7.2 Electromagnetic Induction
7.2.1 Faraday’s Law
1831 Michael Faraday
Experiment 1: Lorentz force
Experiment 2: electric field induced
by the changing magnetic field
Experiment 3: electric field induced
by the changing magnetic field
v
v
change field
Conclusion: A changing magnetic field induces an electric field. Æ universal flux rule
r r
r
dΦ
d r
ε = ∫ E ⋅ dl = ∫ ∇ × E ⋅ nˆda = −
= − ∫ B ⋅ nˆ da
dt
dt
r
∂ r
∇× E = − B
∂t
(
)
All three experiments come to the same flux rule that leads Einstein to the special
theory of relativity.
Nature abhors a change in flux.
Lenz’s law: a handy rule to express the current flow under a change of magnetic
fields.
Example: A long cylindrical magnet of length L and radius a carries a uniform
magnetization M parallel to its axis. It passes at constant velocity v through a
circular wire ring of slightly larger diameter. Graph the emf induced in the ring, as a
function of time.
r
r
Solenoidal: K = M × nˆ = Mφˆ
M
r
r
B ⋅ L + 0 ⋅ L = µ0 ML Æ B = µ0 M
Φ = µ0 Mπa 2 , ε = −
∆Φ
∆t
Example: The “jumping ring demonstration”. If you wind a solenoidal coil around
an iron core (the iron is there to beef up the magnetic field), place a metal ring on top,
and plug it in, the ring will jump several feet in the air. Why?
What is induction heating?
Fig 1 Induction heating is a noncontact heating method
Fig 2 Heat energy (E) produced in an electric circuit is equal to I2 _ R.
Induction heating (Fig. 1) is a noncontact heating method; one in which an electrically conductive material
(typically a metal) is heated by an alternating magnetic field. Invisible lines of force are created by a work coil
when a current flows through it, the result of which is an induced current in the conductive workpiece. Heating
results due to the Joule effect and, to a lesser degree, magnetic hysteresis (i.e., power loss other than by eddy
currents in a magnetic material caused by reversals of the magnetic field). Joule’s Law (Fig. 2) states that the rate
at which heat energy is produced in any part of an electric circuit is measured by the product of the square of the
current (I) times the resistance (R) of that part of the circuit.
Ref: http://www.industrialheating.com/CDA/ArticleInformation/features/BNP__Features__Item/0,2832,124816,00.html
7.2.2 The Induced Electric Field
Static charge
r ρ
Æ Electric field (Coulomb’s law) ∇ ⋅ E =
ε0
r
r
∂B
Changing magnetic fields Æ Electric field (Faraday’s law) ∇ × E = −
∂t
r
r
r
r
∂B
ÅÆ ∇ × B = µ0 J (Ampere’s law)
(Faraday’s law) ∇ × E = −
∂t
r
r
∇⋅E = 0 Å Æ ∇⋅B = 0
when ρ = 0 ,
r
∂B
in exactly the same way as
Faraday-induced electric fields are determined by −
∂t
r
magnetostatic fields are determined by µ0 J .
The tricks associated with Ampere’s law can be adopted for Faraday’s law.
r
r r
r r
r
dΦ
∂B
∫ B ⋅ dl = µ0 I enclosed ,
∫ E ⋅ dl = − dt ( ∇ × E = − ∂t )
Example: A uniform magnetic field B(t), pointing straight up, fills the shaded circular
region in the figure. If B is changing with time, what is the induced electric field?
B
r r
dΦ
d
2
∫ E ⋅ dl = 2πsE = − dt = − dt (πs B(t ))
E=−
s dB ˆ
φ
2 dt
E
Example: A line charge λ is glued onto the rim of a wheel of radius b, which is then
suspended horizontally, so that it is free to rotate. In the central region, out to radius
a, there is a uniform magnetic field B0, pointing up. Now someone turns the field off.
What happens?
B0
a
r r
dΦ
dB
b
= −πa 2
Faraday’s law: ∫ E ⋅ dl = −
dt
dt
r
r
r
r r r
E Æ F = qE Æ N = r × F Æ N = Iα
∫ Ndt = ∫ Idω = Iω
N = ∫ b ⋅ (λdlE ) = ∫ bλEdl = −πa 2bλ
dB
Æ
dt
0
2
2
∫ Ndt = −πa bλ ∫ dB = λπa bB0 = Iω
B0
It’s the electric field that did the rotating.
Example: An infinitely long straight wire carries a slowly varying current I(t).
Determine the induced electric field, as a function of the distance s from the wire.
B=
µ0 I (t )
2πs
µ0 I
µ LI ⎛ S ⎞
ds = 0 ln⎜⎜ ⎟⎟
2πs
2π
⎝ S0 ⎠
S0
S
Φ = L∫
l
S0
induced Iind parallel to original I(t) Æ induced Eind parallel to I(t)
like linear magnetic media (para- or diamagnetic material)
r r
µL
dΦ
dI
∫ E ⋅ dl = ε (s0 )L − ε (s )L = − dt = − 20π (ln s − ln s0 ) dt
⎛ µ0 dI
⎞
ln s + K ⎟ zˆ
⎝ 2π dt
⎠
ε (s ) = ⎜
7.2.3 Inductance
Mutual Inductance:
S
I(t)
The mutual inductance is determined if the geometrical configuration between the two
looks is given.
The magnetic field of loop1 is:
loop 2
r r r
loop 1
r r µ I dl × (r − r )
B1 (r ) = 0 1 ∫ 1 r r 3 1
4π
r − r1
(
)
r
r r
r r
r r
dΦ
at loop 2 Æ Φ = ∫ B1 (r2 ) ⋅ nˆ2 da2 = ∫ ∇ 2 × A1 (r2 ) ⋅ nˆ2 da2 = ∫ A1 (r2 ) ⋅ dl2
dt
r r
r
r r µI
d
l
µ
I
d
l
⋅ dl
A1 (r ) = 0 1 ∫ r 1r Æ Φ = 0 1 ∫ ∫ r1 r 2
4π r − r1
4π
r − r1
ε =−
Define: Φ at _ loop 2 = M 21I1
µ
Neumann formula: M 21 = 0
4π
r r
dl1 ⋅ dl2
∫ ∫ rr − rr1
Conclusion:
1. M21 is purely geometrical quantity
2.
Φ2
Φ1
M 21 = M 12 ( Φ at _ loop1 = M 12 I 2 and Φ at _ loop 2 = M 21I1 )
loop 2
loop 1
I
Flux is the same when
the current running through loop1 and through loop2 is equal in magnitude.
Example: A short solenoid (length l and radius a with n1 turns per unit length) lies on
the axis of a very long solenoid (radius b with n2 turns per unit length). Current I
flows in the short solenoid. What is the flux through the long solenoid?
B from _ long _ solenoid = µ0 n2 I
Φ through _ short _ solenoid = πa 2 µ 0 n2 In1l
M 12 = M 21 = µ0πa 2 n1n2l
Mutual inductance:
Suppose that you vary the current in loop1, the induced emf in loop 2 is:
Φ at _ loop 2 = MI1 , ε 2 = − M
Self-Inductance:
dI1
dt
Φ = LI , ε = − L
dI
self-inductance L with the unit of henries (H)
dt
Example: Find the self-inductance of a toroidal coil with rectangle cross section (inner
radius a, outer radius b, height h), which carries a total of N turns.
B 2πr = µ0 NI , B =
µ0 NI
2πr
r =b
µ0 N 2 Ih ⎛ b ⎞
µ0 NI
µ0 N 2 Ih ⎛ b ⎞
ln
=
LI
,
dr
Φ = N ∗h ∫
=
L=
ln⎜ ⎟
⎜ ⎟
2πr
2π
2π
⎝a⎠
⎝a⎠
r =a
The emf is in a direction as to oppose any change in current Æ back emf
Transient solution and steady state solution of a circuit:
Suppose a current flowing around a loop when someone suddenly cuts the wire. The
current drops “instantaneously” to zero. This generate a whopping back emf, for
although I may be small, dI/dt is enormous. That’s why you often draw a spark
when you unplug an iron or toaster.
Example: Suppose a battery (which supply a constant emf ε 0 ) is connected to a
circuit of resistance R and inductance L.
ε 0 − IR − L
What current flows?
dI
=0
dt
initial condition I=0
R
dI
dI
dt d (ε 0 − RI )
= − dt ,
= ,
L = ε 0 − IR ,
(ε 0 − IR ) L (ε 0 − IR )
L
dt
I = I (t )
∫ d [ln(ε
I =0
0
− RI )] = − ∫
R
dt
t =0 L
t
R
R
− t⎞
− t
⎛ ε 0 − RI ⎞
ε 0 ⎛⎜
R
L
⎟
⎜
I
1
e
Æ = ⎜ − L ⎟⎟
ln⎜
⎟ = − L t Æ ε 0 − RI = ε 0e
R⎝
ε
0
⎠
⎝
⎠
τ=
L
is called time constant
R
7.2.4 Energy in Magnetic Fields
Calculate energy of magnetic field from the self-inductance in a circuit.
I
energy stored in the inductor (magnetic field inside)
energy stored in the battery
energy consumed by the resistor (heating)
energy stored in the self-inductance of the circuit
(magnetic field)
1
dI
dW
= −εI = L I Æ W = LI 2
2
dt
dt
r r
r
dΦ
ε =−
, Φ = LI , Φ = ∫ B ⋅ nˆda = ∫ A ⋅ dl
dt
r
r
1
1
1 r r 1 r r
W = I ∗ LI = IΦ = I ∫ A ⋅ dl = ∫ A ⋅ I dl (self-inductance Æ I = Idl )
2
2
2
2
r
r
1 r r
W = ∫ A ⋅ J dτ , relate to magnetic field Æ ∇ × B = µ0 J
2
r
r r
r r
r
r
r
1
W=
A
⋅
∇
×
B
d
τ
∇
⋅
A
×
B
=
B
⋅
∇
×
A
−
A
⋅
∇
×
B
,
since
2 µ0 ∫
r
r r
r r
r
1
1
1
2
τ
τ
W=
B
⋅
∇
×
A
−
∇
⋅
A
×
B
d
=
B
d
−
A
× B ⋅ nˆ da
2µ0 ∫
2µ0 ∫
2 µ0 ∫
P=
(
)
( (
))
( (
)
(
(
)
(
))
)
(
(
)
)
Take the integral over the whole space.
1 2
W =∫
B dτ
2µ0
The work done by electric field and magnetic field:
ε
1
ρVdτ = 0 ∫ E 2 dτ
∫
2
2
1 r r
1
W = ∫ J ⋅ A dτ =
B 2 dτ
∫
2
2 µ0
W=
(
)
Example: A long coaxial cable carries current I (the current
flows down the surface of the inner cylinder, radius a, and back
along the outer cylinder, radius b). Find the magnetic energy
stored in a section of length l.
b
I
a
µI
1
1 ⎛ µ0 I ⎞
1
µ 0 I 2l ⎛ b ⎞
2
=
=
B
d
τ
l
2
π
rdr
ln⎜ ⎟
B = 0 φˆ , W =
⎜
⎟
2µ0 ∫
2 µ0 ⎝ 2π ⎠ ∫ r 2
4π
2πr
⎝a⎠
2
Example (7.58): A certain transmission line is constructed from two thin metal
“ribbons”, of width w, a very small distance h << w apart. The current travels down
one strip and back along the other. In each case it spreads out uniformly over the
surface of the ribbon.
(a) Find the capacitance per unit length, C.
E
(b) Find the inductance per unit length, L.
(c) What is the product LC, numerically?
B
σ
V
Q
V
Q wε 0
=
Æ
Æ C=
=
(a) E = , E =
h
ε0
lwε 0 h
lV
h
(b) B = µ0 K Æ u B =
µ0 I 2
µ0
2
K 2 Æ U B=
µ0
2
K 2 hwl =
1 2
LI
2
1
L' µ h
L' I 2 Æ L = = 0
2 w
2
l
w
(c) LC = ε 0 µ0 in vacuum or LC = εµ in media
hl =
Exercise: 7.12, 7.13, 7.17, 7.20, 7.24, 7.25, 7.27
I