UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 23
Prof. Steven Errede
LECTURE NOTES 23
Eddy Currents in Conductors
Conductors, by definition, contain “free” electrons – i.e. the electrons are free to move around
inside the metal, but in fact are (weakly) bound to the metal by the work function ϕW of the metal
(SI units = eV (i.e. Joules)). Due to internal thermal energy associated with the metal being at
finite temperature, the “free” electrons can “evaporate” from the metal via thermionic emission
(from the high-side tail of the thermal energy distribution of the free electrons in a metal). Thus
thermionic emission of electrons is intimately related to black body /thermal radiation.
Thermal energy distribution of “free” electrons in a metal:
ϕW
ne, number
density of
electrons
0
Thermionic emission of electrons from a
thermal
metal occurs when U e
> ϕW
U ethermal ( Joules or eV )
U ethermal = 32 k BT
The free electrons in the metal have mean/average thermal kinetic energies of
U ethermal =
1
3
me ve2 = k BT
2
2
In the presence of a uniform applied external magnetic field Bext = Bo zˆ the free electrons in
metal move in circular orbits in a plane ⊥ Bext = Bo zˆ (ignoring/neglecting scattering effects in the
metal). The (mean/average) momentum of each electron is pe = me ve = qBo R where R =
(mean/average) radius of curvature of the circular orbit:
R=
Note that
me ve
q Bo
ve =
=
me
q Bo
3k BT
me
3k BT
1
3me k BT
=
me
qBo
and
ve =
3k BT
.
me
1.2 × 105 m s for T = 300K (see P435 Lect. Notes 21, page 9), using:
me = 9.1×10−31 kg
q = 1.6 × 10−19 Coulombs = |e|
Bo = 1 Tesla
k B = 1.38 ×10−23 J K
T = 300K
This corresponds to a radius of curvature of R = 6.65 × 10−7 m = 0.665μ m
a Bo = 1 Tesla magnetic field!
0.7 μ m for
© 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 23
Prof. Steven Errede
This result (of course) assumes no scattering of the free electrons in the conducting metal while it
is making one orbit of cyclotron motion:
zˆ, B = Bo zˆ
(
)
Fe = −e ve × B = eE ′
Δτ =
R
= −e ve Bo ρˆ
{radially inward}
n.b. vz = 0 ( here )
C = 2π R
C
2π R
=
= mean cyclotron orbit time
ve
ve
ŷ
ϑ
e−
ve
x̂
me
In general vz ≠ 0 , thus free electrons will additionally move up/down || to ẑ -axis in a helical/spiraling
motion as shown in the figure below, since the electron’s motion in z is unaffected by the presence of
the externally-applied magnetic field:
The magnetic dipole moment me associated with a free electron in a cyclotron orbit is:
= A⊥
⎛ e ⎞ 2
e ve
1
1
π R 2 ( − zˆ ) = e ve R ( − zˆ ) = − e ve Rzˆ
me = IA⊥ = I π R 2 ( − zˆ ) = ⎜⎜
⎟⎟ π R ( − zˆ ) =
2
2
2π R
⎝ Δτ ⎠
1
Note that induced/resulting magnetic dipole moment, me = − IA⊥ zˆ = − e ve Rzˆ points opposite
2
direction of applied magnetic field.
⇒ Free electrons in metal/conductor in the presence of Bext have diamagnetic properties.
In reality the mean free path of a free electron in a metal, λmf ρ (= average/mean distance
between scatterings/collision) in most metals is much smaller than C = 2π R
T = 300K and Bo = 1 Tesla).
In copper metal for example:
Cu
Cu
λmfp
= 3.9 × 10−8 m 0.04 μ m = 40nm / = 400Þ (!!) Thus λmfp
Cu
i.e. λmfp
2
0.04 μ m
4.2 μ m (for
C = 2π R
4.2 μ m ,
1
C in copper (for T = 300K and Bo = 1 Tesla).
100
© 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
Where λmfp =
σ c me ve
ne e
2
Fall Semester, 2007
. For copper σ c cu
with ve =
Lecture Notes 23
5.95 ×107 Siemens and necu
3k BT
me
Prof. Steven Errede
8.5 ×1028 m3
1.2 × 105 m s for T = 300K
In this regime the path of free electrons is more complex due to the intrinsic scattering processes
extant in the metal, but the free electrons still travel (on average) in circular and/or helical paths.
Again, it is important to point out that static magnetic fields perform/carryout/ do NO work on
charged particles. Thus, for a constant (i.e. time-independent) magnetic-field, e.g. B = Bo zˆ , no
net energy is deposited and/or removed from the metal conductor by the constant magnetic field.
All a constant magnetic field B = Bo zˆ does is change/rearrange the nature of the six-dimensional
phase space associated with the free electrons ( re , pe ) = ( x, y, z , px , p y , pz ) by introducing
correlations in (x & y), and (px & py) due to the induced cyclotron-type motion of the free
electrons. The block of metal remains in thermal equilibrium.
In e.g. a sheet of copper metal, the free electron number density necu
8.5 ×1028 m3 . If the
copper metal sheet lies in the x–y plane, with Bext = Bo zˆ , then the net magnetic dipole moment is
Ne
mnet = ∑ me = mnet ( − zˆ ) with macroscopic magnetization Μ = mnet volume = mnet ( − zˆ ) volume
i =1
due to the free electrons only.
An effective surface K e current flows (only) on the periphery of the metal sheet – due to
cancellation of the nearest neighbor induced dipole currents, as shown in the figure below:
© 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
Fall Semester, 2007
Lecture Notes 23
Prof. Steven Errede
Again, there are NO power losses here with a static magnetic field because B does NO work if
B = constant.
∂B ( r , t )
However for time-varying magnetic fields, because of Faraday’s Law: ∇ × E ( r , t ) = −
∂t
∂B ( r , t )
∂Φ m ( t )
ida⊥ = −
or: ∫ ∇ × E ( r , t )ida⊥ = ∫ E ( r , t )id = − ∫
= ε mf ε ( t )
S⊥
C
S⊥
∂t
∂t
∂B ( r , t )
a time-varying magnetic field
induces an electric field E ( r , t ) which (depending on
∂t
∂B ( r , t )
the sign of
(increasing/decreasing) either accelerates/decelerates the free electrons in
∂t
the conducting metal.
∂B ( t ) ∂Bo ( t )
=
zˆ and for simplicity, if ve is in the x–y plane,
∂t
∂t
i.e. vez = 0 then the free electron’s cyclotron orbit radius, R remains constant:
⇒ If B ( t ) = Bo ( t ) zˆ and
me
m v
R = e e = constant
eB
(
d ve ( t )
dt
dB ( t )
e
dt
)
or: me
d ve ( t )
dt
=e
dB ( t )
dt
The (average) kinetic energy gain (or loss) per cyclotron orbit is:
ΔKEe ( t ) = q * ( ε mf ε e ( t ) ) = −e * ( ε mf ε e ( t ) ) = +e
∂Φ m ( t )
∂B ( t )
=e
A⊥ where A⊥ = π R 2
∂t
∂t
The (average) rate of a free electron’s kinetic energy gain (or loss) per cyclotron orbit is thus:
The power gain/loss per cyclotron orbit: P ( t )e =
but:
τ rev
τ rev
∂B ( t )
* (π R 2 )
∂t
=
e
τ rev
C
2π R
=
ve
ve
∂B ( t )
* π R2
∂t
Thus: Pe ( t ) =
2π R
e
Therefore:
ΔKEe ( t )
(
)
∂B ( t )
1
1
ve = e ve R
But: me = − e ve Rzˆ (from above)
2
2
∂t
∂B ( t )
∂B ( t ) ∂ ⎡⎣ − me i B ( t ) ⎤⎦ ∂U emag ( t )
1
Pe ( t ) = e ve R
= − me i
=
=
2
∂t
∂t
∂t
∂t
Note: By Lenz’s Law, Pe is always positive for diamagnetic conducting materials. Then the
macroscopic power: Pmacro ( t ) = N e Pe ( t ) = ne * volume * Pe ( t ) corresponds to the macroscopic
induced EMF ε macro and (net) macroscopic current, Imacro flowing in the conducting material!!!
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
Fall Semester, 2007
Lecture Notes 23
Prof. Steven Errede
2
2
Then we see that Pmacro ( t ) = ε macro ( t ) I macro ( t ) = I macro
( t ) Rmaterial = ε macro
( t ) R material
= Joule heating / power losses in the metal.
The (net) macroscopic current I macro ( t ) flows around periphery of material, as a surface
current K macro ( t ) , then I macro ( t ) = K macro ( t ) t⊥ , where t⊥ = perpendicular thickness of the
conducting material (see above figure).
Note that this surface current K macro ( t ) , which is created by an externally-created ∂B ( t ) ∂t
in the metal is associated with real power losses/real power dissipation because of the
acceleration/deceleration of free electrons in the metal (which is a time-irreversible process).
Such induced macroscopic currents in conductors/metals, caused by an induced macroscopic
d Φm (t )
∂B ( t ) macro
=−
A⊥
EMF ε macro ( t ) = −
are known as Eddy currents.
∂t
∂t
Eddy currents e.g. in transformers are harmful because they sap power from the transformer:
power input = (power output + Eddy current power)
thus: power output = (power input – Eddy current power)
Since Eddy current power winds up as heat, the transformer will (eventually) get hot –
possibly so hot it could be destroyed, if it has not been designed properly!
Eddy currents in metals can also be used beneficially e.g. to cook food by induction heating!!
⇒ Place food in metal container in proximity to intense alternating B -field. Eddy current
power losses (e.g. in transformers) can be dramatically reduced by laminating the transformer
core.
↑
∂B ( t ) ∂Bo ( t )
=
zˆ
∂t
∂t
increasing (here)
© 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
For a Solid-Core Transformer:
solid
core
The induced macroscopic ε mf ε macro
(t ) = −
Eddy
Lecture Notes 23
Prof. Steven Errede
solid
solid
∂Φ m ( t )
∂B ( t ) core
core
=−
A⊥
A
where ⊥ = ( L *W )
∂t
∂t
The power dissipation in the solid core of this transformer is:
solid
solid
solid
solid
Eddy
Eddy
Eddy
Eddy
core
core
core
Pmacro
( t ) = ε macro
( t ) I macro
( t ) = I 2core
macro ( t ) Rsolid
core
⎛ ∂B ( t ) ⎞ 2 core
ε
( t ) ⎜ ∂t ⎟ A ⊥
⎠
=
=⎝
Rsolid
Rsolid
solid
2 core
macro
Eddy
core
P
core
⎛ ∂B ( t ) ⎞ 2 core ⎛ ∂B ( t ) ⎞ 4
⎜
⎟ A⊥
⎜
⎟ L
solid
∂t ⎠
∂t ⎠
⎝
⎝
core
t
A
=
=
if ⊥ = L2 (i.e. L = W for a square core)
()
Rsolid
Rsolid
2
solid
core
macro
Eddy
2
core
2
core
For a Laminated Core Transformer:
Divide the solid core into N lam individual laminations, each of width ΔW . Coat each lamination
with a very thin layer of electrically-insulating material (e.g. varnish or epoxy). Then the total
with of all laminations stacked back together is W = N lam ΔW .
↑
∂B ( t ) ∂Bo ( t )
=
zˆ
∂t
∂t
increasing (here)
1
A⊥core
N Lam
The resistance of each lamination is now increased relative to the resistance of the whole core:
RLam = N Lam Rcore (i.e. Rcore = RLam N Lam - for N Lam laminations electrically connected in parallel).
Each (now insulated) lamination has a cross-sectional area A⊥Lam = LΔW =
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
Fall Semester, 2007
Lecture Notes 23
Prof. Steven Errede
The induced macroscopic EMF associated with each lamination is:
Lam
ε mf ε macro
(t ) = −
Eddy
solid
∂Φ m ( t )
∂B ( t ) Lam
=−
A⊥ where A⊥Lam = L * ΔW = L * (W N Lam ) = A⊥core / N Lam
∂t
∂t
The corresponding Eddy current power loss associated with each lamination is:
solid
solid
⎛ ∂B ( t ) ⎞ 2 Lam ⎛ ∂B ( t ) ⎞ 2 core
⎛ ∂B ( t ) ⎞ 2 core
2
2 Lam
N Lam
ε macro ( t ) ⎜
⎟ A⊥
⎜
⎟ A⊥
⎜
⎟ A⊥
∂t ⎠
∂t ⎠
1 ⎝ ∂t ⎠
1
Eddy
⎝
⎝
PLam ( t ) =
=
=
= 3
= 3 Psolid ( t )
RLam
RLam
N Lam Rsolid
N Lam
Rsolid
N Lam core
2
2
2
core
core
Since there are N lam laminations (now) making up the transformer core, then the total Eddy
current power loss of the laminated transformer core is N lam times the power loss for one
lamination, i.e.:
solid
⎛ ∂B ( t ) ⎞ 2 core
A
⎜
⎟
⊥
1 ⎝ ∂t ⎠
1
Tot
PLam ( t ) = N Lam PLam ( t ) = 2
= 2 Psolid
N Lam
Rsolid
N Lam core
core
2
core
Thus, we see that the Eddy current power loss in a transformer decreases as the square of the
number of laminations (compared to no laminations) (i.e. the Eddy current power loss decreases
as the square of A⊥Lam )
n.b. Ferrite cores used in transformers have high resistance (e.g. compared to iron cores) and
also have good magnetic permeability ⇒ the use of ferrite materials in transformer cores can
reduce Eddy current losses even further, by a factor of Riron R ferrite 1 !!! The use of ferrite
materials for transformer cores is most common/most useful for low-power/small-signal
applications.
Today, there exist various kinds of magnetic field sensors – e.g. which utilize magnetoresistive effects (B-field dependent resistance!) such as Giant Magneto-Resistance (GMR)
sensors, and/or electron Spin-Dependent Tunnelling (SDT) devices as well as Superconducting
Quantum Interference Devices (SQUIDs) which are also very sensitive to magnetic fields. These
devices are used in all kind of applications to detect small variations in magnetic fields.
One important application is Eddy current sensing – used for detecting structural flaws in
critical conducting materials. The state-of-the-art of Eddy current sensing is now such that
imaging capability with ~ 100 micron resolution has been achieved. Please see/read handout on
Eddy current sensing for more information.
© 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 23
Prof. Steven Errede
Energy Stored in Magnetic Fields
In electrical circuits containing (one or more) inductors, work must be done against the back
∂I ( t )
∂Φ ( t )
= − L free
in order to get a free current, Ifree to flow in the circuit.
ε mf ε L ( t ) = − m
∂t
∂t
Suppose we have a simple electrical circuit consisting of a battery (which supplies a constant
ε mf , ε o ), an on/off switch, an inductor, (with inductance, L) and a resistor (of resistance, R) as
show in the figure below:
Then: ΔVTot =
∫
C
E id = 0 ⇐ Kirchoff’s Voltage Law: The sum of potential differences
N
around a closed circuit (mesh) = 0, i.e.
∑ ΔV
i =1
i
= 0.
⇒ E is a conservative field associated with a conservative force F = qE
Then:
ΔVbattery + ΔVinductor ( t ) + ΔVresistor ( t ) = 0
Where: ΔVbattery
= ( ε o − 0 ) = constant, ≠ fcn ( t )
ΔVinductor ( t ) = (VR ( t ) − ε o )
ΔVresistor ( t ) = ( 0 − VR ( t ) )
Then:
or:
ΔVbattery + ΔVinductor ( t ) +ΔVresistor ( t ) = 0
= ( ε o − 0 ) + (VR ( t ) − ε o ) + ( 0 − VR ( t ) ) = 0
ε o = − (VR ( t ) − ε 0 ) + VR ( t )
=ΔVinductor ( t )
=− ε L ( t ) Back
ε mf
( )
8
=ΔVresistor ( t )
= I free ( t ) R
(by Ohm's Law)
© 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
∴ ε 0 = −ε L ( t ) + I free ( t ) R = constant, but ε L ( t ) = − L
∂I free ( t )
∴ ε0 = +L
or: L
∂t
∂I free ( t )
∂t
Lecture Notes 23
Prof. Steven Errede
∂I free ( t )
∂t
+ I free ( t ) R
+ RI free ( t ) = ε o ⇐ 1st order linear inhomogeneous differential equation (solution
is = general solution of homogeneous differential equation +
particular solution for inhomogeneous equation (imposed by
initial conditions and/or final conditions)
First, solve the homogeneous differential equation:
L
∂I free ( t )
∂t
+ RI free ( t ) = 0
dI free ( t )
⎛ L ⎞ dI free ( t )
⎛R⎞
+ I free ( t ) = 0 or:
= − ⎜ ⎟ I free ( t )
⇒ ⎜ ⎟
dt
⎝ r ⎠ dt
⎝t ⎠
The general solution to this homogeneous differential equation is of the form:
I free ( t ) = I ofree e
Then:
dI free ( t )
dt
−t
τ
+ C where C = constant of integration.
−t
1
= − I ofree e τ
τ
⎛L⎞
⇒ τ ≡ ⎜ ⎟ = characteristic time constant (SI units = seconds)
⎝R⎠
Now solve the inhomogeneous differential equation:
L
∂I free ( t )
∂t
+ RI free ( t ) = ε o but I free ( t ) = I ofree e
−t
+C
τ
−t
⎛ L ⎞ ⎛ R ⎞ −t
ε
ε
Thus: − ⎜ ⎟ ⎜ ⎟ I 0 e τ + I 0 e τ + C = o ⇒ integration constant C = o
R
R
⎝ R ⎠⎝ L ⎠
∴ The solution of this inhomogeneous differential equation is: I free ( t ) = I ofree e
However we don’t (yet) know explicitly what I
free
o
−t
τ
+
εo
R
is…
∴ Impose the initial condition at time t = 0: I free ( t = 0 ) = 0 .
i.e. initially no current flows through the circuit when the switch is closed at t = 0
(this is due to the back EMF in the inductor – i.e. Lenz’s law!!!)
Thus at t = 0, we see that I free ( t = 0 ) = I ofree e −0 +
=1
εo
R
= I ofree +
εo
R
= 0 ⇒ I ofree = −
εo
R
∴ The specific solution to the inhomogeneous differential equation (here) is:
−t
ε −t ε
ε
I free ( t ) = − o e τ + o = o 1 − e τ
R
R R
(
)
© 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 23
Prof. Steven Errede
Thus, for this circuit, the free current flowing in this circuit as a function of time is:
(
−t
⎛ε ⎞
I free ( t ) = ⎜ o ⎟ 1 − e τ
⎝R⎠
)
( )
where τ ≡ L R
The free current I free ( t ) flowing in circuit vs. time, t is shown in the figure below:
⇐
n.b. this electrical current flows
through all components
⇐
(i.e. battery, inductor,
resistor, switch, wires…)
1τ
2τ
3τ
0
The voltage (aka potential difference) across the resistor, R vs. time, t:
ΔVR ( t ) = I ( t ) * R (by Ohm's Law) =
(
ΔVR ( t ) = ε 0 1 − e
−t
τ
)
εo
R
(
* R 1− e
−t
τ
)
( )
where τ ≡ L R
The voltage (aka potential difference) across the inductor, L vs. time, t:
(
−t
⎛ε ⎞
Since: I free ( t ) = ⎜ o ⎟ 1 − e τ
⎝R⎠
ΔVinductor ( t ) = −ε L ( t ) = + L
) where τ ≡ ( L R )
∂I free ( t )
1τ
∂t
⎛ ⎛ε
= L ⎜−⎜ o
⎜ ⎝R
⎝
2τ
0
10
⎞ ⎛ R ⎞ −t τ
⎟⎜ ⎟e
⎠⎝ L ⎠
⎞
−t
−t
⎟ = −ε o e τ i.e. ΔVinductor ( t ) = −ε o e τ
⎟
⎠
3τ
t
-5.0% ε o
-13.5% ε o
ΔVinductor ( t )
Thus:
-36.8% ε o
← = −ε 0 = back ε mf !! ⇒
At t = 0 the back ε mf across the inductor
(= −ε o ) exactly cancels Vbattery = ε 0 , but
inductor can’t sustain opposing it forever!
© 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 23
Prof. Steven Errede
Kirchoff’s Voltage Law: ΔVbattery ( t ) = −ΔVinductor ( t ) − ΔVresistor ( t ) (Volts)
εo
= +ε o e
−t
(
+ εo 1− e
τ
−t
τ
)=ε
o
(Volts)
The voltage (potential difference) across the battery = voltage (potential difference) across
[inductor & resistor] = constant, independent of time.
The instantaneous electrical power stored in the inductor is:
PL ( t ) = −ε L ( t ) I free ( t ) = + LI free ( t )
dI free ( t )
dt
(Watts)
The instantaneous electrical energy stored in the inductor is:
t
t
dI free ( t ′ )
t
1
WL ( t ) = ∫ P ( t ′ ) dt ′ = ∫ LI free ( t )
dt ′ = L ∫ I free ( t ′ ) dI free ( t ′ ) = LI 2free ( t )
0
0
0
dt ′
2
1
1
WL ( t ) = LI 2free ( t ) (Joules) {n.b. analog of WC ( t ) = C ΔV 2 ( t ) for capacitors!}
2
2
Thus, we see that the instantaneous electrical power stored in the inductor is:
PL ( t ) =
dI free ( t )
dWL ( t )
= −ε L ( t ) I free ( t ) = + LI free ( t )
(Watts)
dt
dt
Recall that the magnetic flux in an inductor is: Φ m ( t ) = LI free ( t ) (Webers or Tesla-m2)
However: Φ m ( t ) = ∫ B ( r , t )ida⊥ = ∫
S⊥
S⊥
( ∇ × A ( r , t ) )ida = ∫
⊥
C
A ( r , t )id = LI free ( t )
1 2
1
1
LI free ( t ) = ⎡⎣ LI free ( t ) ⎤⎦ I free ( t ) = ⎡ ∫ A ( r , t )id ⎤ I free ( t )
⎦
2
2
2⎣ C
1
1
1
= I free ( t ) ⎡ ∫ A ( r , t )id ⎤ = ∫ A ( r , t )i I free ( t ) d = ∫ A ( r , t )i I free ( t ) d
C
C
⎣
⎦
2
2
2 C
WL ( t ) =
Then:
(
)
(
)
Thus, more generally, for any magnetic vector potential, A ( r , t ) with its corresponding
filamentary/line free current I free ( t ) , or surface free current density K free ( t ) , or volume free
current density J free ( t ) we see that the magnetic energy stored in such systems can be written as:
(
)
1
1
A ( r , t )i I free ( r , t ) d = I free ( t ) ∫ A ( r , t )id
∫
C
C
2
2
1
Wmag ( t ) = ∫ A ( r , t )i K free ( r , t ) da⊥ (Joules)
2 S⊥
1
Wmag ( t ) = ∫ A ( r , t )i J free ( r , t ) dτ (Joules)
2 v
Wmag ( t ) =
(
(
(Joules)
)
)
Note from the last formula above that the energy density associated with an inductor is:
1
umag ( t ) = A ( r , t )i J free ( r , t ) (Joules/m3)
2
© 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 23
Prof. Steven Errede
Using Ampere’s law (in differential form): ∇ × B ( r , t ) = μo J Tot ( r , t ) , then for non-magnetic
media we see that J Tot ( r , t ) = J free ( r , t ) (only) whereas for magnetic media, in general
J Tot ( r , t ) = J free ( r , t ) + J Bound ( r , t ) .
Wmag ( t ) =
1
2 ∫v
(
(
)
1
A ( r , t )i J free ( r , t ) dτ as
2 ∫v
1
A ( r , t )i J Tot ( r , t ) dτ =
A ( r , t )i∇ × B ( r , t ) dτ
2μo ∫v
Then we can more generally write Wmag ( t ) =
)
(
)
Integrate the RHS term of the above relation by parts, and use the relation:
∇i A × B = Bi ∇ × A − Ai ∇ × B
i.e.
∴
( ) ( ) ( )
Ai( ∇ × B ) = Bi( ∇ × A ) − ∇i( A × B ) = Bi B − ∇i( A × B ) = B
1 ⎡
W (t ) =
B ( r , t ) dτ − ∫ ∇i( A ( r , t ) × B ( r , t ) ) dτ ⎤
⎦
2μ ⎣ ∫
2
(
− ∇i A × B
)
2
mag
o
v
v
Using the divergence theorem on the 2nd term:
1 ⎡
Wmag ( t ) =
B 2 ( r , t ) dτ − ∫ A ( r , t ) × B ( r , t ) ida ⎤ (n.b. surface S encloses volume v)
∫
v
S
⎣
⎦
2 μo
(
)
The volume of integration v = entire volume occupied by the current density J Tot ( r , t ) . Any
integration volume larger than this is perfectly fine also, so integrating over all space for a
localized/finite volume current distribution J Tot ( r , t ) is also fine. Then for an infinite integration
volume v with accompanying infinite enclosing surface S, we see that the surface integral on the
RHS of the above equation vanishes, i.e. ∫ A ( r , t ) × B ( r , t ) ida = 0 .
S∞
(
)
Then the magnetic energy associated with any system is given by:
(
)
1
A ( r , t )i J Tot ( r , t ) dτ (Joules) or equivalently by:
all
2 ∫space
space
1
2
Wmag ( t ) = ∫ all umag ( r , t ) dτ =
all B ( r , t ) dτ
∫
2μo space
space
(Joules)
1
1
B ( r , t )i B ( r , t ) dτ = ∫ all B ( r , t )i H ( r , t ) dτ
=
all
2μo ∫space
2 space
Wmag ( t ) = ∫ all umag ( r , t ) dτ =
(
)
(
)
The corresponding magnetic energy density is given by:
1
A ( r , t )i J Tot ( r , t ) (Joules/m3) or equivalently by:
2
1 2
1
1
umag ( r , t ) =
B (r ,t ) =
B ( r , t )i B ( r , t ) = B ( r , t )i H ( r , t ) (Joules/m3)
2μo
2 μo
2
umag ( r , t ) =
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
Lecture Notes 23
Prof. Steven Errede
We can see that these are directly analogous to the electric field case:
The energy associated with the electric field in any system is given by:
1
all ( ρTot ( r , t ) V ( r , t ) ) dτ (Joules) or equivalently by:
2 ∫space
space
1
Welect ( t ) = ∫ all uelect ( r , t ) dτ = ε o ∫ all E 2 ( r , t ) dτ
2 space
space
(Joules)
1
1
= ε o ∫ all E ( r , t )i E ( r , t ) dτ = ∫ all E ( r , t )i D ( r , t ) dτ
2 space
2 space
Welect ( t ) = ∫ all uelect ( r , t ) dτ =
(
)
(
)
The corresponding electric energy density is given by:
1
ρTot ( r , t ) V ( r , t ) (Joules/m3) or equivalently by:
2
1
1
1
uelect ( r , t ) = ε o E 2 ( r , t ) = ε o E ( r , t )i E ( r , t ) = E ( r , t )i D ( r , t ) (Joules/m3)
2
2
2
uelect ( r , t ) =
Electric and Magnetic Energies & Energy Densities in Macroscopic Matter – Linear Media:
In linear dielectric materials, ε o (free space) → ε (linear medium), with D ( r , t ) = ε E ( r , t )
In linear magnetic materials, μo (free space) → μ (linear medium), with H ( r , t ) = B ( r , t ) μ
The electric and magnetic energy densities in the volume v (only) of the linear media are:
1
1
1
dielectric
uelect
( r , t ) = ε E 2 ( r , t ) = ε E ( r , t )i E ( r , t ) = E ( r , t )i D ( r , t ) (Joules/m3)
2
2
2
1
1
1
magnetic
umag
( r , t ) = B 2 ( r , t ) = B ( r , t )i B ( r , t ) = B ( r , t )i H ( r , t ) (Joules/m3)
2μ
2μ
2
The corresponding electric and magnetic energies associated with the volume v (only) of the
linear media are:
1
dielectric
dielectric
Welect
( t ) = ∫v uelect
( r , t ) dτ = ε ∫v E 2 ( r , t ) dτ
2
(Joules)
1
1
= ε ∫ E ( r , t )i E ( r , t ) dτ = ∫ E ( r , t )i D ( r , t ) dτ
2 v
2 v
1
magnetic
magnetic
Wmag
( t ) = ∫v umag
( r , t ) dτ = ∫v B 2 ( r , t ) dτ
2μ
(Joules)
1
1
B ( r , t )i B ( r , t ) dτ = ∫ B ( r , t )i H ( r , t ) dτ
=
2 μ ∫v
2 v
(
)
(
)
(
)
(
)
The total electric/magnetic energy densities and electric/magnetic energies (i.e. integrated over
all space must (obviously) use ε o and μo for these integrals in the region(s) outside of the
volume v of the media.
© 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 23
Prof. Steven Errede
Griffiths Example 7.13:
A long coaxial cable carries a steady free current Ifree down the surface of the inner cylinder
(of radius a) and then returns along the surface of the outer cylinder (of radius b), as shown in the
figure below:
I free
ρ
b
a
C
I free
ẑ
Determine the magnetic energy density stored in a section of this coaxial cable, e.g. associated
with a length of this cable.
Use e.g. Ampere’s circuital law
∫ B ( r )i d
C
enclosed
= μo ITot
to determine B inside and outside the
cable, take contour(s) as shown above.
For ρ < a, I
enclosed
Tot
B(ρ)
therefore B ( ρ < a ) = 0 .
=0
enclosed
= I free therefore B ( a ≤ ρ ≤ b ) =
For a ≤ ρ < b, ITot
μo
I freeϕˆ .
2πρ
ρ
enclosed
For ρ > b, ITot
(net ) = 0 therefore B ( ρ > b ) = 0 .
Since the magnetic field B ( a ≤ ρ ≤ b ) =
a
b
μo
I freeϕˆ is non-zero only in the region a ≤ ρ < b , then
2πρ
the magnetic energy density will also be non-zero only in this same region:
umag ( ρ ) =
1
2 μo
1
B2 ( ρ ) =
2 μo
2
=
umag ( ρ )
B ( ρ )i B ( ρ )
μ
1 ⎡ μo ⎤ 2
I free = 2o 2 I 2free
⎢
⎥
2 μo ⎣ 2πρ ⎦
8π ρ
(Joules/m3)
ρ
The magnetic energy associated with a finite length
Wmag =
1
2μo
1
∫ B ( r ) dτ = 2μ ∫
2
v
= 2 π I 2free
14
z =0
o
μo
8π 2
ρ =b
∫ρ
=a
z=
ρ dρ
ρ
2
dz ∫
=
ρ =b
ρ =a
of the coaxial cable is:
ρd ρ ∫
ϕ = 2π
ϕ =0
a
b
B 2 ( r ) dϕ
( )
μ o 2 ρ =b d ρ μ o 2
=
I free ∫
I free ln b
a
ρ =a ρ
4π
4π
© 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
Thus: Wmag =
μo 2
I free ln b
a
4π
( )
Fall Semester, 2007
Lecture Notes 23
Prof. Steven Errede
(Joules)
The magnetic energy per unit length associated with this coaxial cable is:
wmag ≡ Wmag
=
μo 2
I free ln b
a
4π
( )
(Joules/meter)
Note that the magnetic energy in this coaxial cable is also Wmag = WL =
1 2
LI free , thus we see that
2
the inductance L associated with a finite length of this cable is:
Wmag = WL =
( )
μ
1 2
LI free = o I 2free ln b
a
2
4π
⇒ L=
( )
μo
ln b
a
2π
(Henrys).
The inductance per unit length of this coaxial cable is therefore:
L ≡
L
=
μo
ln b
a
2π
( )
(Henrys/meter).
Note that the inductance per unit length L ≡ L
magnetic permeability of free space μo .
has the same SI units (Henrys/meter) as the
Magnetic Forces:
Example: The Force on a Linear Magnetic Core Exerted by a Long Solenoid
Suppose we have a long solenoid of length and radius R (thus cross-sectional area A⊥ = π R 2 )
with NTot turns (thus n = NTot turns per unit length) carrying a steady free current I. What
happens when we insert a linear magnetic material (with constant magnetic permeability μ )
having the same length and radius into the bore of the long solenoid? Is there a force acting on it,
e.g. analogous to that associated with inserting a linear dielectric between the plates of a parallelplate capacitor?
Cross-Sectional View of Long Solenoid:
Magnetic Core
Δz
{partially inserted into
bore of long solenoid}
Δz
ẑ
z
z=0
z=z
© 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 23
Prof. Steven Errede
The infinitesimal difference in the magnetic energy associated with moving the magnetic core
inward into the bore of the long solenoid an infinitesimal distance Δz (ignoring/neglecting any/all
end effects) is given by:
ΔWmag ( Δz ) = Wmag ( z + Δz ) − Wmag ( z )
1
2
( μ − μo ) ∫v H solin 2 ( r ) dτ
in
Where (using Ampere’s circuital law for the H -field for the long solenoid): H sol
( r ) = nI free zˆ
The infinitesimal change in the volume due to the infinitesimal displacement Δz is ΔV = A⊥ Δz .
∴ ΔWmag ( Δz )
1
2
( μ − μo ) n 2 I 2free A⊥ Δz
Then the net force acting on the magnetic core is:
Fmag ( Δz ) =
ΔWmag ( Δz )
Δz
=
zˆ
I free = constant
dWmag ( Δz )
dz
1
2
zˆ
( μ − μo ) n 2 I 2free A⊥ zˆ
I free = constant
( μ − μo ) > 0 (i.e. paramagnetic materials) we see that the direction of the
magnetic force acting on the magnetic core ( + ẑ ) is such that it wants to suck the magnetic core
Thus, as long as
into the bore of the long solenoid, analogous to what we saw in the case of a linear dielectric
partially inserted into the gap of a parallel-plate capacitor. Note however, for diamagnetic core,
since ( μdia − μo ) < 0 , a diamagnetic magnetic core is repelled from the bore of the long solenoid!
Electric and Magnetic Pressure:
We have seen for the electrostatics case, that the energy density uelect ( r ) (SI units of Joule/m3)
also has the same units as that for pressure, P ( r ) (Newtons/m2), and in fact the presence of an
energy density u ( r ) can/will exert a pressure P ( r ) on material if present at the point r .
If there are no linear dielectrics and/or linear magnetic materials present, then:
1
1
1
uelect ( r , t ) = Pelect ( r , t ) = ε o E 2 ( r , t ) = ε o E ( r , t )i E ( r , t ) = E ( r , t )i D ( r , t ) (J/m3 or N/m2)
2
2
2
1 2
1
1
umag ( r , t ) = Pmag ( r , t ) =
B (r ,t ) =
B ( r , t )i B ( r , t ) = B ( r , t )i H ( r , t ) (J/m3 or N/m2)
2 μo
2 μo
2
If there are linear dielectrics and/or linear magnetic materials present, then:
1
1
1
dielectric
dielectric
uelect
( r , t ) = Pelect
( r , t ) = ε E 2 ( r , t ) = ε E ( r , t )i E ( r , t ) = E ( r , t )i D ( r , t ) (J/m3 or N/m2)
2
2
2
1
1
1
magnetic
magnetic
umag
( r , t ) = Pmag
( r , t ) = B 2 ( r , t ) = B ( r , t )i B ( r , t ) = B ( r , t )i H ( r , t ) (J/m3 or N/m2)
2μ
2μ
2
16
© 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 23
Prof. Steven Errede
Example: The Magnetic Pressure Inside a Long Air-Core Solenoid
For simplicity, assume steady free current I free flowing in windings of long air-core solenoid.
in
From Ampere’s circuital law: Bsol
( r ) = μo nI free zˆ .
Note that:
inside
usol
( r ) = Psolinside ( r ) =
1
2 μo
in 2
Bsol
(r ) =
1
μo n 2 I 2free = constant > 0 (i.e. is a positive quantity).
2
Cross-Sectional View of Long Air-Core Solenoid:
in
Bsol
( r ) zˆ points out of page:
outside
outside
umag
( r ) = Pmag
(r ) = 0
1
μo n 2 I 2free = constant > 0
2 μo
2
points radially outward
in
usol
( r ) = Psolin ( r ) =
1
in 2
Bsol
(r ) =
The magnetic pressure/magnetic energy density inside the solenoid pushes/exerts a radialoutward force on the solenoid!
The net magnetic force acting on the long air-core solenoid of radius R and length is:
Fsolnet = Psolinside ( r = R ) Asol ρˆ =
1
μo n 2 I 2free ⋅ 2π R ρˆ = μoπ n 2 I 2free R ρˆ (radially outward)
2
Example: The Magnetic Pressure Inside an Air-Core Toroid
inside
Similar to the long, air-core solenoid, except Btoroid
(r ) =
inside
inside
Then: utoroid
( r ) = Ptoroid
(r ) =
1
2μo
in 2
Btoroid
(r ) =
2
2
μo NTot I free
≠ constant > 0
8π 2 ρ 2
inside
inside
@ inner radius: utoroid
( ρ = a ) = Ptoroid
( ρ = a) =
inside
toroid
@ outer radius: u
( ρ = b) = P
inside
toroid
μo N tot I free
ϕˆ for a ≤ ρ ≤ b .
2π
ρ
( ρ = b) =
1
2μo
1
2 μo
2
2
μo NTot I free
8π 2 a 2
Force at inner
radius is radially
inward
2
2
μo NTot I free
( ρ = b) = 2
b2
8π
Force at outer
radius is radially
outward
in 2
Btoroid
( ρ = a) =
in 2
toroid
B
© 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 23
Prof. Steven Errede
Example: The Magnetic Pressure Associated with a Thin Cylindrical Current-Carrying Tube
A steady free sheet current K free ( r ) = K o zˆ (Amperes/meter) flows down the surface of a
long, very thin cylindrical conducting metal tube of radius a as shown in the figure below:
ẑ
K free ( r ) = K o zˆ
a
CK
CB
The (total) free current flowing down the surface of the conducting tube is:
I free = ∫ K free ( r )id
CK
{Since d
⊥
⊥
= ∫ K o zˆid
CK
2π
⊥
= ∫ K o zˆ iadϕ zˆ = K o a ∫ dϕ = 2π aK o ≡ I o
CK
0
= adϕ zˆ using the arc length formula " S = Rθ " }
Then using Ampere’s law for B :
∫
CB
enclosed
B ( r )id = μ o ITot
we see that:
enclosed
= 0 ⇒ B ( ρ < a) = 0
For ρ < a : ITot
tube
tube
⇒ umag
( ρ < a ) = Pmag
( ρ < a) =
1 2
Btube ( ρ < a ) = 0
2 μo
enclosed
= Io ⇒ B ( ρ ≥ a ) =
For ρ ≥ a : ITot
tube
tube
⇒ umag
( ρ ≥ a ) = Pmag
( ρ ≥ a) =
B(ρ)
μo I o
ϕˆ
2π ρ
ρ
a
μ I2
1 2
Btube ( ρ ≥ a ) = o2 o2
2 μo
8π ρ
Thus {here}, since the magnetic field/magnetic energy density is only non-zero outside the long,
thin conducting tube, the magnetic pressure creates a force on the tube which is radially inward.
The net force acting on the thin conducting tube is:
tube
mag
F
( ρ = a ) = P ( ρ = a ) ⋅ Atube
tube
mag
μo I o2
μo I o2
= 2 2 ⋅ 2π a =
8π a
4π a
n.b. = length of
a)
the tube (
The net force per unit length acting on the thin conducting tube is:
F
18
( ρ = a) ≡
tube
Fmag
( ρ = a)
μo I o2
=
4π a
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.