Electric Charge Conservation

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LECTURE NOTES 1 CONSERVATION LAWS

Conservation of energy E, linear momentum p , angular momentum L and electric charge q are (all) of fundamental importance in electrodynamics (n.b. this is also true for all of the fundamental forces of nature, and thus is true for weak, strong, EM and gravitational charges {= mass, m for gravity}) – these are all absolutely conserved quantities (and separately so), both microscopically and hence macroscopically, as well as locally and globally (i.e. the entire universe)!

Electric Charge Conservation

Previously (i.e. last semester in P435), we discussed electric charge conservation:

Electric charge contained in volume v at time t: Q

free

( ) t = ∫

v

ρ

free

( ) r t d ^, τ Electric current flowing out of volume v

through closed bounding surface S at time t:

free

( )

free

( ) ^,

I t = ∫

S

J r t da ⁱ

{Global} Conservation of electric charge:

free

( )

^free

( )

free

( ) ^,

S

dQ t

I t J r t da

= dt = − ∫ ⁱ

However:

free

( )

free

( ) ^,

v

dQ t r t

dt t d

ρ τ

= ∂

∫ ∂ . Using the divergence theorem on RHS of above eqn.

we obtain:

^free

( ) ^,

free

( ) ^,

v v

r t d J r t d

t

ρ τ τ

∂ = − ∇

∫ ∂ ∫ ⁱ ⇐ Integral form of the Continuity Equation.

This relation must hold for any arbitrary volume v associated with the enclosing surface S;

hence the integrands in the above equation must be equal – and thus we obtain the Continuity Equation (in differential form) expressing {local} electric charge conservation :

( ) ^, ( ) ^,

free

r t J r t

t ρ

∂ = −∇

∂ i ⇐ Differential form of the Continuity Equation.

n.b. The continuity equation doesn’t explain / shed any light on why electric charge is conserved – it just explains how electric charge is conserved!!

Volume, v

Enclosing surface, S ρfree

J

free Surface area element,

da

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Poynting’s Theorem and Poynting’s Vector S We know that the work required to assemble a static charge distribution is:

( ) ₂

^o ²

( ) ^, ₂

^o

( ( ) ( ) ^, ^, ) ¹ ₂ ( ^{( ) ( )} ^, ^, )

E v v v

W t = ε ∫ E r t d τ = ε ∫ E r t E r t ⁱ d τ = ∫ D r t E r t ⁱ d τ Likewise, the work required to get electric currents flowing, e.g. against a back EMF is:

( ) ₂ ¹

²

( ) ^, ₂ ¹ ( ( ) ( ) ^, ^, ) ¹ ₂ ( ^{( ) ( )} ^, ^, )

M v v v

o o

W t B r t d τ B r t B r t d τ H r t B r t d τ

μ μ

= ∫ = ∫ ⁱ = ∫ ⁱ

Thus the total energy, U

EM

stored in EM field(s) is (by energy conservation) = total work done:

( ) ( ) ( ) ( ) ( ) ¹

²

( ) ^, ¹

²

( ) ^, ( ) ^,

EM tot EM E M

2

o EM

v v

o

U t W t W t W t W t ε E r t B r t d τ u r t d τ

μ

⎛ ⎞

= = = + = ⎜ + ⎟ =

⎝ ⎠

∫ ∫

( ) ( ) ^, ¹

²

( ) ^, ¹

²

( ) ^,

2

EM v EM v o

o

U t u r t d τ ε E r t B r t d τ

μ

⎛ ⎞

= = ⎜ + ⎟

⎝ ⎠

∫ ∫

Where u

_EM

= total energy density: ( ) ^, ¹

²

( ) ^, ¹

²

( ) ^,

2

EM o

o

u r t ε E r t B r t

μ

⎛ ⎞

= ⎜ + ⎟

⎝ ⎠ (SI units: Joules/m

³

)

Suppose we have some charge density ^ρ ( ) ^{r t} ^, and current density J r t configuration(s) ( ) ^,

that at time t produce EM fields E r t and ( ) ^, B r t . In the next instant dt, i.e. at time t + dt, the ( ) ^,

charge moves around. What is the amount of infinitesimal work dW done by EM forces acting on these charges / currents, in the time interval dt ?

The Lorentz Force Law is: ^{F r t} ( ) ^, ⁼ ^{q E r t} ( ( ) ( ) ( ) ^, ⁺ ^{v r t} ^, ^× ^{B r t} ^, )

The infinitesimal amount of work dW done on an electric charge q moving an infinitesimal distance d = vdt in an infinitesimal time interval dt is:

( ) (

^{. . to}^!!!

)

0

n b v

dW F d q E v B d qE vdt q v B vdt

⊥

=

= i = + × i = i + × i = qE vdt i

But: q

free

( ) r t ^, = ρ

free

( ) r t d ^, τ and: ρ

free

( ) ( ) r t v r t ^, ^, = J

free

( ) r t ^,

(n.b. magnetic forces do no work!!)

SI units:

Joules SI units:

Joules

SI units:

Joules Linear Dielectric Media

Linear Magnetic Media

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The (instantaneous) rate at which (total) work is done on all of the electric charges within the volume v is:

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

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( )

, , , , , , ,

, , , using : , ,

, , , but :

v v v free

free free free

v

free free

v

dW t F r t d r t dt F r t v r t q r t E r t v r t dt

r t d E r t v r t q r t r t d

E r t r t v r t d J r

ρ τ ρ τ

ρ τ

= = =

= =

=

∫ ∫ ∫

∫

i i i

i

i ( ) ^, t = ρ

free

( ) ( ) r t v r t ^, ^,

∴ ^{dW t} ( )

_v

( ^{E r t J} ( ) ^,

^free

( ) ^{r t} ^, ) ^d ^{P t} ^{( )}

dt = ∫ ⁱ τ = = instantaneous power (SI units: Watts) The quantity E r t J ( ) ^, i

free

( ) r t ^, is the (instantaneous) work done per unit time, per unit volume – i.e. the instantaneous power delivered per unit volume (aka the power density).

Thus: ^{P t} ( ) ^{dW t} ( )

_v

( ^{E r t J} ( ) ^,

^free

( ) ^{r t} ^, ) ^d

dt τ

= = ∫ ⁱ (SI units: Watts = Joules

sec ) We can express the quantity ( ^{E J} ⁱ

^free

) in terms of the EM fields (alone) using the Ampere- Maxwell law (in differential form) to eliminate J

_free

.

Ampere’s Law with Maxwell’s Displacement Current correction term (in differential form):

( ) ^,

^o

{

^free

( ) ^,

^D

( ) ^, }

^o ^free

^{( )} ^,

^o ^o

^{E r t} ^{( )} ^,

B r t J r t J r t J r t

μ μ μ ε ^∂ t

∇× = + = +

∂

Thus:

^free

( ) ^, ¹ ( ( ) ^, )

^o

^{( )} ^,

o

E r t

J r t B r t

ε t μ

= ∇× − ∂

∂

Then:

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

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

1 ,

, , , ,

1 ,

, , ,

free o

o

o o

E r t

E r t J r t E r t B r t

t E r t E r t B r t E r t

t μ ε

μ ε

⎧ ∂ ⎫

⎪ ⎪

= ⎨ ⎪ ⎩ ∇× − ∂ ⎬ ⎪ ⎭

= ∇× − ∂

∂

i i

Now: ^∇ ⁱ ( ^{E B} ^× ) ⁼ ^B ⁱ ( ^∇× Ê ) ( ⁻ Ê ⁱ ^∇× ^B ) Griffiths Product Rule #6 (see inside front cover) Thus: Ê ⁱ ( ^∇× ^B ) ( ⁼ ^B ⁱ ^∇× Ê ) ^{− ∇} ⁱ ⁽ ^{E B} ^× ⁾

But Faraday’s Law (in differential form) is: ( ) ( ) ^,

, B r t

E r t

t

∇× = − ∂

∂

∴ ^E ( ^B ) ^B ^B ( ^{E B} )

t

∇× = − ∂ − ∇ ×

i i ∂ i

However: ^B ^B ¹ ₂ ( ) ^{B B} ¹ ₂ ( ) ^B

²

t t t

∂ = ∂ = ∂

∂ ∂ ∂

i i and similarly: Ê Ê ¹ ₂ ( ) ^{E E} ¹ ₂ ( ) Ê

²

t t t

∂ = ∂ = ∂

∂ ∂ ∂

i i

(4)

Therefore:

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

( ) ( ) ( ( ) ( ) )

2 2

1 1 1

, , , , , ,

2 2

1 1 1

, , , ,

2

free o

o

o o

E r t J r t B r t E r t B r t E r t

t t

E r t B r t E r t B r t

t μ ε

ε μ μ

∂ ∂

⎧ ⎫ ⎧ ⎫

= − ⎨ ⎩ − ∂ − ∇ × ⎬ ⎭ − ⎨ ⎩ ∂ ⎬ ⎭

⎛ ⎞

= − ∂ ⎜ + ⎟ − ∇ ×

∂ ⎝ ⎠

i i

i Then:

( ) ( ) ( ( ) ( ) )

2 2

, ,

1 1 1

, , , ,

2

v free

v o v

o o

P t dW t E r t J r t d

dt

d E r t B r t d E r t B r t d

dt

τ

ε τ τ

μ μ

= =

⎛ ⎞

= − ⎜ + ⎟ − ∇ ×

⎝ ⎠

∫

∫ ∫

i

Apply the divergence theorem to this term, get:

Poynting’s Theorem = “Work-Energy” Theorem of Electrodynamics:

( ) ( )

_v

¹ ₂

^o ²

( ) ^, ¹

²

( ) ^, ¹

_S

( ( ) ( ) ^, ^, )

o o

dW t d

P t E r t B r t d E r t B r t da

dt dt ε τ

μ μ

⎧ ⎛ ⎞ ⎫

⎪ ⎪

= = − ⎨ ⎜ + ⎟ ⎬ − ×

⎪ ⎝ ⎠ ⎪

⎩ ⎭

∫ ∫ ⁱ

Physically, ¹

²

( ) ^, ¹

²

( ) ^,

2

^v

ε

^o

E r t

_o

B r t d τ μ

⎛ ⎞

⎜ + ⎟

⎝ ⎠

∫ = instantaneous energy stored in the EM fields

( ) ( )

( ^{E r t} ^{, and} ^{B r t} ^, ) within the volume v (SI units: Joules) Physically, the term ¹

_S

( ( ) ( ) ^, ^, )

o

E r t B r t da

− μ ∫ × ⁱ = instantaneous rate at which EM energy is carried / flows out of the volume v (carried microscopically by virtual (and/or real!) photons across the bounding/enclosing surface S by the EM fields E and B − i.e. this term represents/is the instantaneous EM power flowing across/through the bounding/enclosing surface S

(SI units: Watts = Joules sec ).

Poynting’s Theorem says that:

The instantaneous work done on the electric charges in the volume v by the EM force is equal to the decrease in the instantaneous energy stored in EM fields ( E and B ), minus the energy that is instantaneously flowing out of/through the bounding surface S .

We define Poynting’s vector: ( ) ^, ¹ ( ( ) ( ) ^, ^, )

o

S r t E r t B r t

≡ μ × = energy / unit time / unit area, transported by the EM fields ( E and B ) across/through the bounding surface S

n.b. Poynting’s vector S has SI units of Watts / m

²

– i.e. an energy flux density.

(5)

Thus, we see that: ^{P t} ( ) ^{dW t} ( ) ^dU

^EM

( ) ^t

_S

^{S r t da} ( ) ^,

dt dt

= = − − ∫ ⁱ

where ^{S r t da} ( ) ^, ⁱ = instantaneous power (energy per unit time) crossing/passing through an infinitesimal surface area element da = nda ˆ , as shown in the figure below:

ˆn ˆn = outward pointing unit normal vector (everywhere ⊥ to surface S )

da EM energy flowing out of volume v through enclosing surface S

ˆz

ˆy ϑ

ˆx

Volume v

Enclosing surface S Poynting’s vector:

¹

S ≡

_μo

E B × = Energy Flux Density (SI units: Watts / m

²

)

The work W done on the electrical charges contained within the volume v will increase their mechanical energy – kinetic and/or potential energy. Define the (instantaneous) mechanical energy density u

mech

( ) r t ^, such that:

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

mech

free

du r t

E r t J r t

dt = i Hence: ^dU _dt

^mech

⁼ _∫

_v

( ^{E r t J} ( ) ^, ⁱ

^free

( ) ^{r t} ^, ) ^d ^τ

Then: ^{P t} ( ) ^{dW t} ( ) ^dU

^mech

^d

_v

^u

^mech

( ) ^{r t d} ^,

_v

( ^{E r t J} ( ) ^,

^free

( ) ^{r t} ^, ) ^d

dt dt dt τ τ

= = = ∫ = ∫ ⁱ

However, the (instantaneous) EM field energy density is:

( ) ^, ¹

²

( ) ^, ¹

²

( ) ^,

2

EM o

o

u r t ε E r t B r t

μ

⎛ ⎞

= ⎜ + ⎟

⎝ ⎠ (Joules/m

³

)

Then the (instantaneous) EM field energy contained within the volume v is:

( ) ( ) ^,

EM v EM

U t = ∫ u r t d τ ^(Joules)

(6)

Thus, we see that: _dt ^d _∫

_v

( ^u

^mech

( ) ^{r t} ^, ⁺ ^u

^EM

( ) ^{r t} ^, ) ^d ^τ ^{= −} _∫

_S

^{S r t da} ( ) ^, ⁱ ^{= − ∇} _∫

_v

( ⁱ ^{S r t} ( ) ^, ) ^d ^τ

The integrands of LHS vs. {far} RHS of the above equation must be equal for each/every space- time point ( ) ^{r t} ^, within the source volume v associated with bounding surface S. Thus, we obtain:

The Differential Form of Poynting’s Theorem: u

mech

( ) r t ^, u

EM

( ) r t ^, S r t ( ) ^,

t

∂ ⎡ ⎣ + ⎤ ⎦ = −∇

∂ i

Poynting’s theorem = Energy Conservation “book-keeping” equation, c.f. with the Continuity equation = Charge Conservation “book-keeping” equation:

The Differential Form of the Continuity Equation: ( ) ^{r t} ^, ^{J r t} ( ) ^,

t ρ

∂ = −∇

∂ i

Since

^mech

( ) ^, ( ) ^,

free

( ) ^,

u r t

E r t J r t t

∂ =

∂ i , we can write the differential form of Poynting’s theorem as:

( ) ^,

free

( ) ^,

^EM

( ) ^, ( ) ^,

u r t

E r t J r t S r t

t

+ ∂ = −∇

i ∂ i

Or: ( ) ^,

free

( ) ^,

^EM

( ) ^, ( ) ^, ⁰

u r t

E r t J r t S r t

t

+ ∂ + ∇ =

i ∂ i

Poynting’s Theorem / Poynting’s vector S r t represents the (instantaneous) flow of EM ( ) ^,

energy in exactly the same/analogous way that the free current density J

free

( ) r t represents the ^, (instantaneous) flow of electric charge.

In the presence of linear dielectric / linear magnetic media, if one is ONLY interested in FREE charges and FREE currents, then:

^u

^EM^free

( ) ^{r t} ^, = ¹ ₂ ( E r t D r t ( ) ( ) ^, i ^, + B r t H r t ( ) ( ) ^, i ^, ) ^{D r t} ( ) ^, ⁼ ^ε ^{E r t} ( ) ^, ε ε =

o

( ¹ + χ

e

)

^{S r t} ( ) ^, =

μ¹

^{E r t} ( ) ( ) ^, × ^{B r t} ^, = ^{E r t} ( ) ^, × ^{H r t} ( ) ^, ^{B r t} ( ) ^, ⁼ ^μ ^{H r t} ( ) ^, μ μ =

o

( ¹ + χ

m

)

Using the Divergence

theorem

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Griffiths Example 8.1:

Poynting’s vector S , power dissipation and Joule heating of a long, current-carrying wire.

When a steady, free electrical current I (≠ function of time, t) flows down a long wire of length L a (a = radius of wire) and resistance ^R ( ⁼ ^L ^{π σ} ^a

² ^C

) , the electrical energy is dissipated as heat (i.e. thermal energy) in the wire.

Electrical power dissipation: P = Δ ⋅ = V I I R

²

I

Long wire of

resistance R

I

Battery L a

(or power supply) V

1

Δ V V

2

Free Current Density: J

_free

= σ

_C

E = ( I π a

²

) ( z ˆ Amps m

²

) Longitudinal Electric Field:

^free

ˆ Volts/m ( )

C

J V

E z

σ L

= = Δ

Potential Difference: Δ = − V V

1

V

2

0 ( ) ( > Volts )

n.b. The {steady} free current density J

_free

( = σ

_C

E = I π a

²

) and the longitudinal electric field

( ) ^ˆ

E = Δ V L z are uniform across (and along) the long wire, everywhere within the volume of the wire ( ^ρ ^< ^a ) ^. ⇒ Thus, this particular problem has no time-dependence…

From Ampere’s Law: ( )

₂

^ˆ

2

inside o

I

B a

a

ρ μ ρ ϕ

< = π ρ = x

²

+ y

²

in cylindrical coordinates

{ ∫

C

B r d ( ) ⁱ = μ

^{o encl}

I } ^B

^outside

⁽ ^ρ ^≥ ^a ⁾ ⁼ ₂ ^μ _πρ

^o

^I ^ϕ ^ˆ ^(Tesla)

n.b. for simplicity’s sake, we have approximated the finite length wire by an ∞-length wire.

This will have unphysical, but understandable consequences later on….

Poynting’s Vector: ( ) ¹ ( ) ( )

o

S r E r B r

= μ × ^B ( ) ^ρ

( ) ( ) ( )

ˆ

2

ˆ ˆ

2

ˆ

2 2

inside

V I V I

S a z

a L a L

ρ

ρ ϕ ρ

π π

Δ ⋅

=−

Δ ⋅

< = × = −

Poynting’s vector S oriented radially inward for ρ < . a 0 ρ = a ρ

( ) ⁰

outside

S ρ > a = {because ^E ( ^ρ ^> ^a ) ^{= !!!}} ⁰

I

( )

E ρ

ρ V L Δ

ρ = a 0

( )

B

in

ρ < a varies

linearly with ρ ^B

^out

( ^ρ ^≥ ^a )

varies as 1 ρ 2

o

I a μ

π ˆ

E + z B + ϕ ˆ S − ρ ˆ

1

S ≡

_μo

E B ×

a ˆz

Δ V

(8)

Note the following result for Poynting’s vector evaluated at the surface of the long wire, i.e. @ ρ = : a

( ) ( ) ^ˆ

2

inside

V I

S a

ρ aL ρ

π

= = Δ ⋅ − (SI units: Watts / m

²

)

Since ^E

^outside

( ^ρ ^≥ ^a ) ^{= :} ⁰ ^S

^outside

( ^ρ ⁼ ^a ) = ⇒ ∃ a discontinuity in ⁰ S at ρ = !!! a

( )

S ρ

2 V I π aL

Δ ⋅ ( )

₂

( ) ^ˆ

₂

^ˆ

2 2

long wire

V I V I

S a

a L a L

ρ ρ

ρ ρ ρ

π π

Δ ⋅ Δ ⋅

≤ = − = −

0 ρ = a ρ

Now let us use the integral version of Poynting’s theorem to determine the EM energy flowing through an imaginary Gaussian cylindrical surface S of radius ρ < and length H a L :

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

( ) ( ) ( ) ( ( ) )

, , ,

mech

mech free

v v

EM

S v EM v

dW t dU d

P t u r t d E r t J r t d

dt dt dt

dU t d

S r t da u r t d S r t d

dt dt

τ τ

= = = =

= − − = − − ∇

∫ ∫

∫ ∫ ∫

i

i i

Since this is a static/steady-state problem, we assume that dU

EM

( ) t dt = ⁰ . da

2

ˆx

3 3

ˆ ˆ, , z n da +

Gaussian Surface S ˆz

ϑ

da

1

H da

₃

ˆy

V

1

Δ V V

2

Then for an imaginary Gaussian surface taken inside the long wire ( ρ < ): a

0 1

wire S wire endcapLHS

P S da S da

=

= − ∫ ⁱ = − ∫ ⁱ

( ) ² ²

1 1

0

2 3

ˆ ˆ

cyl RHS

surface endcap

da da da da z

S da S da

ρ

=

= − =

− ∫ ⁱ − ∫ ⁱ

( )

3 3 ˆ

da=da +z

( ^ˆ )

S − ρ is ⊥ to da

1

( − z ˆ ) ; ^S ( ⁻ ^ρ ^ˆ ) is anti- to da

2

( + ρ ˆ ) ; ^S ( ⁻ ^ρ ^ˆ ) is ⊥ to da

3

( + z ˆ )

Only surviving term is:

( ) ( )

2 ² 0² 2

ˆ ˆ

2

( 2 )

2²

2 2

z H

wire cyl z H

V I V I

P S da d dz H V I

a H a H a

ϕ π ϕ

ρ ρ

ρ ρ ρ ρ ϕ ρ ρ πρ

π π

=+ =

=− =

⎛ ⎞

Δ ⋅ Δ ⋅

⎛ ⎞ ⎛ ⎞

= − = − ⎜ − ⎟ = ⎜ ⎟ = Δ ⋅ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∫ ⁱ ∫ ∫

a ρ

1 1

ˆ z n da ˆ, ,

−

2 2

ˆ ˆ , , n da ρ

+

S − ρ ˆ

(9)

Thus: P

wire

( ) V I

²

a ρ _{= Δ ⋅ ⎜ ⎟} ^{⎛ ⎞} ρ

⎝ ⎠ (Watts) P

wire

( ) ρ

2

V I a

⎛ ⎞ ρ Δ ⋅ ⎜ ⎟

⎝ ⎠ And: P

wire

( ρ = a ) = Δ ⋅ V I (Watts)

0 ρ = a ρ This EM energy is dissipated as heat (thermal energy) in the wire – also known as Joule heating of the wire. Since P

wire

( ) ρ ∝ ρ

²

^, note also that the Joule heating of the wire occurs primarily at/on the outermost portions of the wire.

From Ohm’s Law: Δ = ⋅ V I R

_wire

where R

_wire

= resistance of wire = ρ

_C^wire

L A

_⊥^wire

= L σ

_C^wire

A

_⊥^wire

( )

² ²

wire wire

P I R

a ρ = − ^{⎛ ⎞} ⎜ ⎟ ρ

( )

²

⎝ ⎠

wire wire

P ρ = a = − I R

Now let us repeat the use of the integral version of Poynting’s theorem to determine the EM energy flowing through an imaginary Gaussian cylindrical surface S of radius ρ ≥ and a length H L .

We expect that we should get the same answer as that obtained above, for the ρ < Gaussian a cylindrical surface. However, for ρ ≥ , a ^S

^outside

( ^ρ ^> ^a ) = , because ⁰ ^E

^outside

( ^ρ ^> ^a ) ^{= !!!} ⁰

Thus, for a Gaussian cylindrical surface S taken with ρ ≥ we obtain: a

_wire _wire

0 P = − ∫

S

S ⁱ da = ^!!!

What??? How can we get two different P

_wire

answers for ρ < vs. a ρ ≥ ??? This can’t be!!! a

⇒ We need to re-assess our assumptions here…

It turns out that we have neglected an important, and somewhat subtle point...

The longitudinal electric field ^E ^{= Δ} ( ^{V L z} ) ^ˆ formally/mathematically has a discontinuity at ρ = : a

i.e. The tangential ( ˆz ) component of E is discontinuous at ρ = . a

Power losses in wire show up / result in Joule heating of wire. Electrical energy is converted into heat

(thermal) energy – At the microscopic level, this is due to kinetic energy losses associated with the ensemble of individual drift/conduction/free electron scatterings inside the wire!

Joule Heating of current- carrying wire

( )

E ρ

^free ²

C C

J I a V

L π

σ σ

= = Δ

a ρ

ρ =

0 0

(10)

Formally/mathematically, we need to write the longitudinal electric field for this situation as:

( )

^free

¹ ( )

^free

¹ ( ) ^ˆ

C C

J J

E ρ ρ a ρ a z

σ σ

= ⎡ ⎣ − Θ − ⎤ ⎦ = ⎡ ⎣ − Θ − ⎤ ⎦

where the Heaviside step function is defined as: ( ) ^{0 for}

1 for a a

a ρ ρ

ρ

⎧ <

Θ − ≡ ⎨ ⎩ ≥ as shown below:

Furthermore, note that: ^Θ ( ) ^x ⁼ ∫

_−∞^x

^δ ( ) ^{t dt} and that: ^d ( ) ^x ( ) ^x

dx Θ = δ , where ^δ ( ) ^x is the Dirac delta function.

Now, in the process of deriving Poynting’s theorem (above), we used Griffith’s Product Rule # 6 to obtain ^E ⁱ ( ^∇× ^B ) ( ⁼ ^B ⁱ ^∇× ^E ) ^{− ∇} ⁱ ( ^{E B} ^× ) , and then used Faraday’s law (in differential form)

E B t

∇× = −∂ ∂ and then used ^B ^B ¹ ₂ ( ) ^{B B} ¹ ₂ ( ) ^B

²

t t t

∂ ∂ ∂

= =

∂ ∂ ∂

i i and Ê Ê ¹ ₂ ( ) ^{E E} ¹ ₂ ( ) Ê

²

t t t

∂ ∂ ∂

= =

∂ ∂ ∂

i i

with u

EM

=

¹²

( ε

o

E

²

+

_μ¹o

B

²

) to finally obtain:

( ) ( )

,

mech

mech free

v v

EM

S v EM v

dW t dU d

P t u d E J d

dt dt dt

dU t d

S da u d S r t d

dt dt

τ τ

= = = =

= − − = − − ∇

∫ ∫

∫ ∫ ∫

i

i i

So here, in this specific problem, what is ∇× E ???

In cylindrical coordinates, the only non-vanishing term is:

( ) ( ) ( )

ˆ

^free

1 ˆ

^free

ˆ

^free

ˆ

z

C C C

J J a J B

E E a a

t

ϕ ρ ϕ ρ ϕ δ ρ ϕ

ρ ρ σ σ ρ σ

⎧ ⎫ ∂Θ −

∂ ∂ ⎪ ⎪ ∂

∇× = − ∂ = ∂ ⎨ ⎪ ⎩ − ⎡ ⎣ − Θ − ⎤ ⎦ ⎬ ⎪ ⎭ = + ∂ = − = − ∂

In other words:

0 for ˆ for 0 for

free

C

a B J

E a

t

a ρ

ϕ ρ

σ

ρ

⎧ <

⎪ ⎧ ⎫

∂ ⎪ ⎪ ⎪

∇× = − ∂ = ∞ ⋅ ⎨ ⎪ ⎨ ⎪ ⎩ ⎬ ⎪ ⎭ =

⎪ >

⎩

( ^ρ ^a )

Θ −

ρ ρ = a

0 0

1

(11)

Thus, {only} for ρ > integration volumes, we {very definitely} need to {explicitly} include a the δ -function such that its contribution to the integral at ρ = is properly taken into account! a

( ) ( )

(

² ²

)

1 1

2

2 2

1 1

2 2

1

o

v EM S

v o S

o v v S

o v

dW t d

P t u d S da

dt dt

d E B d S da

dt

d d

E d B d S da

dt dt

dE dB

E d B d S da

dt dt

E dE d B

dt

μ

τ

ε τ

ε τ τ

ε τ

= = − −

= − + −

= − − −

= − + ∇

∫ ∫

∫ ∫ ∫

∫

i i

i

i i i

i i

( ) ^ˆ

v S

free

o v v S

o C

Ed S da

dE J

E d B a d S da

dt

τ

ε τ δ ρ ϕ τ

μ σ

× −

= − + − −

∫ ∫

∫ ∫ ∫

i

i i i

For this specific problem: dE dt = and for 0 ρ > , a ( )

¹

( ) ( )

0

o

0 S ρ a

_μ

E ρ a B ρ a

=

> = > × > = . Thus for ρ > : a

( )

^free

( ) ^ˆ ²

^free

( ) ²

o C v o C

J J

P t B δ ρ a ϕ τ d π aL B ρ a π a

μ σ μ σ

= ∫ ⁱ − = = =

^free

o

J L μ

o

C

μ σ 2

I π a

free

C

J σ I L

= ⋅

But:

^free

ˆ

C

J V

E z

σ L

= = Δ , and thus, finally we obtain, for ρ > : a ^{P t} ( ) ^V

L

= Δ I L ⋅ = Δ ⋅ , V I

which agrees precisely with that obtained earlier for ρ < : a ^{P t} ( ) ^{= Δ ⋅} ^{V I} ^!!!

For an E&M problem that nominally has a steady-state current I present, it is indeed curious that

( ) ^ˆ

free

C

J B

E a

δ ρ ϕ t σ

∇× = − = − ∂

∂ is non-zero, and in fact singular {at ρ = }! The singularity is a a

consequence of the discontinuity in E on the ρ = surface of the long, current-carrying wire. a

The relativistic nature of the 4-dimensional space-time world that we live in is encrypted into

Faraday’s law; here is one example where we come face-to-face with it!

(12)

Let’s pursue the physics of this problem a bit further – and calculate the magnetic vector potential A r inside ( ) ( ^ρ ^< ^a ) and outside ( ^ρ ^> ^a ) the long wire…

In general, we know/anticipate that {here}: ^{A r} ( ) ( ) ^{J r} ^{+ since:} ^z ^ˆ ( ) ( )

4

o v

A r μ J r d τ π

^′

′ ′

= ∫ _r

where r = r ≡ − r r′ .

We don’t need to carry out the above integral to obtain A r − a simpler method is to use ( ) ( ) ( )

B r = ∇× A r in cylindrical coordinates. Since A r ( ) = A r z

z

( ) ^ˆ (only, here), the only non- zero contribution to this curl is: ( ) A r

z

( ) _ˆ

B r ϕ

ρ

= − ∂

∂ .

For ρ < : a ( ) ( )

2

ˆ 1 ˆ ˆ

2 2

o z

o

A a

B a I J

a μ ρ ρ

ρ ϕ μ ρϕ ϕ

π ρ

∂ <

< = = = −

∂ ⇒ ( ) 1 ˆ

2

^o

A a

ρ J z ρ μ ρ

∂ <

∂ = −

For ρ ≥ : a ( ) ˆ 1

2

1 ˆ ( ) ˆ

2 2

o z

o

A a

B ρ a μ I ϕ μ Ja ϕ ρ ϕ

πρ ρ ρ

∂ ≥

≥ = = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − ∂ ⇒ ( ) 1

2

1 ˆ

2

^o

A a

Ja z

ρ μ

ρ ρ

∂ ≥ ⎛ ⎞

= − ⎜ ⎟

∂ ⎝ ⎠

Using ρ = as our reference point for carrying out the integration {and noting that as in the a case for the scalar potential ^{V r} ( ) , we similarly have the freedom to e.g. add any constant vector to ^{A r }:} ( )

( ) ¹ ^ˆ ¹ ¹ (

² ¹²

) ^ˆ ¹ (

² ¹²

) ^ˆ

2

^o

2

^o

2 4

^o

A ρ < a = − μ J ∫ ρ ρ d z = − μ J ρ − c z = − μ J ρ − c z

( )

² ²

(

2

)

1 1 1

ˆ ˆ

2

^o

ln 2

^o

A ρ a μ Ja d ρ z μ Ja ρ c z

ρ

≥ = − ⎛ ⎞ ⎜ ⎟ = −

∫ ⎝ ⎠

where c and

₁

c are constants of the integration(s).

₂

Physically, we demand that ^A ( ) ^ρ be continuous at ρ = , thus we must have: a

( ) ¹ ₄

^o

(

² ¹²

) ^ˆ ¹ ₂

^o ²

^ln ⁽

²

⁾ ^ˆ

A ρ = a = − μ J a − c z = − μ Ja a c z

Obviously, the only way that this relation can be satisfied is if c

₁

= c

₂

= ± , because then a ^A ( ^ρ ⁼ ^a ) ⁼ ⁰

{n.b. ^{ln 1} ( ) ⁼ ^ln ^e

⁰

⁼ ⁰ ^}.

Additionally, we demand that ^{A r} ( ) ( ) ^{J r} + , hence the physically acceptable solution is ^z ^ˆ

1 2

c = c = − , and thus the solutions for the magnetic vector potential a A r for this problem are: ( ) ( ) ¹ ₄

^o

(

² ²

) ^ˆ ¹ ₄

^o

(

² ²

) ^ˆ

A ρ < a = − μ J ρ − a z = + μ J a − ρ z

( ) ¹

²

^ln ( ) ^ˆ ¹

²

^ln ( ) ^ˆ

2

^o

2

^o

A ρ ≥ a = − μ Ja ρ − a z = + μ Ja ρ a z

(13)

^A ( ) ^ρ

0 ρ = a ρ

Note that: ( ) ¹ ^ln ( ) ^ˆ

2

^o

A ρ ≥ a = μ J ρ a z has a {logarithmic} divergence as ρ → ∞ , whereas:

( ) ( ) ¹

²

¹ ^ˆ ⁰

2

^o

B ρ A ρ μ Ja ϕ

ρ

→ ∞ = ∇× → ∞ = ⎛ ⎞ ⎜ ⎟ →

⎝ ⎠

This is merely a consequence associated with the {calculationally-simplifying} choice that we made at the beginning of this problem, that of an infinitely long wire – which is unphysical.

It takes infinite EM energy to power an infinitely long wire… For a finite length wire carrying a steady current I, the magnetic vector potential is mathematically well-behaved {but has a correspondingly more complicated mathematical expression}.

It is easy to show that both of the solutions for the magnetic vector potential ^A ( ^ρ

^<^≥

^a ) ^satisfy

the Coulomb gauge condition: ^∇ ⁱ ^{A r} ( ) ⁼ ⁰ , by noting that since ^A ( ^ρ

^<^≥

^a ) ⁼ ^A

^z

( ^ρ

^<^≥

^{a z} ) ^ˆ ^are

functions only of ρ , then in cylindrical coordinates: ^∇ ⁱ ^A ( ^ρ

^<^≥

^a ) ^{= ∂} ^A

^z

( ^ρ

^<^≥

^a ) ^{∂ =} ^z ⁰ ^.

Let us now investigate the ramifications of the non-zero curl result associated with Faraday’s law at ρ = for the A -field at that radial location: a

( ) ^ˆ

free

C

J B

E a

δ ρ ϕ t σ

∇× = − = − ∂

∂

Since A

^z

ˆ

B A ϕ

ρ

= ∇× = − ∂

∂ {here, in this problem}, then:

( )

_z

_ˆ

^free

₍ ₎ _ˆ

C

A A J

B a

t t t ϕ δ ρ ϕ

ρ σ

∂ ∇× ⎛ ∂ ⎞

∂ ∂

= = − ⎜ ⎟ = − −

∂ ∂ ∂ ⎝ ∂ ⎠ or:

^z ^free

( )

C

A J

t δ ρ a

ρ σ

⎛ ∂ ⎞

∂ ⎜ ⎟ = −

∂ ⎝ ∂ ⎠

Then: ( )

( )

free free

z

C C

a

J J

A a d a

t

ρ

δ ρ ρ ρ

σ σ

≡ Θ −

∂ = − = Θ −

∂ ∫ ^or:

^free

⁽ ⁾ ^ˆ

C

A J

t ρ a z

σ

∂ = Θ −

∂ .

( )

A

in

ρ < a

varies as 1−(ρ/a)

²

( )

A

out

ρ ≥ a

varies as ln(ρ/a)

( ) ⁰

¹₄ o ²

A = μ Ja

(14)

Now, recall that the {correct!} electric field for this problem is:

( )

^free

¹ ( ) ^ˆ

C

J

E ρ ρ a z

= σ ⎡ ⎣ − Θ − ⎤ ⎦

However, in general, the electric field is defined in terms of the scalar and vector potentials as:

( ) ( ) ( ) ^,

, , A r t

E r t V r t

t

= −∇ − ∂

∂

Since {here, in this problem}:

^free

( ) ^ˆ

C

A J

t ρ a z

σ

∂ = Θ −

∂ , we see that:

^free

ˆ

C

J

V z

−∇ = σ

and hence {in cylindrical coordinates} that: ( )

^free

C

V z J z

= − σ , then:

( )

ˆ ˆ ˆ

free free free

C C C

J J J

V z z z z z

z σ σ z σ

⎛ ⎞

∂ ⎜ ⎟ ∂

−∇ = + = =

⎜ ⎟

∂ ⎝ ⎠ ∂

.

Note that the {static} scalar field ( )

^free

C

J

V z z

= − σ pervades all space, as does ^A ( ^ρ

^<≥

^a ) + . ^z ^ˆ Explicitly, due to the behavior of the Heaviside step function ^Θ ( ^ρ ⁻ ^a ) we see that the electric

field contribution

^free

( ) ^ˆ

C

A J

t ρ a z

σ

∂ = Θ −

∂ is:

0 for ˆ for

free

C

a

A J

t z a

ρ σ ρ

⎧ <

∂ ⎪

∂ = ⎨ ⎪ ≥

⎩

.

Explicitly writing out the electric field in this manner, we see that:

( ) ( ) ( ) ^ˆ 0 ^ˆ for

ˆ ˆ 0 for

free free

C C

free free

C C

J J

z z a

A a

E a V a

t J J

z z a

ρ σ σ ρ

ρ ρ

σ σ ρ

<≥

< <

≥ ≥

⎧ ⎪ + = <

∂ ⎪

= −∇ − ∂ = ⎨ ⎪

− = ≥

⎪ ⎩

Thus, for ρ ≥ we see that the a ^{− ∂} ^A ( ^ρ ^≥ ^a ) ∂ contribution to the E -field outside the wire ^t {which arises from the non-zero ∇× E of Faraday’s law at ρ = } exactly cancels the a

( )

V ρ a

−∇ ≥ contribution to the E -field outside the wire, everywhere in space outside the wire,

despite the fact that ^A ( ^ρ ^≥ ^a ) varies logarithmically outside the wire!!!!

(15)

The long, current-carrying wire can thus also be equivalently viewed as an electric flux tube:

( ) ¹ ⁽ ⁾ ^ˆ

E S

E da J

free

σ

C S

ρ a z da I σ

C

Φ = ∫ ⁱ = ∫ ⎡ ⎣ − Θ − ⎤ ⎦ ⁱ =

The electric field E is confined within the tube ( = the long, current carrying wire) by the

( )

A ρ a t

− ∂ ≥ ∂ contribution arising from the Faraday’s law effect on the ρ = boundary of the a flux tube, due to the {matter geometry-induced} discontinuity in the electric field at ρ = ! a

The ^{∇× =} ^E ( ^J

^free

^{σ δ ρ}

^C

) ⁽ ⁻ ^a ⁾ ^ϕ ^ˆ ^{= − ∂ ∂} ^{B t} effect at ρ = also predicts a non-zero “induced” EMF a in a loop/coil of wire: ε = −∂Φ ∂ . The magnetic flux through a loop of wire is:

_m

t

loop

m C

A d

S

B da B A

_⊥

Φ = ∫ ⁱ = ∫ ⁱ ⋅ ^where ^A

^⊥^loop

is the cross-sectional area of a loop of wire {whose plane is perpendicular to the magnetic field at that point}. Note further that the width, w of the coil only needs to be large enough for the coil to accept the B t ∂ ∂ contribution from the δ-function at ρ = . Then, here a in this problem, since the magnetic field at the surface of the wire is oriented in the ϕ ˆ -direction, and:

( ) ^ˆ

free

C

B J

t δ ρ a ϕ

σ

∂ = − −

∂ , then we see that:

^m ^loop ^free ^loop

( )

C

J A

B A a

t t

ε δ ρ

σ

⊥

⋅

⊥

∂Φ ∂ ⋅

= − = − = −

∂ ∂

For a real wire, e.g. made of copper, how large will this EMF be – is it something e.g. that we could actually measure/observe in the laboratory with garden-variety/every-day lab equipment???

A number 8 AWG (American Wire Gauge) copper wire has a diameter D = 0.1285” = 0.00162 m (~ 1/8” = 0.125”) and can easily carry I = 10 Amps of current through it.

The current density in an 8 AWG copper wire carrying a steady current of I = 10 Amps is:

( )

⁶

(

²

)

8 2 2 2

4 4 10

4.8 10 0.001632

AWG

I I

J Amps m

a D

π π π

⋅ ⋅

= = = ×

The electrical conductivity of {pure} copper is: σ

C^Cu

= ^{5.96 10} ×

⁷

( Siemens m ) .

If our “long” 1/8” diameter copper wire is L = 1 m long, and if we can e.g. make a loop of ultra- fine gauge wire that penetrates the surface of the wire and runs parallel to the surface, then if we approximate the radial delta function ^{δ ρ} ( ⁻ â ) ât ^ρ = as ~ a narrow Gaussian of width â

~ 10 Å 1 10

9

w = nm =

⁻

m (i.e. ~ the order of the inter-atomic distance/spacing of atoms in the copper lattice { 3.61 Å }), noting also that the delta function ^{δ ρ} ( ⁻ ^a ) has physical SI units of inverse length (i.e. m

^-1

) and, neglecting the sign of the EMF, an estimate of the magnitude of the “induced” EMF is:

( )

8 8

loop

AWG AWG

Cu Cu

C

J A J L w

ε δ ρ a

σ

^⊥

⋅ ⋅ ⋅

= −

_Cu

C

σ ^⋅ w

( )

6 2

8

7

4.8 10

1 80 !!!

6 10

AWG Cu C

Amps m

J L m mV

Siemens m σ

⎛ × ⎞

⎛ ⎞ ⎜ ⎟

= ⎜ ⎝ ⎟ ⎠ ⋅ ⎜ ⎝ × ⎟ ⎠ ⋅

This size of an EMF is easily measureable with a modern DVM…

(16)

Using Ohm’s Law: V = ⋅ I R , note that the voltage drop V

_drop

across a L = 1 m length of 8 AWG copper wire with I = 10 Amps of current flowing thru it is:

1

1 8

Cu

m C wire

drop m wire AWG

V I R I L J A

A ρ

⊥

= ⋅ = ⋅ = ( ⋅ )

Cu wire

C

L σ A

_⊥

⋅

⁸^AWG_Cu _Cu

!!!

C

J L ε

= σ ⋅ =

In other words, the “induced” EMF, ^ε ⁼ ( ^J

^free

^⋅ ^A

^⊥^loop

^{σ δ ρ}

^C

) ⁽ ⁻ ^a ⁾ in the one-turn loop coil of length L {oriented as described above} is precisely equal to the voltage drop ^V

^drop

⁼ ( ^J

^free

^σ

^C

) ^⋅ ^L

along a length L of a portion of the long wire with steady current I flowing through it, even though the 1-turn loop coil is completely electrically isolated from the current-carrying wire!!!

This can be easily understood... Using Stoke’s theorem, the surface integral of ∇× E can be converted to a line integral of E along a closed contour C bounding the surface of integration S ; likewise, a surface integral of B t ∂ ∂ = ∇× ∂ ∂ can be converted to a line integral of A t A t ∂ ∂ along a closed contour C bounding the surface of integration S:

( )

^m

S C S S C

B A A

E da E d da da d

t t t t

ε = ∫ ∇× ⁱ = ∫ ⁱ = − ^∂Φ ∂ = − ∫ ^∂ ∂ ⁱ = − ∫ ^⎛ ⎜ ⎝ ∇× ^∂ ∂ ^⎞ ⎟ ⎠ ⁱ = − ∫ ^∂ ∂ ⁱ

n.b.: 0

C

−∇ V d ≡

∫ ⁱ

Then for any closed contour C associated with the surface S that encloses the Faraday law

∇× E δ -function singularity at ρ = , e.g. as shown in the figure below: a

the “induced” EMF ε can thus also be calculated from the line integral

C

E d

∫ ⁱ taken around the closed contour C. From the above discussion(s), the electric field inside (outside) the long current-carrying wire is E

_in

= J σ

_C

( ^E

^out

⁼ ⁰ ) , respectively {n.b. ⇒ tangential E is

discontinuous across the boundary of a {volume} current-carrying conductor!}. Then:

Electric Charge Conservation

LECTURE NOTES 1 CONSERVATION LAWS