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UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
LECTURE NOTES 21
Electrodynamics
Electromotive Force — Ohm’s Law
In order to make a free electrical current flow in matter, one somehow has to push (and/or
pull) on electric charges – i.e. exert a force on them. Note however, that the speed at which the
electrical charges move in a conducting medium depends on the detailed microscopic nature of
that conducting medium.
For most electrically conducting materials, the volume free current density, J free ( r ) is linearly
proportional to the force per unit charge, f ( r ) ≡ F ( r ) Q Note that J free ( r ) f ( r ) = F ( r ) Q .
(
J free ( r ) = σ c f ( r ) = σ c F ( r ) Q
)
where the constant of proportionality σ c ≡ electrical conductivity of the material.
SI units of electrical conductivity σ c :
Volume free current density: J free ( r )
Amperes
F (r )
Q
meter 2
( m)
A
2
dQ
( Amperes )
dτ
1 Ampere = 1 Coulomb/sec
I=
( )
Newtons
N
C
Coulomb
A 2
J (r )
2
σc =
= m = C
N − m2 − s
N
f (r )
C
Force per unit charge: f ( r ) ≡
F (r ) ⎛ N ⎞
F ( r ) ⎛ Newtons Volts ⎞
=
⎜ ⎟ also has SI units of: E ( r ) =
⎜
⎟
Q ⎝C⎠
Q ⎝ Coulomb meter ⎠
Amps m 2 ⎛ Amps ⎞
σc =
=⎜
⎟ m
Volts m ⎝ Volt ⎠
SI units symbol = Ω
V − V ⎛ Volts
ΔV
⎞
= 2 1 ⎜
≡ Ohms ⎟
Define resistance: R ≡
I
I
⎝ Amps
⎠
But: f ( r ) =
⇒ SI units of electrical conductivity: σ c = Ohms −1 m 1 Ω ≡ 1V A
1 Ohm ≡ 1Volt Amp
1
Define electrical resistivity of material: ρ c ≡
SI units of ρ c = Ohm-m
σc
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
1
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
−1
In some circles (EE-types) 1 Ohm −1 ≡ “mho” (Ohm spelled backwards) σ c = ohm m = mho m
1 Ampere
Note also that (esp. in Europe): 1 Siemens ≡
Volt
⎛ Amps ⎞
∴ SI units of electrical conductivity: σ c = ⎜
⎟ m = Siemens m
⎝ Volt ⎠
∴ 1 Siemens ≡ 1 Ohm = 1 mho
Some typical values of resistivity ρc ( Ω − m ) for various materials are listed in the table below:
Any force capable of driving free electrical charge will produce
⎛ F (r ) ⎞
J free ( r ) = σ c f ( r ) = σ c ⎜
A m2 )
⎜ Q ⎟⎟ (
⎝
⎠
(
)
Usually, in E&M this force is: FToT ( r ) = qE ( r ) + q v ( r ) × B ( r ) = FE ( r ) + Fm ( r )
(
)
If B ( r ) = 0 , or: B ( r ) ≠ 0, but v ( r ) × B ( r ) is very small, such that FE ( r )
Fm ( r )
then (usually):
Ohm’s Law:
J free ( r ) = σ c E ( r ) = E ( r ) ρc note that: E ( r ) J free ( r )
(
)
{In plasmas, where v ( r ) may not be small, Fm ( r ) Q = v ( r ) × Bext ( r ) must be included…}
2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Inside a conducting wire, if a steady free electric current Ifree flows inside the wire:
I free = ∫ J free ( r )ida⊥ = ∫ J free ( r )inˆ⊥ da⊥ = J free A⊥ (if J free is uniform)
S⊥
S⊥
e.g. A⊥ = π R 2 for a wire of radius R.
I free , J free , E , nˆ⊥
I free , J free , E
R
The electric field E inside the electrical conductor exists only when the free current Ifree is
flowing. Einside = 0 when Ifree = 0 J free = 0 (i.e. electrostatics case)
(
)
Now a steady Ifree can flow in an electrical conductor when an electrostatic potential difference,
ΔV is applied across the two ends of the conductor, e.g. using a battery:
ΔV
I free , J free , E
+
•
−
•
cylindrical electrical conductor (wire)
with cross sectional area A⊥ = π R 2
L
+
−
Ideal wire
Ideal wire
Battery
(Ideal voltage source)
If the conducting wire has a uniform cross-section and material of the wire is uniform,
homogenous, isotropic and linear then the free current density will also be uniform, thus:
I free = J free A⊥ = (σ c E ) A⊥ = σ c A⊥ E
But for this same (uniform) wire, E = ΔV L = constant for uniform, homogeneous, isotropic and
linear conducting wire of cross-sectional area, A⊥ . {n.b. what this last relation is actually saying
is that Laplace’s equation ∇ 2V ( r ) = 0 is operative inside the wire, not Poisson’s equation
∇ 2V ( r ) = − ρ free ( r ) / ε o , i.e. ρ free ( r ) = 0 inside the wire!!!}
⎛ ΔV ⎞
∴ I free = σ c A⊥ ⎜
⎟
⎝ L ⎠
We can now define the DC resistance of the conducting wire as: R =
⎛ L ⎞
ΔV
1 ⎛ L ⎞
= ⎜
⎟ = ρc ⎜
⎟
I free σ c ⎝ A⊥ ⎠
⎝ A⊥ ⎠
SI units of DC resistance = Ohms = Volts/Ampere
= 1/SI units of 1/DC resistance = Siemens = Amperes/Volt
1 Ohm = 1/Siemens
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
3
UIUC Physics 435 EM Fields & Sources I
Ohm’s Law: I free = ΔV R
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
I free (Amps)
slope = 1
Linear relationship
R
I free vs. ΔV
ΔV (Volts)
y = mx + b, b = 0
n.b.
y-intercept ( I free ) = 0
y = mx + b
⎛1⎞
I = ⎜ ⎟ ΔV + 0
⎝R⎠
Ohm’s Law: I free = ΔV R
DC Resistance:
or:
ΔV = I free R
1
R = ρc ⎛⎜ L ⎞⎟ = ⎛⎜ L ⎞⎟ Ohms
⎝ A⊥ ⎠ σ c ⎝ A⊥ ⎠
Griffiths Example 7.2: Calculate the DC resistance of a conducting material of conductivity σ c
sandwiched between two long, superconducting cylinders (inner radius a, outer radius, b) with
electrostatic potential difference ΔV ≡ Vb − Va between the two superconducting cylinders.
{n.b. the superconducting material has no DC resistance}:
∃ a radial electrostatic field between superconductors: E ( ρ ) =
E (ρ) =
λ
ρˆ (in cylindrical coordinates)
2πε o ρ
λ = electric charge/unit length on inner cylinder
λ
ρˆ
2πε o ρ
ρ = x 2 + y 2 (in cylindrical coordinates)
b
a
•
•
L
a
dϕ
adϕ
L >> b,a
I free = ∫ J free ( ρ )ida⊥ = σ c ∫ E ( ρ )ida⊥
s⊥
da⊥ = adϕ L ρˆ
← evaluated at inner radius
σ c λ 2π ⎛ 1 ⎞
σλ
σ λL
ρˆ iadϕ L ρˆ = c 2 π L = c
⎜
⎟
∫
2πε o 0 ⎝ a ⎠
εo
2 π εo
σ λL
= c
Note also that λ L = Qinner
εo
I free =
I free
4
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Now:
ΔV = Vb − Va = − ∫ E ( ρ )id
a
b
ΔV = − ∫
Fall Semester, 2007
Lecture Notes 21
where d = d ρρ
a
λ ⎛1⎞
λ ⎛1⎞
λ
⎜ ⎟ ρˆ id ρρˆ = − ∫b
⎜ ⎟=−
2πε o ⎝ ρ ⎠
2πε o ⎝ ρ ⎠
2πε o
a
b
Prof. Steven Errede
a
dρ
b
ρ
∫
a
λ
λ
λ
ΔV = −
n(ρ ) = −
[ ln a − ln b] = +
[ln b − ln a ]
2πε o
2πε o
2πε o
b
ΔV =
( )
λ
ln b
a
2πε o
Ohm’s Law (here): ΔV = IR
⇒
∴ R=
( )
σ λL
λ
ln b = c R
a
εo
2πε o
εo
λ
1
2π ε o
( a) σ
ln b
c
λL
=
2πσ c L
⎧ 1
⎫
ΔV = IR = ⎨
ln b ⎬ I
a
⎩ 2πσ c L
⎭
1
ln b
R=
a
2πσ c L
( )
( )
( a)
ln b
λ=
Note also that (here):
εoI
σcL
Qinner = λ L =
εoI
σc
Ohm’s Law: J free ( r ) = σ c E ( r )
Take the divergence of both sides – if the free current is steady (i.e. no time dependence), then:
∂ρ free ( r , t )
∇i J free ( r ) = 0
{Continuity equation: ∇i J free ( r , t ) = −
= 0 for steady currents.}
∂t
⇒ ∇i E ( r ) = 0 also
But: ∇i E ( r ) = 0 ⇒ −∇ 2V ( r ) = ρ free ( r ) ε o = 0 (i.e. Poisson’s equation ⇒ Laplace’s equation)
⇒ ρ free ( r ) = 0 inside an electrical conductor carrying a steady free current, Ifree.
⇒ Any unbalanced electric charge must reside on the surface of the conductor!!!
Within the conductor (carrying steady free current Ifree): ∇i E ( r ) = 0 or − ∇i∇V ( r ) = ∇ 2V ( r ) = 0
i.e. Laplace’s Equation: ∇ 2V ( r ) = 0
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
5
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Thus for a simple cylindrical conductor of length L and radius R ( L
Prof. Steven Errede
R) :
V ( z = L) = 0
V ( z = 0 ) = +Vo
ΔV = V ( z = L ) − V ( z = 0 ) = −Vo
conventional
I free
R
J free = σ c E ,
I free , J free , E , + zˆ
E = + ( ΔV L ) zˆ
L
∂ 2V ( z )
d 2V ( z )
Then the solution to Laplace’s equation (in one dimension) ∇ V ( r ) →
→
=0
∂z 2
dz 2
with the boundary conditions V ( z = 0 ) = +Vo and V ( z = L ) = 0 has a solution of the form:
2
⎛ L−z⎞
V ( z ) = Vo ⎜
⎟ and ΔV = V ( z = L ) − V ( z = 0 ) = −Vo
⎝ L ⎠
dV ( z )
V
zˆ = + ⎛⎜ o ⎞⎟ zˆ uniform (=constant)
The corresponding electric field is: E ( z ) = −∇V ( z ) = −
dz
⎝ L⎠
J free ( z ) = σ c E ( z ) = +
The free current density is thus:
σ cVo
L
zˆ = +
σ cVo
L
zˆ
See J.D. Jackson, Am. J. Phys. Vol. 64, p. 855 (1996) for details of calculating the surface charge
distribution ρ ( r = R ) that is associated with producing the longitudinal E -field.
ΔV
I free , J free , E
+
•
−
•
cylindrical electrical conductor (wire)
with cross sectional area A⊥ = π R 2
L
+
−
Ideal wire
Ideal wire
Battery
(Ideal voltage source)
I free ≡ I conventional
Note that: I e− = I physical
= − I conventional
6
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Ohm’s Law: I = ΔV R
Fall Semester, 2007
R=
ρc L
A⊥
Lecture Notes 21
Prof. Steven Errede
for a straight wire of length L
and cross sectional area A⊥
ΔV
I=
ρc L
A⊥
with ρ c =
⎛σ A ⎞
I = ⎜ c ⊥ ⎟ ΔV
⎝ L ⎠
I = GΔV
Conductance, G =
σ c A⊥
L
1
σc
ρ c = resistivity ( Ω − m )
or σ c = 1
−1 −1
ρc where: σ c = conductivity ( Ω m )
( = Siemens m )
Define conductance, G ≡ 1 = 1
resistance
R
SI units of conductance = Siemens =
Amps
Volt
for a straight wire of length L and cross sectional area A⊥
All the physics (at microscopic scale) is contained within resistivity ρc ( Ω − m ) (or equivalently,
conductivity, σ c (Siemens/m) ).
⇒ free electrons inside a conducting wire are accelerated by longitudinal E -field:
−eE
F = qE = −eE = ma ⇒ ae =
me
However, electrons elastically scatter off of the atoms inside the conductor – loose kinetic energy to
the atoms (which ultimately winds up as heat) – electrons drift through conductor with an average
terminal velocity of:
1
vD = a τ
2
where τ = mean/average time between successive collisions =
λmfp
vthermal
and λmfp = mean free path (average distance between successive collisions)
vthermal = average thermal speed of electrons
Thus:
vD =
λmfp
1
1
a τ = a
2
2 vthermal
If there are n atoms/molecules per unit volume and a fraction f of free electrons per atom/molecule
Then: J free = ( nf ) q vD = nq q vD
where nq = number density of free electrons = # of free electrons/unit volume = nf
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
7
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
⇒ Conventional current density:
J free = nq q vD
nq = nf =
# of free electrons
unit volume
n=
# of atoms/molecules
# of free electrons
f =
unit volume
atom/molecule
q = charge of conventional charge carrier = +e
vD = drift/terminal/average velocity =
aconventional =
λmfp
1
a
2 vthermal
qconv E +eE
=
me
me
λmfp
λmfp q 2
1
1
= nq
E
∴ J free = nq q vD = nq q a
2 vthermal
2 me vthermal
Macroscopic power dissipated as heat in a conductor:
Joule Heating Law: Power, P = ΔVI = I 2 R = ΔV R (SI Units: Watts = Joules/sec)
2
1 Watt = 1 Volt-Amp = 1 Amp2 – Ohm = 1
Volt 2
= 1 Joule/sec
Ohm
Heat dissipated in conductor is known as Joule heat.
The electrical conductivity σ c (or equivalently the resistivity ρc = 1 σ c ) are intimately connected to
the heat capacity CV (at constant volume) of the conducting material.
Statistical mechanics (microscopic physics) CV ~ T2 (T = absolute temperature)
ρc ~ CV ~ T 2 also! For e.g. conducting metals (copper, etc.) (See UIUC P401 conductivity
experiment for copper)
ρc (T )
~T2
T
8
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Comments on the Microscopic Aspects of
Electrical Conduction in Metals, (i.e. good conductors)
In any metal, the “free” electrons exist as a “gas” of electrically charged particles inside the
bulk material of the conductor. If the entire block of metal is in thermal equilibrium with its
surrounding local environment, then the “free” electron “gas” is also in thermal equilibrium.
The equipartition (of energy) theorem says that for each degree of freedom associated with such
particles, there is associated 12 k BT of thermal energy associated with it (in equilibrium), where
Boltzmann’s constant, k B = 1.381 x 10-23 Joules/Kelvin degree.
A “free” electron in a metal has 3 kinetic degrees of freedom - one each for vx , v y and vz
(the xˆ , yˆ and zˆ - components of the free electron’s mean (or average) thermal velocity, ve )
Then the mean (or average) kinetic energy of a free electron is: KEe = 12 me ve2
ve2 = ve ive = vx2 + v y2 + vz2
or: KEe = 12 me vx2 + 12 me v y2 + 12 me vz2
KEe = 12 me vx2 + 12 me v y2 + 12 me vz2 = KExe + K ye + KEze
By the equipartition theorem:
⇒ KEe = Ethermal = Exthermal + E ythermal + Ezthermal = 12 k BT + 12 k BT + 12 k BT = 32 k BT
∴
1
2
me ve2 = 32 k BT
The average thermal speeds of “free” electrons in metals are therefore: vethermal =
3k BT
me
Thus, for T = 300K ( room temperature) and me = 9.1 x 10−31 kg, typical thermal electron
speeds at room temperature are:
vethermal (T
300 K ) =
3 ×1.381×10−23 × 300
9.1×10−31
1.17 ×105 meters
sec
~ 1.2 ×105 meters
sec
However, as we have discussed previously, this is not the speed at which electrons conduct
electrical power e.g. when flowing through a wire.
The “free” conduction electrons scatter off of atoms (and each other) in the metal, with a mean
free path, λmfp ≡ average distance between successive collisions / scatterings. The average time
between successive collisions / scatterings is:
τc =
λmfp
vethermal (T )
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
9
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Free electrons inside a conducting metal wire with a potential difference ΔV across the ends of
the wire are accelerated by the resulting E -field:
F = qE = −eE = ma ⇒ ae =
−eE
me
But because the electrons scatter off of the atoms (and each other) inside the conductor, they
(rapidly) lose this kinetic energy, they get re-accelerated, gaining kinetic energy again, then loose
it in another scattering, this process repeats itself over and over....
Thus electrons achieve an average drift/terminal velocity of:
vD =
λmfp
1
1
a τ c = a thermal
2
2 ve
(T )
Imagine now that we have a “co-moving” gas of free electrons drifting down a conducting wire
with mean drift velocity vD . The number density of free electrons # 3 is ne . The steady,
m
macroscopic free current Ifree is due the collective flow of these free electrons moving as a gas
through the conductor.
( )
Let’s consider a “block” of this co-moving free electron gas as it flows along a wire. The length
L of this “block” of co-moving free electron gas is the distance the block moves as a unit in one
where Δt = 1 second.
second. The drift velocity vD = L
Δt
L = distance block of
co-moving electron
gas moves in 1 second
Now the steady, macroscopic free current that flows in the wire Ifree is defined as the amount of
free electric charge crossing an imaginary boundary (see figure above) per unit time interval:
ΔQ freee ΔN q
=
where ΔN q = number of free electric charges (each with charge q)
I free =
Δt
Δt
Now
i.e.
or:
10
ΔN q
Δt
= rate of free electrons crossing an imaginary boundary (see figure above)
ΔN q
dN q ( t )
=
thus:
I free ( t ) =
dQ free ( t )
qdN q ( t )
dN q ( t )
= qRq ( t )
dt
dt
dt
Δt
dt
dN q ( t )
I (t )
=
= # free electrons crossing imaginary boundary/second
Rq ( t ) = free
q
dt
Rq =
=
=q
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
If the number density of free electrons in the wire is ne ( # m3 ) then the volume of the block of
co-moving charges that moves a distance L per second as a group / en-masse is: V = A⊥ * L
Contained within this volume are ΔN q free electrons. Then ne = ΔN q V
Thus: ne =
ΔN q
V
=
ΔN q
A⊥ L
= ne =
( ΔN
q
Δt )
A⊥ ( L Δt )
=
Rq
A⊥ vD
=
I free
A⊥ q vD
Now the free current density is: J free = I free A⊥
∴ ne =
J free
q vD
J free = ne q vD
or:
Let us calculate the free electron number density ne for e.g. copper:
Avogadro’s number NA = 6.022 x 1023 atoms per gram-mol
Copper has an atomic weight of ACu = 63.54
i.e. → 63.54 grams of copper = 6.022 x 1023 atoms of copper
N
6.022 ×1023
∴ 1 gram of copper contains ηecu ≡ A =
= 9.477 ×1021 copper atoms
63.54
ACu
Nominally, atomic copper has an electron valency of 1, thus in a copper metal, for every copper
atom, there is associated with it one “free” electron. Thus, in one gram of copper, there are
9.477 x 1021 free electrons.
Thus: ηecu = 9.477 ×1021 free electrons / gram of copper, i.e. ηe ≡
NA
A
The mass density of copper is ρcu = 8.95 grams / cm3
Then the number density of free electrons in copper is given by:
necu = ηecu ∗ ρcu = 9.477 ×1021 electrons
∗ 8.95 grams 3
gram
cm
−2
cu
22
3
ne = 8.482 ×10 electrons cm
1 cm = 10 m → 1 cm3 = 10−6 m3
necu = 8.482 ×1028 electrons m3
Suppose we have a copper wire carrying a steady free current of Ifree = 1 Ampere, the copper
wire e.g. has a cross-sectional area of A⊥ = 1 mm2 = 10−6 m2.
Then:
vD =
J free
ne q
=
I free A⊥
ne q
q = 1.602 x 10−19 Coulombs
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
11
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
1.0 10−6
vD =
= 7.359 ×10−5 m
= 73.6 μ m
28
−19
sec
sec
8.482 ×10 ∗1.602 ×10
1.17 ×105 m
vthermal
μ
m
s 109 !!!!
vD = 73.6
≈
Note:
−5 m
sec !!!!
vD
7.36 ×10
s
In one hour ( Δt = 60 sec/min x 60 mm/hr = 3600 sec) free electrons drifting in a copper
conductor will have traveled a distance of: ΔL = VD Δt = 0.265 m = 26.5 cm ≈ 10.5 inches !!!!
This spectacularly slow drift velocity vD ~ 75 μ m sec doesn’t square at all with our everyday
experience, e.g. of turning on a light – which is incredibly fast – essentially instantaneous, on
human-perceived time-scales (e.g. ~ a few milliseconds) (fast LEDs – light emitting diodes – can
turn on/off in a few nano-seconds (10−9 sec)).
Note that vD ~ 75 μ m sec << c = 3 x 108 m/s (speed of light in a vacuum)
Typical electrical signals in conducting wires can easily propagate at ~ 50% speed of light c
i.e. v prop 1.5 × 108 m s
Thus vD does not correspond in any way to vprop!!!
Now, from the continuity equation
But:
But:
∴
∇i E ( r ) =
εo
∂ρ free ( r )
∂t
∴ ∇i J free ( r ) = σ c ∇i E ( r ) = −
J free ( r ) = σ c E ( r )
1
∇i J free ( r ) = −
ρTOT ( r ) =
1
εo
∂ρ free ( r )
∂t
ρ free ( r ) (here)
∂ρ ( r , t )
σc
ρ free ( r , t ) = − free
⇐ 1st order linear differential equation
∂t
εo
Solution is:
ρ free ( r , t ) = ρo ( r , t = 0 ) e−t (ε
o
σc )
where ρ o ( r , t = 0 ) = initial free electron charge density at time t = 0.
Define charge relaxation time constant:
τ r ≡ ε o σ = time free charge density ρ free ( t ) falls to 1/e = e−1 = 0.368 of its initial value at t = 0.
c
Conductivity for copper: σ
cu
c
−1
= 5.8 ×10 Ω m
7
−1
A m2 A
(= Siemens) =
=
=
V −m
E
V m
J free
ε o = 8.85 ×10−12 Farads m = C V − m
C V −m
C
ε
S.I. Units of τ r : ⎛⎜ o ⎞⎟ =
=
= sec
⎝ σc ⎠ A V − m C s
12
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
8.85 ×10−12
= 1.526 ×10−19 sec
∴ τ ( theory ) =
7
5.8 ×10
cu
r
Compare τ rcu to that for e.g. iron: τ rFe
Lecture Notes 21
Prof. Steven Errede
Very fast charge relaxation time!!!
10−18 sec , water (poor conductor) τ rH 2O ~ 10−8 sec ,
and glass (insulator) τ rglass ~ 2 sec
The measured value of the charge relaxation time for copper is actually τ rcu ( meas )
4 ×10−14 sec
(still very fast), but differs by ~ 4-5 orders of magnitude from the theoretical prediction – eeek!!!
{The reality is: conductivity, σ c is frequency-dependent - i.e. σ c = σ c ( f ) ≠ σ c ( f = 0 Hz ) .}
∴σ c itself is a function of the charge relaxation time of material τ r because frequency f = 1
τ
where τ = period of oscillation.
Finally, let us consider the mechanical kinetic energy / second = mechanical power associated
with our “block” of co-moving drift electrons in the copper wire:
ΔN q
1
2
= rate of electrons
Each free electron has kinetic energy KEe = me vD and Rq =
Δt
2
I free
crossing imaginary boundary = # electrons crossing imaginary boundary per second =
q
∴ Mechanical power associated with a “block” of co-moving electrons in a copper wire of cross
sectional area A⊥ = 1 mm2 and carrying a steady free current Ifree = 1 Ampere is:
⎛ I free ⎞ 1
cu 2
PCue = Rq * KEe = ⎜
⎟ me vD
mech
⎝ q ⎠2
1.0
⎛1⎞
−31
−6 2
−20
Pcue =
⎟ 9.1×10 × ( 73.6 ×10 ) = 1.54 ×10 Watts!!
−19 ⎜
1.602 ×10 ⎝ 2 ⎠
mech
Thus, it is clear from this (albeit crude/simplistic) calculation that the electrical power
transported in conducting wires (Watts → 100’s Watts → Kilowatts and more) very definitely is
NOT associated with the mechanical kinetic energy of the co-moving, “glacial-speed”
( vD ~75 μm/sec) conduction electrons!!!
vprop = characteristic speed electrical signals propagate along a wire: v prop
50% c = 1.5 × 108 m / s
1
3
If we consider the energy density in the electric field: uE ( r ) = ε E 2 ( r ) ≈ ε o E 2 ( r )
2
2
J (r )
E ( r ) = free
and we assume ε cu ≈ 3ε o (typical value ??)
σc
1
The power in the static E-field is: Pem = ε E 2 A⊥ v prop
2
2
3 J free
3
I2
ε o 2 A⊥ v prop = ε o 2 2 A⊥ v prop
2 σc
2 σ c A⊥
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
13
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
3
I2
12
−12
Pem = ε o 2 v prop = 1.5 ∗ 8.85 ×10
1.5 ×108 Watts
2
7
6
−
2 σ c A⊥
( 5.8 ×10 ) (10 )
Pem ≈ 5.9 ×10−13 Watts, ≈ 3.84 x 107 larger than Pcue kinetic power, but still surprisingly very
vD ).
small!! (roughly speaking, the main difference is due to v prop
n.b. if ε → ∞ then Pem → ∞ So what does carry the EM power in a wire??
A wire carrying a steady free current Ifree also has a magnetic field associated with it. If we
assume (for simplicity) that the wire has a circular cross-section A⊥ = π a 2 , then Ampere’s
Circuital Law
∫
C
encl
Bid = μo ITot
= μo I encl
free gives:
2
⎛ρ⎞
μo I free ⎜ ⎟
⎝ a ⎠ ϕˆ = ⎛ μo
Bin ( ρ ≤ a ) =
⎜
2πρ
⎝ 2π
Bout ( ρ ≥ a ) =
⎞
⎛ρ ⎞
⎟ I free ⎜ 2 ⎟ ϕˆ
⎝a ⎠
⎠
A⊥ = 1 mm 2 = π a 2
a
μo I
ϕˆ
2πρ
ρ
A⊥
⇒a=
ρ = x 2 + y 2 in cylindrical coordinates
J free =
π
= 0.564 mm
I free
A⊥
= uniform
The magnetic energy density for copper wire (assumed non-magnetic, i.e. μcu = μo ) is:
u
in
mag
2
1 ⎛ μo ⎞ 2 ⎛ ρ ⎞
B ( ρ ≤ a) =
( ρ ≤ a) =
⎜
⎟ I free ⎜ 2 ⎟
2 μo
2 μo ⎝ 2π ⎠
⎝a ⎠
1
2
2
in
2
2
1 2
1 ⎛ μo ⎞ 2 ⎛ 1 ⎞
u ( ρ ≥ a) =
Bout ( ρ ≥ a ) =
I free ⎜ ⎟
2 μo
2 μo ⎜⎝ 2π ⎟⎠
⎝ρ⎠
The power associated with the magnetic field is:
( dA⊥ = 2πρ d ρ )
out
mag
Pmag = v prop ∫ umag ( ρ )idA⊥ = v prop ∫
ρ =a
ρ =0
in
umag
( ρ )idA⊥ + v prop ∫
ρ =∞
ρ =a
out
umag
( ρ )idA⊥
2
∞ 1
⎤
1 ⎛ μo ⎞ 2 ⎡ a ρ 3
= v prop i
I ⎢∫ 4 d ρ + ∫
dρ⎥
⎜
⎟
a ρ
2μo ⎝ 2π ⎠
⎣ 0 a
⎦
μ
⎡1
⎛∞⎞
⎛ ∞ ⎞⎤
= o2 I 2 v prop ⎢ + ln ⎜ ⎟ ⎥ ← the ln ⎜ ⎟ term is (logarithmically) singular!!
8π
⎝a⎠
⎝ a ⎠⎦
⎣4
We shouldn’t be terribly surprised by this infinite-power result, since (for simplicity’s sake) we
used the magnetic field associated with a steady current flowing in an infinitely long wire, which
requires infinite energy to create/assemble. For a finite-length wire, we would get a finite result!
So let’s assume (here) (again for simplicity) that ∃ an effective radial cut-off distance, Λ such
⎛Λ⎞
that e.g. ln ⎜ ⎟ = 100. Then:
⎝a⎠
Λ
⎛Λ⎞
ln ⎜ ⎟ = 100
= e100
Λ = ae100 ≈ 1.5 ×1040 m i.e. Λ > diameter of universe!!
a
⎝a⎠
14
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Then Pmag ≈ 100
Then Pmag
Then: PTOT
μo 2
I free v prop
8π 2
Fall Semester, 2007
μo = 4π ×10−7 N
A2
Lecture Notes 21
I free = 1A
v prop
Prof. Steven Errede
1.5 × 108 = 12 c m/s
1
× 10−7 i 1×1.5 ×108 = 2.39 ×102 Watts!!
2π
e
= Pcu + Pem + Pmag = 1.54 ×10−20 + 5.9 ×10−13 + 2.4 ×102 Watts
100 ×
mech
PTOT ≈ 2.4 × 102 W !! (but depends on cut-off parameter Λ )
ΡTOT 240W formally, PTOT (calculated) is divergent (i.e. infinity) because Pmag is divergent!
Again, the formally divergent result arises simply because we used the B-field associated with a
steady free current flowing in an infinitely long wire.
Thus, B wire carries the vast bulk of electrical power for wires carrying steady/DC currents!
n.b. We have also neglected a very thin layer of static electric charge residing on the outer
surface of the wire that stabilizes the internal E = J free σ c . For real, matter wires this very thin
(positive) electric surface charge has its own radial E -field associated with it, and thus also has
1
Q
an energy density term uem
( ρ ≥ a ) = ε o EQ2 ( ρ ≥ a ) associated with it. Nominally this is a static
2
energy density. We shall see (next semester, in P436) that both this radial electric field and the
azimuthal magnetic field of the wire are in fact primarily responsible for the transport of
electrical power down a physical wire – both fields are needed for this!
The Electromotive Force
ε (aka EMF) – (aka Electromotance)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
15
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
∃ Two forces (per unit charge) driving a free current, Ifree around an electrical circuit:
F
1) the source of electrical power (e.g. a battery, or a power supply) f s = s
and:
Q
2) the electrostatic force, f E = Fs Q = E , which serves to smooth out the flow of current
(dynamically changing via electrostatic charges in the circuit) and to communicate the
influence of the source to distant parts of the circuit.
Then using the principle of linear superposition: fToT = f source + f E = f s + E
Thus we define the electromotive force as: ε ≡
ε≡
∴ ε=
∫
C
∫
C
fToT id =
∫
C
f s id +
∫
C
∫
C
fToT id
( ∫ E id ≡ 0 for electrostatic fields)
E id
C
f s id = contour line integral of force/unit charge (electric field) around closed contour, C
Here, C = closed contour around electrical circuit.
For an ideal battery (i.e. one with no internal resistance) = ideal source, the net/total force on
electrical charges = 0
ideal
= 0 ⇒ E = − fs
Then: fToT
B
B
A
A
The potential difference is ΔV ≡ VB − VA = − ∫ E id = + ∫ f s id
But f s = 0 outside of the source (i.e. here, the source = the battery), so
∫
B
A
f s id =
∫
c
f s id = ε
B
∴ ΔV = ε = − ∫ E i d
A
Sources of
Force
= f source = Esource
Coulomb
e.g. an ideal battery, connected to a simple circuit:
16
physical electric field
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
b
ε = +∫ E
a
source
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
id = ΔV ≡ Vb − Va
+ sign for sources of ε mf (electromotance/electromotive forces)
(e.g. arising from magnetic induction/motional induction)
For physical electric fields obeying E phys = −∇V :
b
ΔV = − ∫ E phys id ≡ Vb − Va
a
Motional EMF’s
I free =
ε
R
=
ΔV vo Bo L
=
R
R
An “ideal” conducting rod of length L moves with uniform velocity v = vo xˆ in a uniform external
magnetic field Bext = Bo zˆ with the axis of the conducting rod ŷ axis as shown in the figure above.
Here:
f source = f mag =
Fm
= v × Bext = vo xˆ × Bo zˆ = vo Bo ( xˆ × zˆ ) = vo Bo ( − yˆ ) = −vo Bo yˆ
Q
=− yˆ
Now f mag acts on a +ve charge (Q > 0) in the − ŷ direction
(by the right-hand rule of cross product)
→ +ve charge builds up at point B
∴ −ve charge builds up at point A
→ A potential difference exists (= EMF) across the
ends of the moving rod!!
B
EMF, ε = ΔV = VB − VA = ∫ f mag id = +vo Bo L Volts
A
If the rod, moving with velocity v = vo xˆ slides on “ideal” conducting rails to which a resistor of
ε ΔV vo Bo L
=
resistance R is connected (e.g. at the end of the rails), then a free current I free = =
R
R
R
will flow in the completed electrical circuit as long as the rod continues to move, and remains in a
⊥ B -field.
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
17
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
The instantaneous power generated is:
2
(v B L)
⎛ v B L ⎞ ΔV
= o o
P = ε I = ΔVI = ( vo Bo L ) ⎜ o o ⎟ =
Watts (Joules/sec)
R
R
⎝ R ⎠
2
An alternative way of thinking about this to consider e.g. a rectangular-shaped loop (n.b. the
detailed shape of the loop actually doesn’t matter here!) entirely immersed in an external magnetic
field Bext = Bo zˆ which again is perpendicular to the plane of the rectangular loop:
Cross-Sectional area of rectangular loop = Aloop = L x W (meters2)
ˆ = Bo A
Magnetic flux through the loop = Φ m = ∫ B ida = ∫ B inda
S
S
Φ m = Bo A = Bo LW (Tesla–m2 = Webers)
Now if one side of the rectangular loop is movable and moves such that the cross-sectional area
Aloop = LW of the conducting loop decreases uniformly with time, e.g. the side of length W on the
LHS of the loop moves with velocity v = vo yˆ
L
W
v = vo yˆ
Then the time rate of change of the area of the loop is:
dAloop d
dL
= ( LW ) = W = −voW
( n.b. the − sign indicates that the area decreases with time)
dt
dt
dt
The magnetic flux Φ m through the loop also decreases with time, as the area of the loop decreases:
dA
d Φm d
= ( Bo Aloop ) = Bo loop = − Bo voW = −vo BoW
dt
dt
dt
constant
Φ m = Bo Aloop thus:
This is the same result as we obtained in previous example with the moving rod of length L !!!
(In our present example, the moving wire has length W)
18
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Thus, we see that a conducting loop whose area is changing with time, which is linked by magnetic
field lines/magnetic flux also creates an EMF ε :
d Φm
= + Bo voW (here)
EMF, ε = −
dt
In general, an EMF, ε will be produced whenever
d Φm
≠0
dt
d Φm
= Volts.
dt
B ida so either B = fcn (t) and / or: da = fcn (t) can/will produce
Note that the SI Units of ε = −
Since magnetic flux Φ m = ∫
Φ m ( t ) = fcn ( t ) and thus:
ε =−
d Φm
d
=−
dt
dt
S
( ∫ Bida ) = − ⎢⎢∫ ⎜⎜⎝ dBdt(t ) ⎟⎟⎠ida + ∫ B (t )i dadt ⎥⎥
⎡ ⎛
S
⎣
⎤
⎞
S
S
changing magnetic field
(geometry is fixed / constant)
⎦
changing geometry (motional effect)
( B is fixed / constant)
Another example of a motional EMF, similar to the first example, is that of a conducting metal
rod (stiff wire) of length L rotating in the horizontal x-y plane at constant angular velocity ω in an
applied external uniform magnetic field ⊥ plane of rotation, Bext = Bo zˆ :
L
ω = ω ẑ
v =ω×r
0≤ r ≤
2
v = ω × r = ω zˆ × rrˆ
v = ω rzˆ × rˆ = ω r sin Θϕˆ
zˆ = cos θ rˆ − sin θθ in spherical coordinates
Θ =θ =
cos
π
π
2
here
=0
and
2
∴ ẑ = −θ here
⎛
⎞
v = ω r ⎜ −θ × rˆ ⎟ = ω rϕˆ
⎝
⎠
EMF, ε = ΔV = ∫ f mag id
f mag =
What is f mag ?
sin
π
2
=1
= +ϕ by right-hand rule
F mag
= v × B ext
Q
=0
Note that:
v ( r = 0 ) = 0 = ω r ϕˆ ⇒ f mag ( r = 0 ) = 0
Note also:
v (r =
L
2
) = 12 ω Lϕˆ
⇒ f mag ( r =
L
2
) ≠ 0 = 12 ω LBo (ϕˆ × zˆ )
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
19
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
But ẑ = −θ here and thus ϕˆ × ẑ = ϕˆ × −θˆ = + rˆ by the right-hand rule. Thus f mag ( r =
L
2
) = 12 ω LBo rˆ
∴ at a radial point r along the rod ( 0 ≤ r ≤ L2 ) : f mag ( r ) = ω rBo rˆ ← pushes +ve charge to the outer
ends of rod, and leaves –ve charge at the center of rod.
“Snapshot” Top View:
d = drrˆ
B
EMF, ε = ΔV = VB − VA = ∫ f mag id
A
B
ε = ∫ ω rBo rˆidrrˆ
rˆirˆ = 1
A
L
L
ε = ∫ 2 ω rBo dr = ω Bo ∫ 2 rdr
0
1
2
0
L
ε = ω Bo r 2
0
2
1
= ω Bo L2 (Volts)
8
If the conducting rod rotating in the external magnetic field ( ⊥ plane of rotation) is connected
up to an external circuit, it becomes a power source (but n.b. mechanical power input is required to
keep it running!!) then a free current Ifree flows in the circuit:
Bext = Bo zˆ
v = ω rϕˆ (in x-y plane)
Rotationally-induced ε mf : ε = ΔV = VB − VA = V ( r =
Rotationally-induced free current: I free =
L
2
) − V ( r = 0) =
1
ω Bo L2 Volts.
8
ΔV ω Bo L2
=
Amps, flowing through external load
R
8R
resistor, R.
Pmech = mechanical power required to keep device rotating = electrical power P flowing in circuit
ΔV 2 (ω Bo L )
R=
=
R
64 R
2 2
P = Pmech = ΔV i I free = I
20
2
free
(Watts = Joules/sec)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Griffiths Example 7.4: Consider a conducting metal disk of radius a rotating with
constant angular velocity, ω = ω ẑ in a uniform externally applied magnetic field Bext = Bo zˆ .
Then f mag ( r ) =
Fmag
Q
= v ( r ) × Bext with v ( r ) = ω rϕˆ (from above)
Thus: f mag ( r ) = ω rBo rˆ . Then the ε mf , ε = ΔV = VB − VA = V ( r = a ) − V ( r = 0 )
B
ε = ∫ f mag ( r )id
A
and d = drrˆ
a
a
0
0
ε = ∫ ω rBo dr = ω Bo ∫
a
1
1
1
rdr = ω Bo r 2 = ω Bo a 2 . Thus: ε = ω Bo a 2 Volts
2
2
2
0
Again, if this device (source) is connected to an external circuit: I free =
ε
R
=
ΔV ω B0 a 2
=
Amps
R
2R
Electrical power flowing in this circuit = mechanical power required to keep device rotating:
P = Pmech = ΔV i I free = ε I free = I
2
free
2
ΔV 2 (ω Bo a )
R=
=
R
4R
2
(Watts) (= Joules / sec)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
21
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
⎛ v × B ext ⎞
Thus, we can (literally) view the ⎜
⎟ = E ′ as perceived / “observed” by a moving electric
⎝ Q ⎠
charge Q (with relative velocity v w.r.t. the rest frame in which B ext is defined) as an electric
field E ′ that this electric charge Q interacts with!!
i.e. E ′ =
22
v × Bext
or: Fmag = QE ′ = Qv × B
Q
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Electromagnetic Induction
Faraday’s Law
In 1831 Michael Faraday reported on a series of experiments he carried out:
1) Pulled a loop of wire through a ⊥ B -field, causing a current to flow in an externallyconnected circuit.
Move wire loop to right
2) Moved magnetic field (i.e. moved magnet) holding the loop of wire still, causing the same
current to flow in the externally-connected circuit.
Move magnet to left
3) Changed strength of magnetic field as a function of time keeping both wire loop and magnet
fixed. Again current flowed in externally connected circuit.
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
23
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Faraday’s explanation for the common physics origin of these 3 experiments:
A changing magnetic field induces an electric field
Faraday’s Law in Integral Form:
EMF = ε =
For the electrostatic case:
∂B ( r )
d Φm
= −∫
ida⊥ (if a⊥ = constant vector)
∂t
dt
= 0 ⇒ EMF = ε = ΔV = 0
∫ E ( r )i d
∫ E ( r )i d
=−
C
C
Faraday’s Law in Differential Form: ∇ × E ( r ) = −
∂B ( r )
∂t
{n.b. Obtained from
Faraday’s law in integral
form using Stoke’s theorem.}
Example:
A circular conducting loop of radius a is cut at one point (ϕ = 0 ) . Magnetic flux through this
loop is provided by Bext = Bzˆ as shown in the figure below:
The loop is connected
to an external circuit
(e.g. a single resistor)
If Bext is sinusoidally
time-dependent
e.g. Bext ( t ) = Bo sin ωtzˆ
The magnetic flux through the loop is: Φ m = Bi Aloop = π a 2 Bo sin ωt since (here): Aloop = π a 2 zˆ
The {time-dependent} EMF ε ( t ) induced in the loop is:
(
)
d B ( t )i Aloop
d Φm (t )
d
=−
= − (π a 2 Bo sin ωt ) = −π a 2ω Bo cos ωt
dt
dt
dt
2
ε ( t ) = −π a ω Bo cos ωt Volts
ε (t ) = −
ε (t )
=−
π a 2ω Bo
(π a ω B )
R=
cos 2 ωt
The free current flowing in circuit is: I free ( t ) =
R
R
cos ωt Amps
The instantaneous power flowing in the circuit is:
P ( t ) = ε ( t ) ⋅ I free ( t ) =
24
ε 2 (t )
R
=I
2
free
(t )
2
o
R
2
( ≥ 0)
Watts (= Joules/sec)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Another Example:
A conducting loop of radius a rotates about the ŷ -axis at an angular frequency ω in a fixed/
constant / uniform magnetic field Bext = Bo zˆ as shown in the figure below:
The induced EMF in the rotating loop is:
d Φm (t )
d
ε (t ) = −
=−
Bi A⊥
dt
dt
⎛ dA ( t ) ⎞
= − Bi⎜ ⊥ ⎟ where
⎜ dt ⎟
⎝
⎠
nˆ ( t ) = zˆ cos ωt − xˆ sin ωt
(
)
B = Bo zˆ = constant, but A⊥ ≠ constant
A⊥ ( t ) = A⊥ nˆ ( t ) = π a 2 nˆ ( t )
(for ω = −ω ŷ )
dnˆ ( t )
= −ω zˆ sin ωt − ω xˆ cos ωt
dt
dnˆ ( t )
= − Bo zˆ iπ a 2 [ −ω zˆ sin ωt − ω xˆ cos ωt ] = +π a 2ω Bo sin ωt
Thus: ε ( t ) = − Bi A⊥
dt
Then:
Induced EMF in circuit: ε ( t ) = +π a 2ω Bo sin ωt (Volts)
n.b. This is the same result as we obtained in the previous example - except for an arbitrary phase
sin ωt
shift: − cos ωt
If we cut the loop at one point and connect it up to external circuit with single load resistor (as
before), then:
ε ( t ) π a 2ω Bo sin ωt
=
Free current flowing in the circuit: I free ( t ) =
(Amps)
R
R
Instantaneous power flowing in the circuit:
P ( t ) = ε ( t ) ⋅ I free ( t ) =
ε 2 (t )
R
= I 2free ( t )
(π a ω B )
R=
2
o
R
2
sin 2 ωt
( ≥ 0)
(Watts = Joules/sec)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
25
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Lenz’s Law: An induced free current in an electrical circuit will flow in such a direction so that
the magnetic flux this current produces (tries to) cancel the change in external magnetic flux.
e.g. If the external magnetic flux is increasing through a conducting loop, i.e. d Φ m dt > 0, then
the induced EMF ε generated causes a free current Ifree to flow in the loop, which in turn whose
own magnetic flux opposes the increasing external magnetic flux, to try to keep the total
magnetic flux through the conducting loop constant. Note that normal conducting materials can
never actually achieve this, whereas superconducting loops can!
Lenz’s Law:
Maintain the magnetic flux status quo – do not allow a change in the magnetic flux linking an
electrical circuit (oppose any change to keep magnetic flux constant).
Griffiths Example 7.5 A long cylindrical rod magnet of length L and radius a ( L
a ) has
uniform magnetization Μ = Μ o zˆ parallel to its symmetry axis. The rod magnet passes at constant
velocity v through a circular conducting ring of radius a + δ a as shown in the figure below:
What is the induced EMF ε ( t ) vs. time t?
26
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
The magnetic field of rod magnet = magnetic field of long solenoid of length L with effective
surface bound current K bound = Μ × nˆ surface = Μ oϕˆ (in cylindrical coordinates) and Bin μo Μ
(except near ends where it spreads out (radially)).
When the long rod magnet is infinitely far from the loop, the magnetic flux through the loop
Φ m = Bi Aloop = 0 . As the rod magnet approaches the loop, the magnetic flux threading the loop
μo Μ oπ a 2 (neglect δ a ) and then drops back to zero as the
builds up to a maximum of Φ max
m
long rod magnet again becomes infinitely far from the loop:
d Φm (t )
dt
= −ve of slope of Φ m ( t )
Induced EMF ε ( t ) = −
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
27
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
The Induced Electric Field
Faraday’s Law (in differential form) ∇ × E ( r , t ) = − ∂B ( r , t ) ∂t together with Coulomb’s Law
(in differential form) ∇i E ( r , t ) = ρTOT ( r , t ) ε o tells us that there are two kinds of electric fields:
1) Electrostatic field Estatic due to (static) electric charges (or charge distributions)
(i.e. observer with test charge QT at rest WRT charge distributions producing Estatic )
2) Induced electric fields Einduced due to time varying magnetic fields and/or motional effects
(observer and source rest frames in motion relative to one another):
Time Varying
Magnetic Fields:
∇ × E (r ,t ) = −
EMF ε ( t ) =
∫
C
∂B ( r , t )
Φ m ( t ) = ∫ B ( r , t )idA⊥ ( r )
S⊥
∂t
E ( r , t )i d = −
∂Φ m ( t )
∂ B ( t ) fixed
=−
i A⊥
∂t
∂t
Φ m ( t ) = ∫ B ( r )idA⊥ ( r , t )
S⊥
Relative Motional Effects:
EMF ε ( t ) =
∫ E ( r , t )i d
C
=−
∂Φ m ( t )
∂A ( t )
= − B fixed i ⊥
∂t
∂t
∂B ( r , t )
(Faraday’s Law)
∂t
But B ( r , t ) arises from some kind of time varying current distribution, i.e. via Ampere’s Law:
For time-varying magnetic fields: ∇ × E ( r , t ) = −
∇ × B ( r , t ) = μo J TOT ( r , t ) , along with ∇i B ( r , t ) = 0 (no magnetic charges)
(we need both
divergence & curl to determine/uniquely specify B ( r , t ) )
For pure time-varying magnetic fields (no electrostatic E-field(s)), Gauss’ Law becomes
∇i E ( r , t ) = 0
All of the formalism associated with using Ampere’s Circuital Law:
also applies to Faraday’s Law in integral form: Induced EMF
28
ε=
∫ B ( r , t )i d
C
∫
C
encl
= μo ITOT
(t )
E ( r , t )i d = −
d Φ encl
m (t )
dt
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Griffiths Example 7.7:
A uniform magnetic field Bext ( r , t ) = Bo ( t ) zˆ points straight up, but is changing with time. It fills
a circular region of radius R. What is the induced electric field E ( r , t ) associated with this
changing B(t)?
∫
Faraday’s “Circuital” Law:
C
Φ encl
m (t ) = ∫
n̂⊥
d Φ encl
m (t )
dt
B ( r , t )ida⊥encl = πρ 2 B ( t ) = B ( t ) A⊥encl
E ( r , t )i d = −
S⊥
da⊥encl = nˆ⊥ da⊥encl = nˆ⊥ ρ ′d ρ ′dϕ where nˆ⊥ Bext ( t ) zˆ
ρ
C
Take contour C such that:
E ( ρ,t ) = −
Einduced
ρ
⎛1 2⎞
2
∫ da = ∫ϕ =0 dϕ ∫ρ ′=0 ρ ′d ρ ′ = 2π ⎜⎝ 2 ρ ′ ⎟⎠ 0 = πρ
2 ∂B ( t )
∫C E ( r , t )id = E ( ρ , t ) 2πρ = −πρ ∂t
d where d = d ϕˆ
encl
⊥
ϕ = 2π
ρ ′= ρ
πρ 2 ∂B ( t )
1 ∂B ( t )
ϕˆ = − ρ
ϕˆ
2πρ ∂t
2
∂t
1 ∂B ( t )
ϕˆ
Thus: E ( ρ , t ) = − ρ
2
∂t
1 ∂B ( t )
Define: Eo ( ρ , t ) ≡ − ρ
∂t
2
If the magnetic field is decreasing with time, then E ( ρ , t ) = + Eo ( ρ , t ) ϕˆ
⎛ ∂B ( t )
⎞
< 0⎟
⎜
⎝ ∂t
⎠
If the magnetic field is increasing with time, then E ( ρ , t ) = − Eo ( ρ , t ) ϕˆ
⎛ ∂B ( t )
⎞
> 0⎟
⎜
⎝ ∂t
⎠
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
29
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
ANOTHER ASPECT OF FARADAY’S LAW
Faraday’s Law: ∇ × E ( r , t ) = −
But:
B (r ,t ) = ∇ × A(r ,t )
Thus: ∇ × E ( r , t ) = −
∴
∂B ( r , t )
∂t
E (r ,t ) = −
∂A ( r , t )
∂A ( r , t )
∂
∇ × A ( r , t ) = −∇ ×
or: ∇ × E ( r , t ) = −∇ ×
∂t
∂t
∂t
(
∂A ( r , t )
∂t
)
(Electrodynamics only)
Including electrostatics: ∇ × E ( r ) = 0 ⇒ E ( r ) = −∇V ( r )
E ( r , t ) = −∇V ( r ) −
∂A ( r , t )
∂t
( ∇ × ∇V ( r ) ≡ 0 always )
Recall that the physical SI units of the magnetic vector potential A are:
Newtons
Momentum/Coulomb = kg m/sec/Coulomb = Tesla-meters =
Ampere
From the fundamental definition of force: F =
But: F
Q
=E=
dP
= time rate of change of momentum
dt
1 dP
∂A
=−
Q dt
∂t
From Lenz’s Law
Thus: ε mf ε ( t ) = ΔV ( t ) =
∫ E ( r , t )i d
But we have already seen that Φ m ( t ) =
∴
30
ε mf ε ( t ) = ΔV ( t ) =
= −∫
C
∂A ( r , t )
∂
id = −
∂t
∂t
∫ A ( r , t )i d
∫ E ( r , t )i d
C
C
C
=−
∂
∂t
(∫
C
∫ A ( r , t )i d
C
!!!
A ( r , t )i d
) = − ∂Φ∂t (t )
m
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Magnetic Induction
Example: A long wire carries a quasi-static, but time-dependent free current, I free ( t ) . From
Phys. 435 Lecture Notes 16, the magnetic vector potential A ( r ) associated with long straight
wire of length L is:
⎛μ
A(r ) = ⎜ o
⎝ 4π
( ) ⎭⎪⎬⎪⎥⎥⎦ zˆ
⎡⎛ L ⎞ ⎧⎪
⎞
ρ
⎟ I free ln ⎢⎜ ⎟ ⎨1 + 1 + L
⎠
⎢⎣⎝ ρ ⎠ ⎩⎪
d ′ ⎛ μo
⎞
=⎜
⎟ I free ∫C
r ⎝ 4π
⎠
2
⎫⎤
L
zˆ, I free
ρ
“stuff”
A ( ρ ) = (! x @* ~ ) zˆ ( A is in the ẑ -direction and I free )
For L
⎛μ
ρ : A( ρ ) ⎜ o
⎝ 2π
∂A ( r , t )
⎛ 2L ⎞
⎞
⎟ I free ln ⎜ ρ ⎟ zˆ Then induced E ( r , t ) = − ∂t
⎠
⎝
⎠
Thus: ε mf ε ( t ) = ΔV ( t ) =
∫ E ( r , t )i d
=−
C
∂
∂t
(∫
C
A ( r , t )i d
) = −∫
C
(n.b. E A !!!)
∂A ( r , t )
∂Φ m ( t )
id = −
∂t
∂t
zˆ, I free
Since E A only segments (1) and (3) will contribute to line integral ( d
∴
ε mf ε ( t ) = ΔV ( t ) =
=
∫
∂A ( r , t )
id
C
∂t
and d
4
∂Φ m ( t )
∂
A ( r , t )i d = −
∫
∂t C
∂t
⎛ 2L ⎞⎤
⎛ μ ⎞ ∂I ( t ) ⎡ ⎛ 2 L ⎞
(for L
− ⎜ o ⎟ free
⎢ln ⎜
⎟ − ln ⎜
⎟⎥
⎝ 2π ⎠ dt ⎣ ⎝ ρ1 ⎠
⎝ ρ2 ⎠⎦
∫ E ( r , t )i d
C
2
=−
⎛ μo ⎞ ∂I free ( t ) ⎡ ⎛ ρ 2 ⎞ ⎤
⎢ln ⎜ ⎟ ⎥
⎜
⎟
⎝ 2π ⎠ ∂t ⎣ ⎝ ρ1 ⎠ ⎦
(
(for L
are ⊥ A ).
)
ρ)
ρ)
Therefore:
ε mf ε ( t ) = ΔV ( t ) =
∫ E ( r , t )i d
C
= ⎡⎣ E ( ρ1 , t ) − E ( ρ 2 , t ) ⎤⎦ ≡ ΔE ( t )
⎛ μ ⎞ ∂I free ( t ) ⎛ ρ 2 ⎞
−⎜ o ⎟
ln ⎜ ⎟
⎝ 2π ⎠ ∂t
⎝ ρ1 ⎠
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
31
UIUC Physics 435 EM Fields & Sources I
⎛ μ ⎞ ∂I free ⎛ ρ 2 ⎞
= −⎜ o ⎟
ln ⎜ ⎟
⎝ 2π ⎠ ∂t
⎝ ρ1 ⎠
ε mf ε = ΔE ( t )
with: ΔE ( t ) ≡ ⎡⎣ E ( ρ1 , t ) − E ( ρ 2 , t ) ⎤⎦
i.e. E ( ρ1 , t )
E ( ρ2 , t )
More generally:
E ( ρ,t )
Fall Semester, 2007
⎛μ
−⎜ o
⎝ 2π
⎛μ
−⎜ o
⎝ 2π
Lecture Notes 21
ε mf induced around
Volts ⇐
closed contour, C as
drawn in above figure
⎞ ∂I free ( t ) ⎛ ρ 2 ⎞
ln ⎜ ⎟ zˆ
⎟
⎠ ∂t
⎝ ρ1 ⎠
⎞ ∂I free ( t ) ⎛ 2 L ⎞
ln ⎜
⎟ zˆ
⎟
⎠ ∂t
⎝ ρ1 ⎠
⎛ μ ⎞ ∂I free ( t ) ⎛ 2 L ⎞
−⎜ o ⎟
ln ⎜
⎟ zˆ
⎝ 2π ⎠ ∂t
⎝ ρ2 ⎠
Prof. Steven Errede
Volts/m (for L
ρ)
n.b. ρ 2 > ρ1
∴ E ( ρ1 , t ) > E ( ρ 2 , t )
⎛ μ ⎞ ∂I ( t ) ⎛ 2 L ⎞
− ⎜ o ⎟ free
ln ⎜
⎟ zˆ (for L
⎝ 2π ⎠ ∂t
⎝ ρ ⎠
ρ)
Here, think of /view the contour, C as a complete electrical circuit
(e.g. a square loop of real/physical wire).
Note also the following:
The induced ε mf ε (Volts) associated with the closed contour, C (drawn above) can be written as
follows, using the principle of linear superposition:
ab
bc
da
ε ( t ) = ε (1)
( t ) + ε (2)
( t ) + ε (3)cd ( t ) + ε (4)
( t ) = ε Totabcda ( t )
bc
= ΔV ( t ) = ΔV(1)ab ( t ) + ΔV(2)
( t ) + ΔV(3)cd ( t ) + ΔV(4)da ( t ) = ΔVTotabcda ( t )
Each of the four individual contributions to ε ToT ( ΔVToT ) are associated with the four
corresponding portions of the line integrals
32
∫ E ( r , t )i d
C
= −∫
C
∂A ( r , t )
id .
∂t
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
(1)
bc
bc
ε (2)
= ΔV(2)
2
b
da
da
ε (4)
= ΔV(4)
Lecture Notes 21
Prof. Steven Errede
( E ( ρ ) zˆ ) whereas ( d ( ) = +d zˆ )
= V − V = ∫ E ( ρ )id ( ) = 0 because ( E ( ρ ) zˆ ) whereas ( d ( ) + ρˆ ) i.e. E ⊥ d
= V − V = ∫ E ( ρ )i d
= −E ( ρ )
( E ( p ) zˆ ) whereas ( d ( ) = −d zˆ )
= V − V = ∫ E ( ρ )i d
= 0 because ( E ( ρ ) zˆ ) whereas ( d ( ) − ρˆ ) i.e. E ⊥ d
ab
ε (1)
= ΔV(1)ab = Vb − Va = ∫ E ( ρ1 )id
cd
cd
ε (3)
= ΔV(3)
Fall Semester, 2007
a
c
c
b
d
e
c
a
a
d
d
b
= + E ( ρ1 )
1
1
2
( 2 ) !!
d
2
(3)
2
2
3
(4)
4
( 4) !
( ) ( )
contribution from (3) ( cd ) actually partially cancels the contribution from (1) ( ab ) (because the
Thus, only portions (1) ab and cd contribute to ε ToT ( ΔVToT ) and in fact we see that the
contour integral is going in the opposite direction on (3).
( )
Indeed if ρ 2 = ρ1 , then ε ToT = ΔVToT = 0 i.e. the contribution from (3) cd would exactly cancel
( )
with that from (1) ab . ⇒ this physically corresponds to an area of the loop A⊥ = 0 , and of
course, these two induced voltages would then have to cancel each other.
Then:
ab
bc
da
ε Tot ( t ) = ε (1)
( t ) + ε (2)
( t ) + ε (3)cd ( t ) + ε (4)
( t ) = ε (1)ab ( t ) + ε (3)cd ( t )
bc
Or: ΔVtot ( t ) = ΔV(1)ab ( t ) + ΔV(2)
( t ) + ΔV(3)cd ( t ) + ΔV(4)da ( t ) = ΔV(1)ab ( t ) + ΔV(3)cd ( t )
Thus: ε ToT ( t ) = ΔVToT ( t )
⎛ μ ⎞ ∂I free ( t ) ⎛ 2 L ⎞ ⎛ μo ⎞ ∂I free ( t ) ⎛ 2 L ⎞
−⎜ o ⎟
ln ⎜
ln ⎜
⎟ −⎜
⎟
⎟
⎝ 2π ⎠ ∂t
⎝ ρ1 ⎠ ⎝ 2π ⎠ ∂t
⎝ ρ2 ⎠
⎛ μo
⎝ 2π
ε ToT ( t ) = ΔVToT ( t ) = − ⎜
⎞ ∂I free ( t ) ⎡ ⎛ ρ 2 ⎞ ⎤
ln ⎜
⎟
⎟
⎠ ∂t ⎢⎣ ⎝ ρ1 ⎠ ⎥⎦
(for L
(for L
ρ)
ρ)
This result may seem somewhat disturbing – how can a conducting wire loop (connected to
itself) have a potential difference across it ???
Point 0: The conducting wire loop is a source of ε mf :
n.b. f (2) = f (4) = 0
Point 1: The material of which the loop is made does have finite resistance associated with it.
Point 2: The time varying A ( t ) -field of the wire, driven by the time-varying current in the wire
induces an electric field at ρ1 which is stronger than that at ρ 2 , because ρ 2 > ρ1 .
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
33
UIUC Physics 435 EM Fields & Sources I
Note that E ( ρ )
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
∂I free ( t )
⎛ μ ⎞ ∂I ( t ) ⎛ 2 L ⎞
− ⎜ o ⎟ free
ln ⎜
> 0 (i.e. I free ( t ) increasing with
⎟ zˆ , which for
∂t
⎝ 2π ⎠ ∂t
⎝ ρ ⎠
( )
time) means that E points in the − ẑ direction along both segments of wire - segment (1) ab
( )
and segment (3) cd .
At ρ = ρ1 , the force per unit charge f1 = E ( ρ1 ) = − E ( ρ1 ) zˆ acts on free electrons in (1)
At ρ = ρ 2 , the force per unit charge f3 = E ( ρ 2 ) = − E ( ρ 2 ) zˆ acts on free electrons in (3)
But f (1) and f (3) point in the same direction, however, because ρ1 > ρ 2 then f (1) > f (3) and the
force imbalance is Δf = f (1) − f (3) > 0 since f1 ( ρ1 ) > f (3) ( ρ 2 ) so f (1) “wins out” over f (3) , and
(
)
thus a net current, I free ( t ) will flow around the loop, abcda :
I free ( t ) =
⎛ μ ⎞ ∂I free ( t ) ⎡ ⎛ ρ 2 ⎞ ⎤
ln ⎜
−⎜ o ⎟
⎟
⎝ 2π ⎠ ∂t ⎢⎣ ⎝ ρ1 ⎠ ⎥⎦
ε ToT ( t )
R oop
R oop
(for L
ρ)
Another Example: (Magneto-Motional Effect )
A rectangular loop of dimensions L x W = Area, A is immersed in a ⊥ uniform and constant
external magnetic field, Bext = Bo zˆ of infinite extent. The rectangular loop lies in the horizontal
(z-y) plane:
ẑ
Bext = Bo zˆ
•
ŷ
x̂
Closed contour, C
d
•
n̂
W
c
•
(4)
(3)
•
(2)
⇒ v = vo yˆ
(1)
•
a
• b
L
At time t = 0 the entirety of the rectangular loop suddenly moves to the right ( + ŷ direction) with
constant velocity v = vo yˆ . Show that no net ε mf ε ToT is induced in the rectangular loop.
The total induced ε mf ε ToT ( t ) = ΔVToT ( t ) =
34
∫
c
Em ( r , t )id =
∂
∂t
( ∫ B ( r , t )ida ) = − ∂Φ∂t (t )
m
c
loop
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
First, we examine next-to right-most term in the above equation:
−
∂
∂t
(∫
C
B ( r , t )ida oop
)
B = Bo zˆ = constant, ≠ fcn ( time, t )
and da oop = da oop nˆ
Neither da oop nor nˆ are fcns ( time,t )
∴ −
∂
∂t
( ∫ B ( r , t )ida ) = − ∂Φ∂t ( t ) = 0 (here).
m
i.e. ∃ no change in the (net) amount of magnetic flux, Φ m ( t ) enclosed by the conducting
rectangular loop, i.e. Φ m ( t ) = Bo A = Bo LW = constant.
However, let us now examine (in detail) the term(s) on the left-hand side of this equation:
ε ToT ( t ) = ΔVToT ( t ) =
∫
c
Em ( r , t )id
ab
bc
da
abcda
Again: ε Tot ( t ) = ε (1)
( t ) + ε (2)
( t ) + ε (3)cd ( t ) + ε (4)
( t ) = ε Tot
(t )
bc
= ΔVTot ( t ) = ΔV(1)ab ( t ) + ΔV(2)
( t ) + ΔV(3)cd ( t ) + ΔV(4)da ( t ) = ΔVTotabcda ( t )
= + ∫ Em ( r , t )id
b
a
+ ∫ Em ( r , t )id
c
(1)
b
+ ∫ Em ( r , t )id
d
(2)
c
+ ∫ Em ( r , t )id
a
(3)
d
(4)
Top View:
The Lorentz Force/unit charge =
Fm
Q
= Em = v × Bext ⇐ common to all four sides of
rectangular conducting loop. = E − field in rest frame of charge (which is moving in lab frame)
v = v0 yˆ , Bext = B0 zˆ
Em = v × Bext = v0 B0 yˆ × zˆ = + v0 B0 xˆ
+ xˆ
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
35
UIUC Physics 435 EM Fields & Sources I
Em = Eo xˆ = vo Bo xˆ
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Em points in + x̂ direction with magnitude Eo = vo Bo x
{ + ve charge is pushed to + x̂ , −ve charge is pushed to − x̂ }
Then:
b
ab
ε (1)
= ΔV(1)ab ≡ Vb − Va = ∫ Em id
a
(1)
c
bc
bc
ε (2)
= ΔV(2)
≡ Vc − Vb = ∫ Em id
b
(2)
d
cd
cd
ε (3)
= ΔV(3)
≡ Vd − Vc = ∫ Em id
c
a
da
da
ε (4)
= ΔV(4)
≡ Va − Vd = ∫ Em id
d
(
b
b
a
a
= ∫ Em i( −dxxˆ ) = ∫ −vo Bo ( xˆ i xˆ )dx = −vo Bo w
c
c
b
b
(4)
(
)
= ∫ Em i( −dyyˆ ) = ∫ −vo Bo xˆ i yˆ dy = 0
d
(3)
)
= ∫ Em i( dyyˆ ) = ∫ vo Bo xˆ i yˆ dy = 0
d
c
a
a
c
d
d
= ∫ Em i( dxxˆ ) = ∫ vo Bo ( xˆ i xˆ )dx = + vo Bo w
Thus:
ab
ε (1)
= ΔV(1)ab ≡ Vb − Va = 0
bc
bc
ε (2)
= ΔV(2)
≡ Vc − Vb = −vo Bo w
cd
cd
ε (3)
= ΔV(3)
≡ Vd − Vc = 0
da
da
ε (4)
= ΔV(4)
≡ Va − Vd = + vo Bo w
Thus we see that ∃ a potential difference ΔVx = +vo Bo w in the + x̂ direction and no potential
difference ΔVy = 0 in the yˆ direction. This is because Em = Eo xˆ = vo Bo xˆ
( )
( )
(n.b. ΔV is the same for both sides (2) bc and (4) da ).
36
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
⎛x⎞
Em = +∇Vm ⇒ Vm = ⎜ ⎟ Vo
(Vo ≡ vo Bo w = Eo w )
⎝ w⎠
Because this is a source (like a battery!)
da
bc
completely cancels ε (2)
Thus we see there can be no net current flow here, because ε (4)
( ΔV
da
( 4)
)
bc
completely cancelsΔV(2)
:
ε ToT = ε (1) + ε (2) + ε (3) + ε (4) = 0 − vo Bo w − 0 + vo Bo w = 0
bc
cd
da
= ΔVToT = ΔV(1)ab + ΔV(2)
+ ΔV(3)
ΔV(4)
= 0 − vo Bo w − 0 + vo Bo w = 0
Yet Another Example: (Magnetic Induction)
Let’s consider the same rectangular conducting loop (again in the horizontal (x-y) plane), this
time it is fixed in space, but is immersed in a time-varying magnetic field (of infinite spatial
extent). The rectangular loop has dimensions L×W and cross-sectional area, Aloop = LW.
Bext ( + ) = B ( + ) zˆ (points in ẑ direction in space).
ẑ
Vector area of loop
ŷ
Aloop = Anˆ , nˆ zˆ
x̂
Bext ( t ) = B ( t ) zˆ
Aloop = LWzˆ
Closed contour
C
d
•
d
W
( 4)
= dxxˆ
(3) d
n̂
(4)
( 3)
c
•
− dyyˆ
•
(1) d
(2) d
(1)
= dy yˆ
•
a
( 2)
= − dx xˆ
•
b
L
∂Bext ( t ) ∂B ( t )
zˆ ≠ 0 the magnitude of Bext changes with time, but always points in ẑ direction.
=
∂t
∂t
ε ToT ( t ) = ΔVToT ( t ) =
∫ E ( r , t )id
C
Now Bext ( t ) Area, Aloop
=−
(t )
∂Bext
⇒
∂t
∂
∂t
∫
S⊥
Bext ( r , t )ida⊥ = ∫
S⊥
∂Bext ( r , t )
∂Φ ( t )
i da⊥ = − m
∂t
∂t
Area, Aloop zˆ
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
37
UIUC Physics 435 EM Fields & Sources I
∴
∫
S⊥
Since
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
∂Bext ( t )
∂B ( t )
∂B ( t )
∂B ( t )
Aloop ≠ 0 and Aloop = LW ≠ 0
ida = ext i ∫ da = ext i Aloop = ext
S
⊥
∂t
∂t
∂t
∂t
∂Bext ( t )
≠ 0 and Aloop = LWzˆ ≠ 0 then:
∂t
∂B ( t )
Aloop ≠ 0 .
ε ToT = ΔVToT = − ext
∂t
∂Bext ( t )
Aloop
∂t
∂B ( t )
bc
= ΔVTot ( t ) = ΔV(1)ab ( t ) + ΔV(2)
( t ) + ΔV(3)cd ( t ) + ΔV(4)da ( t ) = ΔVTotabcda ( t ) = − ext Aloop
∂t
ab
bc
da
abcda
ε Tot ( t ) = ε (1)
( t ) + ε (2)
( t ) + ε (3)cd ( t ) + ε (4)
( t ) = ε Tot
(t ) = −
What is the electric field E here in this situation? We know that Bext ( t ) = ∇ × A ( t ) = B ( t ) zˆ
The magnetic vector potential A ( t ) that is needed to produce Bext ( t ) = B ( t ) zˆ is A ( t ) = Aϕ ( t ) ϕˆ
⎤
1⎡∂
ρ Aϕ ( t ) ) ⎥ zˆ = B ( t ) zˆ
(
⎢
ρ ⎣ ∂ρ
⎦
⎤
⎡∂
⎤
1⎡∂
ρ = x2 + y 2
or:
ρ Aϕ ( t ) ) ⎥ = B ( t ) ⇒ ⎢ ( ρ Aϕ ( t ) ) ⎥ = ρ B ( t )
(
⎢
ρ ⎣ ∂ρ
⎦
⎣ ∂ρ
⎦
1 2
1
Integrating both sides: ∫ ∂ ( ρ Aϕ ( t ) ) = ∫ ρ B ( t ) ∂ρ ⇒ ρ Aϕ ( t ) = ρ B ( t ) or: Aϕ = ρ B ( t )
2
2
1
∴ A ( t ) = Aϕ ( t ) ϕˆ = ρ B ( t ) ϕˆ
2
∂A ( t ) ∂Aϕ ( t )
1 ∂B ( t )
Then:
ϕˆ = ρ
ϕˆ
=
2
∂t
∂t
∂t
∂A ( t )
∂A ( t )
1 ∂B ( t )
ϕˆ
= − ϕ ϕˆ = − ρ
Finally: E ( t ) = −
2
∂t
∂t
∂t
∂B ( t )
⇒ E ( t ) is in −ϕ̂ direction (for
> 0)
∂t
In cylindrical coordinates: Bext ( t ) = ∇ × A ( t ) =
Here the origin of coordinates is taken to be at the center of the rectangular loop – assuming that
only the magnetic flux enclosed by the loop matters – the sum total of the contributions from
magnetic flux outside the loop (everywhere else) actually must cancel (tough to prove…)
38
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Then e.g.:
∫ E ( t )i d
b
a
(1)
L
2
L
−
2
=∫
+
⎛ ρ ∂B ( t ) ⎞
ϕˆ ⎟idyyˆ
⎜−
2 ∂t
⎝
⎠
Now: ρ = x 2 + y 2 =
Where: sin ϕ =
Then:
y
ρ
∫ E ( t )i d
=
b
Fall Semester, 2007
( w 2)
2
+ y2
y
( w 2)
2
+y
2
Lecture Notes 21
where
(1)
= dyyˆ
ϕˆ = − sin ϕ xˆ + cos ϕ yˆ
and
and cos ϕ =
= ε (1) ( t ) = ΔV(1)ab ( t ) = ∫
d
Prof. Steven Errede
x
w
=
ρ
2
( w 2) + y 2
+L ⎛
ρ ∂B ( t ) ⎞
2
ϕˆ idyyˆ
−
2
⎜⎜
⎟⎟
⎝ 2 ∂t
⎠
w
x
y
ϕˆ = − sin ϕ xˆ + cos ϕ yˆ with sin ϕ =
and cos ϕ = = 2 where ρ = x 2 + y 2 =
a
(1)
−L
ρ
2
ρ
ρ
a
2
+ y2
⎛ ρ ∂B ( t ) ⎞
⎜−
⎟ cos ϕ dy
−L
2 ∂t ⎠
2 ⎝
Then: ε (1) ( t ) = ΔV(1ab) ( t ) = (Vb ( t ) − Va ( t ) ) = ∫ E ( t )id = ∫
b
( w 2)
+L
2
=VB ( t )
=VA ( t )
⎛ w⎞
+L ⎛
ρ ∂B ( t ) ⎞ ⎜⎝ 2 ⎟⎠
1 ∂B ( t ) ⎛ w ⎞ + L 2
w ∂B ( t ) L
w ∂B ( t ) L
ε (1) ( t ) = ∫ L 2 ⎜ −
=
−
−
dy = −
dy
⎟
⎜
⎟
−
⎟
2 ∂t ⎝ 2 ⎠ ∫− L 2
4 ∂t 2
4 ∂t 2
2 ⎜
⎝ 2 ∂t ⎠ ρ
1 ∂B ( t )
1 ∂B ( t )
1 ∂B ( t )
1 ∂B ( t )
1 ∂B ( t )
A oop −
A oop = −
A oop
ε (1) ( t ) = ΔV(1ab) ( t ) = −
(ω L ) −
(ω L ) = −
8 ∂t
8 ∂t
8 ∂t
8 ∂t
4 ∂t
=VB ( t )
=VA ( t )
=VB ( t )
=VA ( t )
So: ε (1) ( t ) = ΔV(1ab) ( t ) ≡ Vb ( t ) − Va ( t ) = −
=ΔV(1ab) ( t ) = ε (1) ( t )
1 ∂B ( t )
A oop
4 ∂t
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
39
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
1 ∂B ( t )
E (t ) = − ρ
ϕˆ
∂t
2
t = ε (1) ( t ) = ΔV(1ab) ( t ) = −
From symmetry considerations: ε ( 3) ( t ) = ΔV( cd
3) ( )
Prof. Steven Errede
1 ∂B ( t )
A oop
4 ∂t
But we know (from the principle of linear superposition) that:
∂B ( t )
A oop
∂t
1 ∂B ( t )
A oop
∴ ε ( 2) ( t ) + ε ( 4) ( t ) = ε ToT ( t ) − ε (1) ( t ) − ε ( 3) ( t ) = −
2 ∂t
1 ∂B ( t )
A oop
And again from symmetry considerations: ε ( 2) ( t ) = ε ( 4) ( t ) = −
4 ∂t
ε ToT ( t ) = ε (1) ( t ) + ε ( 2) ( t ) + ε (3) ( t ) + ε ( 4) ( t ) = −
This can also be explicitly obtained by doing the remaining integrals:
∫ E ( t )i d
c
b
, ∫ E ( t )i d
d
( 2)
c
, ∫ E ( t )i d
a
( 3)
d
( 4)
obtaining the same result for each of the four sides!!
n.b. This also implies that the EMF induced across the ends of a conducting rod of length L.
But for a conducting (filamentary) rod, Aloop= 0 !!! The EMF in the (filamentary) rod is induced
by E ( t ) = − ∂A ( t ) ∂t !!!
EMF appears across the cut ends of the conducting rectangular loop
∂B ( t )
∂Φ m ( t )
⎛ 1 ∂B ( t )
⎞
A oop ⎟ = −
A oop = −
Now the total EMF, ε ToT ( t ) = 4 × ⎜ −
∂t
∂t
⎝ 4 ∂t
⎠
If the loop closes on itself, how do we get ε ToT ≠ 0 ? Again, the answer is that E = source field
∂B ( t )
∂Φ m ( t )
A oop = −
(Volts) is only directly observable if the loop is cut,
∂t
∂t
∂B ( t )
A oop is directly
e.g. at point a. Then a potential difference ε ToT ( t ) = ΔVToT ( t ) = −
∂t
observable, e.g. using a voltmeter.
and ε ToT ( t ) = −
The situation here with E ( t ) = −
∂A ( t )
1 ∂B ( t )
ϕˆ as the source field is very similar to the
=− ρ
∂t
∂t
2
following one(s):
40
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Induced EMF - Version # 1:
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
41
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 21
Prof. Steven Errede
Induced EMF - Version # 2:
42
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.