Electromagnetism
Physics 15b
Lecture #15
Inductance
Purcell 7.6–7.10
What We Did Last Time
Induction: emf when the magnetic flux in a loop changes
B
1 dΦ
Faraday’s Law E = −
FL q
c dt
–
+

Sign of the emf follows Lenz’s Law:
the induced current opposes the change
of the flux
Differential form: ∇ × E = −
1 ∂B
c ∂t

curl E no longer zero!

Just one last step before completing
Maxwell’s equations
x
+
+
+
+
+
–
–
–
–
–
E
I
q
w
E=
v
vwB
c
1
Today’s Goals
Introduce Inductance
Mutual inductance – Magnetic interaction between two circuits
  hanging current in circuit #1 produces an emf in circuit #2
  Reciprocity theorem
  Self inductance – Interaction
between a circuit and magnetic field
  Changing current in circuit produces
a “back emf” in itself

... and Inductors


Energy stored in inductor L
Circuits with L, C and R
Mutual Inductance
Two wire loops are close to each other

Turn on the switch
 I1 increases
 B increases
 emf E2 on the right loop
loop 1 loop 2
I1
B
E2
dI
1 dΦ 2
Generally true for any pair of circuits
∝− 1
c dt
dt
Define mutual inductance between loop 1 and loop 2:
E2 = −
emf in
loop 2

Unit:
E2 = −M 21
dI1
dt
current
in loop 1
M 21 ≡
1 Φ2
c I1
flux in
loop 2
[emf]
esu cm
sec 2
in CGS
=
=
[current]/[time] esu sec 2
cm
2
Big Loop and Small Loop
Two concentric loops C1 and C2, with radii R1 >> R2


Current I1 on C1 produces B inside C2
From Lecture 12
B=

2π I1
cR1
B
C1
C2
 Biot-Savart
If R2 is really small
I1
2π 2I1R22
Φ 2 = ∫ B ⋅ da 2 ≈ B ⋅ π R =
C2
cR1
2 2
2π R
M 21 = 2 2
c R1
2
2
The “reverse” inductance M12 is much harder to calculate in
this case — but we don’t have to…
Reciprocity Theorem
For any two circuits C1 and C2, related by
dI
dI
E2 = −M 21 1 and E1 = −M12 2
M12
dt
dt
Proof: we can express M12 as
Φ
1
M 21 = 2 =
B ⋅ da 2
cI1 cI1 ∫S2


1
cI1
∫
S2
(∇ × A) ⋅ da 2 =
C2
C1
I1
Use B = ∇×A and apply Stokes
M 21 =
= M 21
1
∫C2 A ⋅ d s2
cI1 
da2
ds1
r12
ds2
Express A using integral of I1 over C1 (Biot-Savart)
M 21 =
I1ds1
1
1
⋅ d s2 = 2

∫

∫
cI1 C2 C1 cr12
c
ds1 ⋅ ds 2
C1
r12
∫ ∫
C2
This is symmetric
between 1 and 2
3
Self Inductance
We don’t really need two circuits to find inductance
Any circuit produces B field
B field changes when the current
I changes
  emf appears on the circuit itself

I

B
Define self inductance L as
E = −L

dI
dt
or
L≡
1Φ
c I
Minus sign comes from Lenz’s law  whenever the current
changes, the circuit tends to resist the change
  E is also known as the
“back emf”
Inductor (a.k.a. Coil)
Wire wound in a coil has a large self inductance
Core can be empty (air-core)
  Ferromagnetic (e.g. iron) core
enhances the inductance
E = −L

emf against I
B
B inside a long air-core coil is
4π NI
B=
c

dI
dt
I increases
Caution: the coil wraps around the same B field N times
Φ = NBA =
4π N 2IA
c
Φ 4π N 2 A
L= =
cI
c 2
Having many
turns help
4
Energy in Inductor
Connect an inductor L to a controllable current source

Increase current from 0 to I0 over time T
I(0) = 0

I(T ) = I0
The current source must flow I against back emf
dI
V =L
dt
That’s work, and takes power

Total work is
T
0
P dt =
–
+

∫
I
∫
T
0
LI
P = IV = LI
dI
dt
V
I0
dI
1
dt = L ∫ dI = LI02
0
dt
2
Inductor L flowing current I holds an energy
LI 2
U=
2
Energy in Magnetic Field
Inductor is nothing but a wire in an empty space

Energy is stored in the empty space itself
Consider the air-core coil again
4π NI
4π N 2 A

We know B =

The energy is
c
U=
and L =
B
c 2
LI 2 4π N 2 AI 2 A 2
1 2
=
=
B =
B × (volume)
2
2
8π
8π
2c 
Generalize: energy stored in any magnetic field
U=
1
8π
∫B
2
dv
cf. We saw earlier for E field U =
1
8π
∫E
2
dv
5
Inductors in Circuits
What does an inductor do in a circuit?

It resists the current change
V = φa − φb = L

dI
dt
φa
φb
I
Compare this with R & C
V = RI
φa
φb
I
V=
Q
C
φa
+Q
−Q
I=
φb
dQ
dt
I
These rules + Kirchhoff determines current
RL Circuit
Let’s consider a simple case: R and L with an emf
Switch is closed at t = 0
  What follows?
Step 1: define the direction of I
Step 2: apply the loop rule

R
L
I
dI
E − L − RI = 0
dt
Step 3: solve the equation I(t) =
+
E!
−
⎛ R ⎞
E
− k exp ⎜ − t ⎟
R
⎝ L ⎠
Step 4: use the initial condition (I(0) = 0)  Why is this true?
I(t) =
⎛ R ⎞⎞
E⎛
1−
exp
⎜⎝ − L t ⎟⎠ ⎟⎠
R ⎜⎝
6
RL Circuit
An RL circuit is a lot like an RC circuit

Current approaches its final value exponentially
E
R
I(t) =
R
− t⎞
E⎛
L
1−
e
⎟
R ⎜⎝
⎠
L
τ =L R
E!
−
t

Asymptotic value = what I would be if L were not there

Time constant
τ=
+
I
R
L
R
LC Circuit
An inductor L and a capacitor C are connected in series
C has initially a charge Q0
  Switch closes at t = 0

Define I and Q
dQ
  Q and I are related: I = +
dt
+Q
I
C
L
-Q
Kirchhoff’s loop rule
Q
dI
− −L =0
C
dt

Q
d 2Q
+L 2 =0
C
dt
d 2Q
Q
=−
2
LC
dt
This is a differential equation for a simple harmonic oscillator
7
Simple Harmonic Oscillator
Mass m is placed on a friction-free floor

Spring pulls/pushes the mass m with force
F = −kx

m
Hooke’s Law
Newton’s law:
d 2x
k
d 2x
=−
x
2
2
m
dt
dt
Simple Harmonic Oscillation
F=m
F
−x
x(t) = A cos(ω t + φ )
ω=
F
k
is the angular frequency
m
x
LC Circuit
An LC circuit is a simple harmonic oscillator

Let’s assume the solution is
Q(t) = A cos(ω t + φ )

d 2Q
Q
=−
2
LC
dt
2
LHS = −Aω cos(ω t + φ )
A
RHS = −
cos(ω t + φ )
LC

+Q
Plug into the differential equation
I
C
L
-Q
ω=
1
LC
Initial conditions (Q(0) = Q0 and I(0) = 0) determine A and φ
⎧⎪ A = Q0
⎨
⎪⎩φ = 0
⎧⎪Q(t) = Q0 cos ω t
⎨
⎩⎪I(t) = −Q0ω sin ω t
8
Charge and Current
Q and I oscillates back
and forth
Their phases are off by
90 degrees

Q0 Q = Q0 cos ω t
ωt
Q lags behind I
−Q0
Period of oscillation
Q0ω
2π
T=
= 2π LC
ω
I = −Q0ω sin ω t
ωt
Frequency of oscillation
f=
1
1
=
T 2π LC
−Q0ω
Why Oscillate?
+Q0
C
−Q0
I=0
L
C discharges
by flowing
current thru L
0
I = +Q0ω
0
I = −Q0ω
L
0
Charge gone, but L
keeps current flowing
Charge gone, but L
keeps current flowing
C
0
C
L
C discharges
by flowing
current thru L
−Q0
C
+Q0
I=0
L
9
Energy
Energy is exchanged between C and L

Maximum values are:
max
C
U
UCmax

L(Q0ω )2
U =
2
1
= ULmax for ω =
LC
Q02
=
2C
max
L
Total energy at any t
2
UC =
Q 2 +Q0
C
2C −Q0
UC = 0 C
2
0
Q
LI
0
+
2C
2
Q 2 cos2 ω t LQ02ω 2 sin2 ω t
= 0
+
−Q0
2C
2
Q2
2
U
=
C
Q
C
2C +Q
= 0 constant
0
2C
U=
I=0
L
UL = 0
I = −Q0w
L
UL =
I=0
L
UL = 0
LI 2
2
Energy
Energy is exchanged between C and L, but the total
remains constant
Q2
0
2C
The situation is
identical to the
mass-spring
oscillator
UC
Q02
2C
UL
m
Q02
2C
UC
UL
ωt
10
Summary
Inductance

Mutual inductance E2 = −M

Self inductance
E = −L
dI
dt
dI1
dt
or
E1 = −M
L=
LI 2
Energy in an inductor U =
2

Energy density in magnetic field
Inductors in a circuit: V = L
dI
dt
dI2
Note: reciprocity
dt
Φ
cI
emf against I
B
u=
B2
8π

RL circuit relaxes exponentially with τ

LC circuit oscillates with angular frequency ω =
I increases
=L R
1
LC
11