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 ϕ

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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 ϕ

W

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:

The free electrons in the metal have mean/average thermal kinetic energies of

thermal

1

2

3 2 2

e e e B

U = m v = k T

In the presence of a uniform applied external magnetic field B

_ext

= B z

_o

ˆ the free electrons in metal move in circular orbits in a plane ⊥ B

_ext

= B z

_o

ˆ (ignoring/neglecting scattering effects in the metal). The (mean/average) momentum of each electron is p

_e

= m v

_e _e

= qB R

_o

where R = (mean/average) radius of curvature of the circular orbit:

3 1

e e e B

3

e B

o o e o

m v m k T

R m k T

q B q B m qB

= = = and

_e

³

^B

e

v k T

= m .

Note that

_e

³

^B

1.2 10

⁵

e

v k T m s

= m × for T = 300K (see P435 Lect. Notes 21, page 9), using:

9.1 10

31

m

e

= ×

⁻

kg 1.6 10

19

q = ×

⁻

Coulombs = |e|

B

_o

= 1 Tesla 1.38 10

23

k

B

= ×

⁻

J K T = 300K

This corresponds to a radius of curvature of R = 6.65 10 ×

⁻⁷

m = 0.665 μ m 0.7 μ m for a B

_o

= Tesla magnetic field! 1

n

e

, number density of electrons

0 ϕ

W

Thermionic emission of electrons from a metal occurs when U

_e^thermal

> ϕ

_W

thermal

( or )

U

e

Joules eV

thermal 3

e 2 B

U = k T

(2)

This result (of course) assumes no scattering of the free electrons in the conducting metal while it is making one orbit of cyclotron motion:

^F

^e

^{= −} ^{e v} (

^e

^× ^B ) ⁼ ^eE′ ²

e e

C R

v v

τ π

Δ = = = mean cyclotron orbit time = − e v B

_e _o

ρ ˆ

{radially inward}

n b v ^{. .} =

z

⁰ ( here )

In general v

_z

≠ , thus free electrons will additionally move up/down || to ˆz -axis in a helical/spiraling 0 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 m associated with a free electron in a cyclotron orbit is:

_e

²

( ) ^ˆ

²

( ) ^ˆ

2

A

e e

e e v

m IA I π R z π R z

τ π

= ⊥

⊥

⎛ ⎞

= = − = ⎜ ⎜ ⎝ Δ ⎟ ⎟ ⎠ − = R π R

²

( ) ^ˆ ¹ ( ) ^ˆ ¹ ^ˆ

2

^e

2

^e

z e v R z e v Rz

− = − = −

Note that induced/resulting magnetic dipole moment, 1

ˆ ˆ

e

2

e

m = − IA z

_⊥

= − e v Rz points opposite direction of applied magnetic field.

⇒ Free electrons in metal/conductor in the presence of B

ext

have diamagnetic properties.

In reality the mean free path of a free electron in a metal, λ

_mf_ρ

(= average/mean distance between scatterings/collision) in most metals is much smaller than C = 2 π R 4.2 μ m (for T = 300K and B

_o

= Tesla). 1

In copper metal for example:

3.9 10

8

0.04 40 / 400

Cu

mfp

m m nm

λ = ×

⁻

μ = = Þ (!!) Thus λ

_mfp^Cu

0.04 μ m C = 2 π R 4.2 μ m ,

i.e. 1

100

Cu

mfp

C

λ in copper (for T = 300K and B

_o

= Tesla). 1

ϑ R

v

e

ˆx

ˆy ˆ ,

_o

ˆ

z B = B z C = 2 π R

m

e

⁻

(3)

Where

_mfp ^c ^e ₂ ^e

e

m v n e

λ = σ . For copper σ

_c^cu

5.95 10 ×

⁷

Siemens and n

_e^cu

8.5 10 ×

²⁸

m

³

with

_e

³

^B

1.2 10

⁵

e

v k T m s

= m × 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 = B z

_o

ˆ , no net energy is deposited and/or removed from the metal conductor by the constant magnetic field.

All a constant magnetic field B = B z

_o

ˆ does is change/rearrange the nature of the six-dimensional phase space associated with the free electrons ( ^{r p}

^e

^,

^e

) ⁼ ( ^{x y z p p} ^{, , ,}

^x

^,

^y

^, ^p

^z

) by introducing

correlations in (x & y), and (p

_x

& p

_y

) 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 n

_e^cu

8.5 10 ×

²⁸

m

³

. If the copper metal sheet lies in the x–y plane, with B

_ext

= B z

_o

ˆ , then the net magnetic dipole moment is

( )

1

ˆ

Ne

net e net

i

m m m z

=

= ∑ = − with macroscopic magnetization Μ = m

net

volume = m

net

( ) − z ^ˆ volume 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:

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Again, there are NO power losses here with a static magnetic field because B does NO work if B = constant.

However for time-varying magnetic fields, because of Faraday’s Law: ( ) ( ) ^,

, B r t

E r t

t

∇× = − ∂

∂

or:

_S

^{E r t da} ( ) ^,

_C

^{E r t d} ( ) ^,

_S

^{B r t} ( ) ^, ^da

^m

( ) ^t ^mf ( ) ^t

t t ε ε

⊥ ⊥ ⊥ ⊥

∂ ∂Φ

∇× = = − = − =

∂ ∂

∫ ⁱ ∫ ⁱ ∫ ⁱ

a time-varying magnetic field ^{B r t} ( ) ^,

t

∂

∂ induces an electric field E r t which (depending on ( ) ^,

the sign of ^{B r t} ( ) ^,

t

∂

∂ (increasing/decreasing) either accelerates/decelerates the free electrons in the conducting metal.

⇒ If ( ) ( ) ( ) ( )

ˆ and

^o

ˆ

o

B t B t

B t B t z z

t t

∂ ∂

= =

∂ ∂ and for simplicity, if v

_e

is in the x–y plane, i.e. 0

ez

v = then the free electron’s cyclotron orbit radius, R remains constant:

( ( ) )

constant ( )

e e e e

d v t

m v m dt

R eB dB t

e dt

= = or:

e

( ) ( )

e

d v t dB t

m e

dt = dt

The (average) kinetic energy gain (or loss) per cyclotron orbit is:

( ) ^* ( ( ) ) ^* ( ( ) )

^m

^{( )} ^{( )}

e e e

t B t

KE t q mf t e mf t e e A

t t

ε ε ε ε ^∂Φ ^∂

_⊥

Δ = = − = + =

∂ ∂ where A

_⊥

= π R

²

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: ( )

_e

( ) ( ) ^* ( )

²

e

rev rev

e B t R

KE t t

P t

π

τ τ

Δ ∂ ∂

= =

but:

_rev

²

e e

C R

v v

τ = π

Thus: ( )

( ) *

e

e B t P t t

∂ π

= ∂

R

2

( )

2 π R

( )

1

e

2

e

v e v R B t t

= ∂

∂ But: 1

2 ˆ

e e

m = − e v Rz (from above)

Therefore: ( ) ¹ ( ) ( ) ( ) ( )

2

mag

e e

e e e

m B t

B t B t U t

P t e v R m

t t t t

⎡ ⎤

∂ ∂ ∂ − ⎣ ⎦ ∂

= = − = =

∂ ∂ ∂ ∂

i i

Note: By Lenz’s Law, P is always positive for diamagnetic conducting materials. Then the

_e

macroscopic power: P

macro

( ) t = N P t

e e

( ) = n

e

^* volume P t ^*

e

( ) corresponds to the macroscopic

induced EMF ε

_macro

and (net) macroscopic current, I

macro

flowing in the conducting material!!!

(5)

Then we see that P

macro

( ) t = ε

macro

( ) t I

macro

( ) t = I

macro²

( ) t R

material

= ε

macro²

( ) 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

EMF ( )

m

( ) ( )

macro

d t B t

t A

t t

ε = − ^Φ = − ^∂

_⊥

∂ ∂ are known as Eddy currents.

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 B t

^o

ˆ

t t z

∂ ∂

↑ =

∂ ∂

increasing (here)

(6)

For a Solid-Core Transformer:

The induced macroscopic

core^solid

( )

m

( ) ( )

core^solid macro

Eddy

t B t

mf t A

t t

ε ε = − ^∂Φ = − ^∂

_⊥

∂ ∂ where ^A

⊥^solid^core

= ( ^{L W} ^* )

The power dissipation in the solid core of this transformer is:

( ) ( ) ( ) ( )

2

( ) ( )

² 2

2

solid core core macro

solid solid solid solid

core core core core Eddy

macro macro macro macro solid

Eddy Eddy Eddy Eddy core solid solid

core core

B t A

t t

P t t I t I t R

R R

ε

^⊥

∂

⎛ ⎞

⎜ ∂ ⎟

⎝ ⎠

= = = =

( )

² 2 4core

( )

²

solid core macro

Eddy solid solid

core core

B t B t

A L

t t

P t

R R

⊥

∂ ∂

⎛ ⎞ ⎛ ⎞

⎜ ∂ ⎟ ⎜ ∂ ⎟

⎝ ⎠ ⎝ ⎠

= = if

²

solid

A

_⊥core

= L (i.e. L = W for a square core)

For a Laminated Core Transformer:

Divide the solid core into N

_lam

individual laminations, each of width Δ W . 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

Δ . W

Each (now insulated) lamination has a cross-sectional area

Lam

1

core Lam

A L W A

⊥

= Δ = N

⊥

The resistance of each lamination is now increased relative to the resistance of the whole core:

Lam Lam core

R = N R (i.e. R

_core

= R

_Lam

N

_Lam

- for N

_Lam

laminations electrically connected in parallel).

( ) ( )

B t B t

^o

ˆ

t t z

∂ ∂

↑ =

∂ ∂

increasing (here)

(7)

The induced macroscopic EMF associated with each lamination is:

Lam

( )

m

( ) ( )

Lam

macro Eddy

t B t

mf t A

t t

ε ε = − ^∂Φ = − ^∂

_⊥

∂ ∂ where A

_⊥^Lam

= L ^* Δ = W L ^* ( W N

Lam

) = A

_⊥^core^solid

^/ N

Lam

The corresponding Eddy current power loss associated with each lamination is:

( ) ( ) ( ) ( ) ( )

( )

2 2 2

2 2 2 2

2

3 3

1 1

solid solid

Lam core core

Lam macro Lam Eddy

Lam solid

Lam Lam Lam solid Lam solid Lam core

core core

B t B t B t

A A N A

t t t t

P t P t

R R N R N R N

ε

⊥ ⊥ ⊥

∂ ∂ ∂

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ∂ ⎟ ⎜ ∂ ⎟ ⎜ ∂ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= = = = =

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.:

( ) ( )

( )

² 2

2 2

1 1

solid core Tot

Lam Lam Lam solid

core Lam solid Lam core

core

B t A

P t N P t t P

N R N

⊥

∂

⎛ ⎞

⎜ ∂ ⎟

⎝ ⎠

= = =

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 R

_iron

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 magneto- resistive 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.

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Energy Stored in Magnetic Fields

In electrical circuits containing (one or more) inductors, work must be done against the back ε mf ( )

m

( )

free

( )

L

I t

t t L

t t

ε = − ^∂Φ = − ^∂

∂ ∂ in order to get a free current, I

free

to flow in the circuit.

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:

_Tot

0 V

C

E d

Δ = ∫ ⁱ = ⇐ Kirchoff’s Voltage Law: The sum of potential differences around a closed circuit (mesh) = 0, i.e.

1

0

N i i

V

=

∑ Δ = ^.

⇒ E is a conservative field associated with a conservative force F qE =

Then: Δ V

battery

+ Δ V

inductor

( ) t + Δ V

resistor

( ) t = 0 Where: Δ V

battery

= ( ε

_o

− 0 ) = constant, ^≠ ^{fcn t} ( )

Δ V

inductor

( ) t = ( V

R

( ) t − ε

o

)

Δ V

resistor

( ) t = ( 0 − V

_R

( ) t )

Then: ( ) ( )

( ) ( ( ) ) ( ^{( )} )

battery inductor resistor

0 0 0 0

o R o R

V V t V t

V t V t

ε ε

Δ + Δ +Δ =

= − + − + − =

or: ( ( ) )

( )

( ) ( )

( )

( ) ( )

resistor inductor

0

(by Ohm's Law)

free L

o R R

V t

I t R

t Back mf

V t V t

ε ε

=Δ =Δ

=− =

= − − +

(9)

∴ ε

0

= − ε

L

( ) t + I

free

( ) t R = constant, but ( )

free

( )

L

I t

t L

ε = − ^∂ t

∂

∴

0

( ) ( )

free

I t

L I t R

ε = + ^∂ t +

∂

or:

^free

( )

free

( )

o

I t

L RI t

t ε

∂ + =

∂ ⇐ 1

^st

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:

^free

( )

free

( ) ⁰

I t

L RI t

t

∂ + =

∂ ⇒

^free

( )

free

( ) ⁰

dI t

L I t

r dt

⎛ ⎞ + =

⎜ ⎟ ⎝ ⎠ or:

^free

( )

free

( )

dI t R

I t

dt t

= −⎜ ⎟ ⎛ ⎞ ⎝ ⎠

The general solution to this homogeneous differential equation is of the form:

I

free

( ) t = I

o^free

e

⁻^t^τ

+ where C = constant of integration. C Then:

free

( ) 1

free ^t

o

dI t

I e dt

τ

= −

−

⇒ L

τ _{≡ ⎜ ⎟} ^{⎛ ⎞} R

⎝ ⎠ = characteristic time constant (SI units = seconds) Now solve the inhomogeneous differential equation:

^free

( )

free

( )

o

I t

L RI t

t ε

∂ + =

∂ but I

free

( ) t = I

o^free

e

⁻^t^τ

+ C Thus: ^L

− R R

L

⎛ ⎞

⎜ ⎟

⎝ ⎠

⁰ ⁰

t t

I e I e C

o

R

τ τ

ε

− −

⎛ ⎞

+ + =

⎜ ⎟

⎝ ⎠ ⇒ integration constant C

^o

R

= ε

∴ The solution of this inhomogeneous differential equation is: I

free

( ) t I

o^free

e

^t ^o

R

τ

ε

=

−

+

However we don’t (yet) know explicitly what I

_o^free

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 ( )

⁰

1

0

^free ^o ^free ^o

0

free o o

I t I e I

R R

ε ε

−

=

= = + = + = ⇒ I

o^free ^o

R

= − ε

∴ The specific solution to the inhomogeneous differential equation (here) is:

^I

^free

( ) ^t ^{= −} ^ε _R

^o

^e

⁻^t^τ

⁺ ^ε _R

^o

⁼ ^ε _R

^o

( ¹ ⁻ ^e

⁻^t^τ

)

(10)

Thus, for this circuit, the free current flowing in this circuit as a function of time is:

^I

^free

( ) ^t ⁼ ^⎛ ^⎜ _⎝ ^ε _R

^o

^⎞ ^⎟ _⎠ ( ¹ ⁻ ^e

⁻^t^τ

) ^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…)

0 1 τ 2 τ 3 τ

The voltage (aka potential difference) across the resistor, R vs. time, t:

V

_R

( ) ( ) t I t * (by Ohm's Law) R

^o

R

Δ = = ε * R ( ¹ ⁻ ^e

⁻^t^τ

)

^Δ ^V

^R

( ) ^t ⁼ ^ε

⁰

( ¹ ⁻ ^e

⁻^t^τ

) ^where τ ≡ ^{( )} ^L R

The voltage (aka potential difference) across the inductor, L vs. time, t:

Since: ^I

^free

( ) ^t ⁼ ^⎛ ^⎜ _⎝ ^ε _R

^o

^⎞ ^⎟ _⎠ ( ¹ ⁻ ^e

⁻^t^τ

) ^where τ ≡ ^{( )} ^L R Thus:

( ) ( ) ( )

inductor

free L

I t

V t t L L

ε ^∂ t

Δ = − = + =

∂

o

R

R L

⎛ ε ⎞

− ⎜ ⎟

⎝ ⎠

⎛ ⎞

⎜ ⎟

⎝ ⎠

t t

e

⁻^τ

ε

o

e

⁻ ^τ

⎛ ⎞

⎜ ⎟ = −

⎜ ⎟

⎝ ⎠ i.e.

inductor

( )

t

V t ε

o

e

⁻ ^τ

Δ = −

1 τ 2 τ 3 τ t 0

-5.0% ε

_o

-13.5% ε

_o

-36.8% ε

_o

( )

inductor

V t

Δ ← = − = ε

₀

back ε mf !! ⇒

At t = 0 the back ε mf across the inductor

(= − ) exactly cancels V ε

_o battery

= ε

₀

, but

inductor can’t sustain opposing it forever!

(11)

Kirchoff’s Voltage Law: Δ V

battery

( ) t = −Δ V

inductor

( ) t − Δ V

resistor

( ) t (Volts) ε

_o

^{= +} ^ε

^o

^e

⁻^t^τ

⁺ ^ε

^o

( ¹ ⁻ ^e

⁻^t^τ

) = (Volts) ^ε

^o

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:

( ) ( ) ( ) ( )

free

( )

L L free free

dI t

P t t I t LI t

ε dt

= − = + (Watts)

The instantaneous electrical energy stored in the inductor is:

( )

₀

( )

₀

( ) ( )

₀

( ) ( ) ¹

²

( )

2

t t free t

L free free free free

dI t

W t P t dt LI t dt L I t dI t LI t

dt

′ ′ ′ ′ ′ ′

= = = =

∫ ∫ ′ ∫

( ) ¹

²

( )

L

2

free

W t = LI t (Joules) {n.b. analog of ( ) ¹

²

( )

C

2 W t = C V Δ t for capacitors!}

Thus, we see that the instantaneous electrical power stored in the inductor is:

( )

L

( ) ( ) ( ) ( )

free

( )

L L free free

dI t

P t dW t t I t LI t

dt ε dt

= = − = + (Watts)

Recall that the magnetic flux in an inductor is: Φ

m

( ) t = LI

free

( ) t (Webers or Tesla-m

²

) However: ^Φ

^m

( ) ^t ⁼ _∫

_S_⊥

^{B r t da} ( ) ^, ⁱ

^⊥

⁼ _∫

_S_⊥

( ^∇× ^{A r t} ( ) ^, ) ⁱ ^da

^⊥

⁼ _∫

_C

^{A r t d} ^{( )} ^, ⁱ ⁼ ^LI

^free

^{( )} ^t

Then:

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ( ) ) ( ^{( )} ^{( )} )

1

2

1 1

2 2 2 ,

1 1 1

, , ,

2 2 2

L free free free C free

free C C free C free

W t LI t LI t I t A r t d I t

I t A r t d A r t I t d A r t I t d

⎡ ⎤

= = ⎣ ⎦ = ⎣ ⎦

⎡ ⎤

= ⎣ ⎦ = =

∫

∫ ∫ ∫

i

i i i

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 ^{( )} ^{( )} ^,

2 2

mag free free

C C

W t = ∫ A r t I ⁱ r t d = I t ∫ A r t d ⁱ (Joules)

( ) ¹ ( ( ) ^, ( ) ^, )

mag

2

free

W t

S

A r t K r t da

⊥ ⊥

= ∫ ⁱ (Joules)

( ) ¹ ( ( ) ^, ( ) ^, )

mag

2

free

W t = ∫

v

A r t J ⁱ r t d τ (Joules)

Note from the last formula above that the energy density associated with an inductor is:

( ) ¹ ( ) ^, ( ) ^,

mag

2

free

u t = A r t J i r t (Joules/m

³

)

(12)

Using Ampere’s law (in differential form): ∇× B r t ( ) ^, = μ

o

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

( ) ^, ( ) ^, ( ) ^,

Tot free Bound

J r t = J r t + J r t .

Then we can more generally write ( ) ¹ ( ( ) ^, ( ) ^, )

mag

2

free

W t = ∫

v

A r t J ⁱ r t d τ ^as

( ) ¹ ( ( ) ^, ( ) ^, ) ¹ ( ^{( )} ^, ^{( )} ^, )

2 2

mag Tot

v v

o

W t A r t J r t d τ A r t B r t d τ

= ∫ ⁱ = μ ∫ ⁱ ∇×

Integrate the RHS term of the above relation by parts, and use the relation:

^∇ ⁱ ( ^{A B} ^× ) ( ⁼ ^B ⁱ ^∇× ^A ) ⁻ ^A ⁱ ( ^∇× ^B )

i.e. ^A ⁱ ( ^∇× ^B ) ⁼ ^B ⁱ ( ^∇× ^A ) ( ^{− ∇} ⁱ ^{A B} ^× ) ⁼ ^{B B} ⁱ ^{− ∇} ⁱ ( ^{A B} ^× ) ⁼ ^B

²

^{− ∇} ⁱ ( ^{A B} ^× )

∴ ( ) ¹

²

( ) ^, ( ( ) ( ) ^, ^, )

mag

2

v v

o

W t B r t d τ A r t B r t d τ

μ ^⎡ ^⎤

= ⎣ ∫ − ∇ ∫ ⁱ × ⎦

Using the divergence theorem on the 2

^nd

term:

( ) ¹

²

( ) ^, ( ( ) ( ) ^, ^, )

mag

2

v S

o

W t B r t d τ A r t B r t da

μ ^⎡ ^⎤

= ⎣ ∫ − ∫ × ⁱ ⎦ (n.b. surface S encloses volume v)

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. _∫

_S_∞

( ^{A r t} ( ) ( ) ^, ^× ^{B r t} ^, ) ⁱ ^da ⁼ ⁰ ^.

Then the magnetic energy associated with any system is given by:

^mag

( )

^all ^mag

( ) ^, ¹ ₂

^all

( ( ) ^,

^Tot

( ) ^, )

space space

W t = ∫ u r t d τ = ∫ A r t J ⁱ r t d τ (Joules) or equivalently by:

( ) ( ) ( )

( ) ( )

( ) ( ^{( ) ( )} )

2

, 1 ,

2 1 1

, , , ,

2 2

mag all mag all

space o space

all all

space space

o

W t u r t d B r t d

B r t B r t d B r t H r t d

τ τ

μ

τ τ

μ

= =

∫ ∫

∫ ⁱ ∫ ⁱ (Joules)

The corresponding magnetic energy density is given by:

( ) ^, ¹ ( ) ^, ( ) ^,

mag

2

Tot

u r t = A r t J i r t (Joules/m

³

) or equivalently by:

( ) ^, ¹

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

mag

o o

u r t B r t B r t B r t B r t H r t

μ μ

= = i = i (Joules/m

³

)

(13)

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

( ( ) ( ) ^, ^, )

elect all elect

2

all Tot

space space

W t = ∫ u r t d τ = ∫ ρ r t V r t d τ (Joules) or equivalently by:

( ) ( ) ( )

( ) ( )

( ) ( ^{( ) ( )} )

2

, 1 ,

2 1 1

, , , ,

2 2

elect all elect o all

space space

o all all

space space

W t u r t d E r t d

E r t E r t d E r t D r t d

τ ε τ

ε τ τ

= =

∫ ∫

∫ ⁱ ∫ ⁱ (Joules)

The corresponding electric energy density is given by:

( ) ^, ¹ ( ) ( ) ^, ^,

elect

2

Tot

u r t = ρ r t V r t (Joules/m

³

) or equivalently by:

( ) ^, ¹

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

elect o o

u r t = ε E r t = ε E r t E r t i = E r t D r t i (Joules/m

³

)

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, μ

_o

(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

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

dielectric elect

u r t = ε E r t = ε E r t E r t i = E r t D r t i (Joules/m

³

) ( ) ^, ¹

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

magnetic

u

mag

r t B r t B r t B r t B r t H r t

μ μ

= = i = i (Joules/m

³

)

The corresponding electric and magnetic energies associated with the volume v (only) of the linear media are:

( ) ( ) ( )

( ) ( )

( ) ( ^{( ) ( )} )

1

2

, ,

2 1 1

, , , ,

2 2

dielectric dielectric

elect v elect v

v v

W t u r t d E r t d

E r t E r t d E r t D r t d

τ ε τ

ε τ τ

= =

∫ ∫

∫ ⁱ ∫ ⁱ (Joules)

( ) ( ) ( )

( ) ( )

( ) ( ^{( ) ( )} )

1

2

, ,

2 1 1

, , , ,

2 2

magnetic magnetic

mag v mag v

v v

W t u r t d B r t d

B r t B r t d B r t H r t d

τ τ

μ

τ τ

μ

= =

∫ ∫

∫ ⁱ ∫ ⁱ (Joules)

The total electric/magnetic energy densities and electric/magnetic energies (i.e. integrated over

all space must (obviously) use ε

_o

and μ

_o

for these integrals in the region(s) outside of the

volume v of the media.

(14)

Griffiths Example 7.13:

A long coaxial cable carries a steady free current I

free

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:

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 ∫

C

B r d ( ) ⁱ = μ

o Tot

I

^enclosed

to determine B inside and outside the cable, take contour(s) as shown above.

For ρ < a I ,

_Tot^enclosed

= therefore 0 ^B ( ^ρ ^< ^a ) ^{= .} ⁰

For a ≤ < ρ b I ,

_Tot^enclosed

= I

_free

therefore ( ) ^ˆ

2

o

B a ρ b μ I

free

ϕ

≤ ≤ = πρ .

For ρ > b I ,

_Tot^enclosed

( net ) = therefore 0 ^B ( ^ρ ^> ^b ) ^{= .} ⁰

Since the magnetic field ( ) ^ˆ

2

o

B a ρ b μ I

free

ϕ

≤ ≤ = πρ is non-zero only in the region a ≤ < , then ρ b the magnetic energy density will also be non-zero only in this same region:

( )

²

( ) ( ) ( )

2

2 2

1 1

2 2

1 2 2 8

mag

o o

free free

o

u B B B

I I

ρ ρ ρ ρ

μ μ

μ πρ π ρ

= =

⎡ ⎤

= ⎢ ⎥ =

⎣ ⎦

i

(Joules/m

³

)

The magnetic energy associated with a finite length of the coaxial cable is:

( )

²

( )

2 2

0 0

1 1

2 2

2

z b

mag v z a

o o

W B r d dz

^ρ

d

^{ϕ π}

B r d

ρ ϕ

τ ρ ρ ϕ

μ μ

= = =

= =

=

∫ ∫ ∫ ∫

π

²

8

o

I

free

μ π

2

ρ

2

d ρ

ρ ₄

²

₄

²

^ln ( )

b b

o o

free free

a a

d b

I I

a

ρ ρ

μ ρ μ

π ρ π

= =

=

∫ ∫

ˆz I

free

I

free

a

b ρ ^C

( )

B ρ

ρ

a b

( )

u

mag

ρ

a b

(15)

Thus: ^W

^mag

₄

^o

^I

²^free

^ln ( ) ^b a μ

= π (Joules)

The magnetic energy per unit length associated with this coaxial cable is:

^w

^mag

^W

^mag

₄

^o

^I

²^free

^ln ( ) ^b a μ

≡ = π (Joules/meter)

Note that the magnetic energy in this coaxial cable is also 1

²

mag L

2

free

W = W = LI , thus we see that the inductance L associated with a finite length of this cable is:

^W

^mag

⁼ ^W

^L

⁼ ¹ ₂ ^LI

²^free

⁼ ₄ ^μ _π

^o

^I

²^free

^ln ( ) ^b _a ^⇒ ^L ⁼ ₂ ^μ _π

^o

^ln ( ) ^b _a (Henrys).

The inductance per unit length of this coaxial cable is therefore:

^L ^{≡ =} ^L ₂ ^μ _π

^o

^ln ( ) ^b _a (Henrys/meter).

Note that the inductance per unit length L ≡ L has the same SI units (Henrys/meter) as the magnetic permeability of free space μ

_o

.

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 N

_Tot

turns (thus n = N

_Tot

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 parallel- plate capacitor?

Cross-Sectional View of Long Solenoid:

Δ z Magnetic Core

{partially inserted into bore of long solenoid}

Δ z

0 z = z = z

ˆz

z

(16)

The infinitesimal difference in the magnetic energy associated with moving the magnetic core inward into the bore of the long solenoid an infinitesimal distance Δ (ignoring/neglecting any/all z end effects) is given by:

( ) ( ) ( ) (

¹2

)

²

( )

in

mag mag mag o v sol

W z W z z W z μ μ H r d τ

Δ Δ = + Δ − − ∫

Where (using Ampere’s circuital law for the H -field for the long solenoid): H

solⁱⁿ

( ) r = nI

free

z ^ˆ The infinitesimal change in the volume due to the infinitesimal displacement z Δ is V Δ = A

_⊥

Δ . z ∴ Δ W

mag

( ) ( Δ z

¹2

μ μ −

o

) n I

² ²free

A

_⊥

Δ z

Then the net force acting on the magnetic core is:

( ) ( ) ( )

¹2

( )

² ²

constant constant

ˆ ˆ ˆ

free free

mag mag

mag o free

I I

W z dW z

F z z z n I A z

z dz μ μ

_⊥

= =

Δ Δ Δ

Δ = = −

Δ

Thus, as long as ( μ μ −

o

) > ⁰ (i.e. paramagnetic materials) we see that the direction of the magnetic force acting on the magnetic core ( ) ⁺ ^ˆz is such that it wants to suck the magnetic core 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 ( μ

dia

− μ

o

) < 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 u

elect

( ) r (SI units of Joule/m

³

) also has the same units as that for pressure, ^{P r} ( ) (Newtons/m

²

), 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

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

elect elect o o

u r t = P r t = ε E r t = ε E r t E r t i = E r t D r t i (J/m

³

or N/m

²

) ( ) ^, ( ) ^, ¹

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

mag mag

o o

u r t P r t B r t B r t B r t B r t H r t

μ μ

= = = i = i (J/m

³

or N/m

²

)

If there are linear dielectrics and/or linear magnetic materials present, then:

( ) ^, ( ) ^, ¹

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

dielectric dielectric

elect elect

u r t = P r t = ε E r t = ε E r t E r t i = E r t D r t i (J/m

³

or N/m

²

)

( ) ^, ( ) ^, ¹

²

( ) ^, ¹ ( ) ( ) ^, ^, ¹ ( ) ( ) ^, ^,

2 2 2

magnetic magnetic

mag mag

u r t P r t B r t B r t B r t B r t H r t

μ μ

= = = i = i (J/m

³

or N/m

²

)

(17)

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.

From Ampere’s circuital law: B

solⁱⁿ

( ) r = μ

o

nI

free

z ^ˆ . Note that:

( ) ( ) ¹

²

( ) ¹

² ²

^constant ⁰

2 2

inside inside in

sol sol sol o free

o

u r P r B r μ n I

= = μ = = > (i.e. is a positive quantity).

Cross-Sectional View of Long Air-Core Solenoid:

The magnetic pressure/magnetic energy density inside the solenoid pushes/exerts a radial- outward force on the solenoid!

The net magnetic force acting on the long air-core solenoid of radius R and length is:

( ) ^ˆ ¹

² ²

² ^ˆ

² ²

^ˆ

2

net inside

sol sol sol o free o free

F = P r = R A ρ = μ n I ⋅ π ρ μ π R = n I R ρ (radially outward) Example: The Magnetic Pressure Inside an Air-Core Toroid

Similar to the long, air-core solenoid, except ( ) ^ˆ

2

tot free

inside o

toroid

B r μ N I ϕ

π ρ

= for a ≤ ≤ . ρ b

Then: ( ) ( ) ¹

²

( )

₂ ² ₂²

^constant ⁰

2 8

Tot free

inside inside in o

toroid toroid toroid

o

N I

u r P r B r μ

μ π ρ

= = = ≠ >

@ inner radius: ( ) ( )

²

( )

2 ² 2²

1 2 8

Tot free

o

N I

u a P a B a

a

ρ ρ ρ μ

μ π

= = = = = =

@ outer radius: ( ) ( )

²

( )

2 ² 2²

1 2 8

Tot free

o

N I

u b P b B b

b

ρ ρ ρ μ

μ π

= = = = = =

( ) ^ˆ

in

B

sol

r z points out of page:

( ) ( ) ¹

²

( ) ¹

² ²

^constant ⁰

2 2

in in in

sol sol sol o free

o

u r P r B r μ n I

= = μ = = >

points radially outward

( ) ( ) ⁰

outside outside

mag mag

u r = P r =

Force at inner radius is radially

inward

Force at outer radius is radially

outward

(18)

Example: The Magnetic Pressure Associated with a Thin Cylindrical Current-Carrying Tube A steady free sheet current K

free

( ) r = K z

o

^ˆ (Amperes/meter) flows down the surface of a long, very thin cylindrical conducting metal tube of radius a as shown in the figure below:

The (total) free current flowing down the surface of the conducting tube is:

( ) ^ˆ ^ˆ ^ˆ

₀²

²

K K K

free C free C o C o o o o

I = ∫ K r d ⁱ

_⊥

= ∫ K z d ⁱ

_⊥

= ∫ K z ad z ⁱ ϕ = K a ∫

^π

d ϕ = π aK ≡ I {Since d

_⊥

= ad z ϕ ˆ using the arc length formula " S = R θ " }

Then using Ampere’s law for B : ( )

B

enclosed o Tot C

B r d = μ I

∫ ⁱ we see that:

For ρ < a I :

_Tot^enclosed

= ⇒ 0 ^B ( ^ρ ^< ^a ) ⁼ ⁰

⇒ ( ) ( ) ¹

²

( ) ⁰

2

tube tube

mag mag tube

o

u ρ a P ρ a B ρ a

< = < = μ < =

For ρ ≥ a I :

_Tot^enclosed

= ⇒ I

_o

( ) ^ˆ

2

o

I

o

B ρ a μ ϕ

≥ = π ρ

⇒ ( ) ( )

²

( )

2 ²2

1 2 8

tube tube o o

mag mag tube

o

u ρ a P ρ a B ρ a μ I

μ π 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.

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 ϕ

LECTURE NOTES 23 Eddy Currents in Conductors