Electrodynamics

UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 21 Prof. Steven Errede


LECTURE NOTES 21
Electrodynamics

Electromotive Force — Ohm’s Law


In order to make a free electrical current flow in matter, one somehow has to push (and/or
pull) on electric charges – i.e. exert a force on them. Note however, that the speed at which the
electrical charges move in a conducting medium depends on the detailed microscopic nature of
that conducting medium.

For most electrically conducting materials, the volume free current density, J free ( r ) is linearly
proportional to the force per unit charge, f ( r ) ≡ F ( r ) Q Note that J free ( r ) f ( r ) = F ( r ) Q .


(
J free ( r ) = σ c f ( r ) = σ c F ( r ) Q )
where the constant of proportionality σ c ≡ electrical conductivity of the material.

SI units of electrical conductivity σ c :
dQ
( Amperes )
Volume free current density: J free ( r ) Amperes
meter 2 ( m)
A 2
I=
dτ
1 Ampere = 1 Coulomb/sec
F (r )
Force per unit charge: f ( r ) ≡
Q
Newtons
Coulomb
N
C ( )
A 2
J (r ) 2
σc = = m = C
f (r ) N N − m2 − s
C

F (r ) ⎛ N ⎞ F ( r ) ⎛ Newtons Volts ⎞
But: f ( r ) = ⎜ ⎟ also has SI units of: E ( r ) = ⎜ = ⎟
Q ⎝C⎠ Q ⎝ Coulomb meter ⎠
Amps m 2 ⎛ Amps ⎞
σc = =⎜ ⎟ m
Volts m ⎝ Volt ⎠
SI units symbol = Ω
ΔV V − V ⎛ Volts ⎞
Define resistance: R ≡ = 2 1 ⎜ ≡ Ohms ⎟
I I ⎝ Amps ⎠

⇒ SI units of electrical conductivity: σ c = Ohms −1 m 1 Ω ≡ 1V A
1 Ohm ≡ 1Volt Amp
1
Define electrical resistivity of material: ρ c ≡ SI units of ρ c = Ohm-m
σc

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 1
2005-2008. All Rights Reserved.

, UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 21 Prof. Steven Errede

−1
In some circles (EE-types) 1 Ohm −1 ≡ “mho” (Ohm spelled backwards) σ c = ohm m = mho m

1 Ampere
Note also that (esp. in Europe): 1 Siemens ≡ Volt

⎛ Amps ⎞
∴ SI units of electrical conductivity: σ c = ⎜ ⎟ m = Siemens m
⎝ Volt ⎠

∴ 1 Siemens ≡ 1 Ohm = 1 mho

Some typical values of resistivity ρc ( Ω − m ) for various materials are listed in the table below:




Any force capable of driving free electrical charge will produce
⎛ F (r ) ⎞
⎜ Q ⎟⎟ (
J free ( r ) = σ c f ( r ) = σ c ⎜ A m2 )
⎝ ⎠
( )
Usually, in E&M this force is: FToT ( r ) = qE ( r ) + q v ( r ) × B ( r ) = FE ( r ) + Fm ( r )


( )
If B ( r ) = 0 , or: B ( r ) ≠ 0, but v ( r ) × B ( r ) is very small, such that FE ( r ) Fm ( r )
then (usually):

Ohm’s Law: J free ( r ) = σ c E ( r ) = E ( r ) ρc note that: E ( r ) J free ( r )

( )
{In plasmas, where v ( r ) may not be small, Fm ( r ) Q = v ( r ) × Bext ( r ) must be included…}


2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.

, UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 21 Prof. Steven Errede

Inside a conducting wire, if a steady free electric current Ifree flows inside the wire:
I free = ∫ J free ( r )ida⊥ = ∫ J free ( r )inˆ⊥ da⊥ = J free A⊥ (if J free is uniform)
S⊥ S⊥

e.g. A⊥ = π R 2 for a wire of radius R.
I free , J free , E , nˆ⊥ R I free , J free , E



The electric field E inside the electrical conductor exists only when the free current Ifree is
( )
flowing. Einside = 0 when Ifree = 0 J free = 0 (i.e. electrostatics case)

Now a steady Ifree can flow in an electrical conductor when an electrostatic potential difference,
ΔV is applied across the two ends of the conductor, e.g. using a battery:

ΔV
+ I free , J free , E −
• • cylindrical electrical conductor (wire)
with cross sectional area A⊥ = π R 2
L
+ −
Ideal wire Ideal wire
Battery
(Ideal voltage source)

If the conducting wire has a uniform cross-section and material of the wire is uniform,
homogenous, isotropic and linear then the free current density will also be uniform, thus:
I free = J free A⊥ = (σ c E ) A⊥ = σ c A⊥ E
But for this same (uniform) wire, E = ΔV L = constant for uniform, homogeneous, isotropic and
linear conducting wire of cross-sectional area, A⊥ . {n.b. what this last relation is actually saying
is that Laplace’s equation ∇ 2V ( r ) = 0 is operative inside the wire, not Poisson’s equation
∇ 2V ( r ) = − ρ free ( r ) / ε o , i.e. ρ free ( r ) = 0 inside the wire!!!}

⎛ ΔV ⎞
∴ I free = σ c A⊥ ⎜ ⎟
⎝ L ⎠
ΔV 1 ⎛ L ⎞ ⎛ L ⎞
We can now define the DC resistance of the conducting wire as: R = = ⎜ ⎟ = ρc ⎜ ⎟
I free σ c ⎝ A⊥ ⎠ ⎝ A⊥ ⎠

SI units of DC resistance = Ohms = Volts/Ampere
= 1/SI units of 1/DC resistance = Siemens = Amperes/Volt

1 Ohm = 1/Siemens



© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 3
2005-2008. All Rights Reserved.

, UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 21 Prof. Steven Errede


Ohm’s Law: I free = ΔV R I free (Amps)

Linear relationship slope = 1
R
I free vs. ΔV
y = mx + b, b = 0 ΔV (Volts)
n.b.
y-intercept ( I free ) = 0 y = mx + b
⎛1⎞
I = ⎜ ⎟ ΔV + 0
⎝R⎠
Ohm’s Law: I free = ΔV R or: ΔV = I free R

1
DC Resistance: R = ρc ⎛⎜ L ⎞⎟ = ⎛⎜ L ⎞⎟ Ohms
⎝ A⊥ ⎠ σ c ⎝ A⊥ ⎠

Griffiths Example 7.2: Calculate the DC resistance of a conducting material of conductivity σ c
sandwiched between two long, superconducting cylinders (inner radius a, outer radius, b) with
electrostatic potential difference ΔV ≡ Vb − Va between the two superconducting cylinders.
{n.b. the superconducting material has no DC resistance}:

λ
∃ a radial electrostatic field between superconductors: E ( ρ ) = ρˆ (in cylindrical coordinates)
2πε o ρ
λ = electric charge/unit length on inner cylinder
λ
E (ρ) = ρˆ ρ = x 2 + y 2 (in cylindrical coordinates)
2πε o ρ
b
a
• • L
a
dϕ adϕ



L >> b,a


I free = ∫ J free ( ρ )ida⊥ = σ c ∫ E ( ρ )ida⊥ da⊥ = adϕ L ρˆ ← evaluated at inner radius
s⊥

σ c λ 2π ⎛ 1 ⎞ σλ σ λL
I free = ∫ ⎜
2πε o 0 ⎝ a ⎠
⎟ ρˆ iadϕ L ρˆ = c 2 π L = c
2 π εo εo
σ λL
I free = c Note also that λ L = Qinner
εo


4 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.