capacitance)
T  m2
L has units of henrys (H): 1 H  1
A
B
L
I
• Archetypal inductor is a long solenoid, just as a pair of parallel
plates is the archetypal capacitor.
A
++++
l
r
N turns
r << l
d
-----
d  A
Calculation
• Long Solenoid:
l
N turns total, radius r, Length l
r
N
r  l  B  m 0 I
l
N turns
For a single turn, A  pr 2    BA  m 0 N Ipr 2
l
The total flux through solenoid is given by:
N2
 B  N  m 0
Ipr 2
l
Inductance of solenoid can then be calculated as:
2
B
N2 2
N
L
 m0
pr  m 0   lpr 2
I
l
 l 
1
Lecture 17, Act 1
• Consider the two inductors shown:
– Inductor 1 has length l, N total turns and
has inductance L1.
– Inductor 2 has length 2l, 2N total turns
and has inductance L2.
– What is the relation between L1 and L2?
(a) L2 < L1
(b) L2 = L1
l
2l
r
r
N turns
(c) L2 > L1
r
2N turns
Lecture 17, Act 1
• Consider the two inductors shown:
– Inductor 1 has length l, N total turns and
has inductance L1.
– Inductor 2 has length 2l, 2N total turns
and has inductance L2.
– What is the relation between L1 and L2?
(a) L2 < L1
(b) L2 = L1
l
2l
r
r
N turns
r
2N turns
(c) L2 > L1
• To determine the self-inductance L, we need to determine the flux
B which passes through the coils when a current I flows: L  B / I.
• To calculate the flux, we first need to calculate the magnetic field B
produced by the current: B = m0(N/l)I
• i.e., the B field is proportional to the number of turns per unit
length.
• Therefore, B1 = B2. But does that mean L1 = L2?
Lecture 17, Act 1
• To calculate L, we need to calculate the
l
2l
flux.
r
r
r
• Since B1=B2, the flux through any
N turns
given turn is the same in each
2N turns
inductor
• There are twice as many turns in inductor 2; therefore the net flux
through inductor 2 is twice the flux through inductor 1! Therefore,
L2 = 2L1.
Inductors in series add (like resistors): Leff  L1  L2
And inductors in parallel
add like resistors in parallel:
1
1 1
 
Leff L1 L2
Inductors in Circuits
General rule: inductors resist change in current
• Hooked to current source
– Initially, the inductor behaves like an open switch.
– After a long time, the inductor behaves like an ideal wire.
• Disconnected from current source
– Initially, the inductor behaves like a current source.
– After a long time, the inductor behaves like an open switch.
2
Preflight 17:
5) At time t = 0 the switch is closed. What is the current through the
circuit immediately after the switch is closed?
a) I = 0
b) I = V/R
c) I = V/2R
Initially, the inductor acts like an open switch, and
no current flows through the branch which contains
the inductor. Hence, I = V/(2R).
6) What is the current through the circuit a long time after the switch is
closed?
a) I = 0
b) I = V/R
c) I = V/2R
After a long time, the inductor acts like an ideal
wire, and there is no voltage drop across it. Hence,
the resistor which is in parallel with the inductor
has no current flowing through it. All the current
in the circuit flows through the inductor, and
I = V/R.
Lecture 17, Act 2
• At t=0 the switch is thrown from position
b to position a in the circuit shown:
2A – What is the value of the current I a
long time after the switch is thrown?
I
a
I
R
b
e
L
R
(a) I = 0
2B
(b) I = e/2R
(c) I = 2e/R
– What is the value of the current I0 immediately after
the switch is thrown?
(a) I0 = 0
(b) I0 = e/2R
(c) I0 = 2e/R
• After a long time the switch is opened.
2C
–What is the value of the current I0 just after the switch is
opened?
(a) I0 = 0
(b) I0 = e/2R
(c) I0 = 2e/R
Lecture 17, Act 2
• At t=0 the switch is thrown from position
b to position a in the circuit shown:
2A – What is the value of the current I a
long time after the switch is thrown?
I
a
I
R
b
e
L
R
(a) I = 0
(b) I = e/2R
(c) I = 2e/R
• A long time after the switch is thrown, the current approaches
an asymptotic value: as t , dI/dt 0.
• As dI/dt  0, the voltage across the inductor 0. Therefore,
I = e/2R.
Lecture 17, Act 2
• At t=0 the switch is thrown from position
b to position a in the circuit shown:
2B – What is the value of the current I0
immediately after the switch is thrown?
(a) I0 = 0
(b) I0 = e/2R
I
a
I
R
b
e
L
(c) I0 = 2e/R
• Just after the switch is thrown, the rate of change of current is
as large as it can be (we had been assuming it was !)
• The inductor limits dI/dt to be initially equal to e/L. The voltage
across the inductor = e; the current, then, must be 0!
• Another way: the moment the switch is thrown, the current tries
to generate a huge B-field. There is a huge change in flux
through coil—an emf is generated to oppose this. Initially, then,
no current flows through no voltage drop across the resistors.
Lecture 17, Act 2
• After a long time the switch is
opened.
2C
–What is the value of the current I0
just after the switch is opened?
(a) I0 = 0
(b) I0 = e/2R
I
a
I
R
b
e
L
(c) I0 = 2e/R
• Just after the switch is thrown, the inductor induces an emf to
keep current flowing: emf = L dI/dt (can be much larger than e)
• However, now there’s no place for the current to go  charges
build up on switch contacts  high voltage across switch gap
•If the electric field exceeds the “dielectric strength”
(~30 kV/cm in air)  breakdown  SPARK!
This phenomenon is used in “flyback generators” to create
high voltages; it also destroys lots of electronic equipment!
RL Circuits, Quantitative
I
a
I
R
• At t=0, the switch is closed and
the current I starts to flow.
b
e
• Loop rule:
e  IR  L
L
dI
0
dt
Note that this equation is identical in form to that for
the RC circuit with the following substitutions:
RC:
ε
Q
dQ
R
0
C
dt
Therefore,
 RC  RC
 RCRL:
  RL 
L
R
RL
1
R
C
QI
RL Circuits
• To find the current I as a
function of time t, we need
to choose an exponential
solution which satisfies the
boundary condition:
a
I
I
R
b
e
L
RL = R
e
dI
(t  )  0  I (t  ) 
R
dt


L
ε
1  e  Rt / L
R
• The voltage drop across the inductor is given by:
dI
VL  L
 εe  Rt / L
dt
• We therefore write: I 
RL Circuit (e on)
I
e

1 e
R
 Rt / L
2L/R
1
2
Sketch curves !

Max = e/R
Q
Current
1
e/R
L/R
f( x ) I
0.5
63% Max at t=L/R
00
0
Voltage on L
VL  L
dI
 ee  Rt / L
dt
Max = e/R
t
3
4
t
3
4
x
t/RC
1 e1
f( xV
) 0.5
L
37% Max at t=L/R
0.0183156 0
0
1
2
RL Circuits
• After the switch has been in
position a for a long time,
redefined to be t=0, it is moved
to position b.
• Loop rule:
IR  L
a
I
I
R
b
e
dI
0
dt
L
• The appropriate initial condition is: I (t  0)  e
R
• The solution then
must have the form:
I 
e
R
VL  L
e  Rt / L
dI
 ee  Rt / L
dt
RL Circuit (e off)
e/R
1 1
Current
I 
e
R
L/R
2L/R
1
2
Sketch curves !
e  Rt / L
f( x ) 0.5
I
Max = e/R
37% Max at t=L/R
0.0183156
0
0
01 0
VL  L
Max = -e
dI
 ee  Rt / L
dt
Q
Voltage on L
t
3
x
4
4
f( x ) V
0.5L
37% Max at t=L/R
-e
0
0
1
2
t
3
4
e on
L/R
1
e/R
2L/R
I
I
f( x ) 0.5
1
1 e1
2
t
3
f( x ) 0.5
I
0
3
0
0
1
2
t
3
Q
dI
VL  L  ee  Rt / L
dt
4
e
1
01 0
f( x )V0.5
L
2
t
3
x
VL  L
1
4
4
dI
 ee  Rt / L
dt
-e0
0
e  Rt / L
R
0
4
x
t/RC
f( xV
) 0.5
L
2L/R
I 
0.0183156
0
L/R
e/R
1 1
e

1  e  Rt / L 
R
00
83156
e off
2
t
3
4
Lecture 17, Act 3
• At t=0, the switch is thrown from
position b to position a as shown:
– Let tI be the time for circuit I to reach
1/2 of its asymptotic current.
– Let tII be the time for circuit II to reach
1/2 of its asymptotic current.
– What is the relation between tI and tII?
(a) tII < tI
(b) tII = tI
(c) tII > tI
I
a
I
R
b
I
e
L
R
a
I
I
L
b
e
II
L
R
Lecture 17, Act 3
• At t=0, the switch is thrown from
position b to position a as shown:
(b) tII = tI
(c) tII > tI
• We must determine the time
constants of the two circuits by writing
down the loop equations.
dI
I: e  I R  L  I R  0
dt
dI
dI
e  L  IR  L
0
II:
dt
dt
I 
L
2R
 II 
2L
R
I
R
b
– Let tI be the time for circuit I to reach
1/2 of its asymptotic current.
– Let tII be the time for circuit II to reach
1/2 of its asymptotic current.
– What is the relation between tI and tII?
(a) tII < tI
I
a
I
e
L
R
a
I
I
L
b
e
II
R
L
This confirms that
inductors in series add!
Energy of an Inductor
• How much energy is stored
in an inductor when a
current is flowing through it?
• Start with loop rule: e  IR  L
dI
dt
• Multiply this equation by I:
εI  I 2 R  LI
I
a
I
R
b
L
e
dI
dt
• From this equation, we can identify PL, the rate at which
energy is being stored in the inductor:
dU
dI
PL 
•
 LI
dt
dt
We can integrate this equation to find an expression for U, the
energy stored in the inductor when the current = I:
U
I
0
0
U   dU   L I dI

1
U  LI 2
2
Preflight 17:
7) At t = 10 hrs the switch is opened, abruptly
disconnecting the battery from the circuit. What
will happen to all the energy stored in the
solenoid?
Energy stored in the inductor: U = 1/2 L I2
When the switch is opened, this energy is dissipated in the resistor.
An inductor doesn’t like change!!!
When the switch is opened, the inductor will try to maintain the current
that was flowing through it before the switch is opened. Since the
battery is disconnected from the circuit, the energy which is necessary
to keep current flowing through the resistor is provided by the
inductor.
Where is the Energy Stored?
• Claim: (without proof) energy is stored in the magnetic field
itself (just as in the capacitor / electric field case).
• To calculate this energy density, consider the uniform field
generated by a long solenoid:
N
l
B  m0 I
l
N2 2
• The inductance L is: L  m 0 l pr
• Energy U:
r
N turns
1 2 1  N 2 2  2 1 2 B2
U  LI   m0 pr  I  pr l
2
2
l
2
m0

• We can turn this into an energy density by dividing by
the volume containing the field:
U
1 B2
u 2 
pr l 2 m 0
Summary
• Self-Inductance
L
B
I
– Self-Inductance for solenoid
– Inductors in series add
N2 2
L  m0
pr
l
Leff  L1  L2
– Inductors in parallel add reciprocally
• Energy of an Inductor
U
1
1 1
 
Leff L1 L2
1 2
LI
2
• Energy stored in the Magnetic Field
1 B2
u
2 m0