Download Chapter 14 Quasi-static approximations A quasi-static system means a system for which

Chapter 14
Quasi-static approximations
∇ ∙ E = ρ /ε0
∇ × E = − ∂B /∂t
∇∙B=0
∇ × B = μ0 j + μ0ε0 ∂E /∂t
∇ ∙ j = − ∂ρ /∂t
For a static system, the red terms are 0.
Then we could calculate the fields using
potentials (or by some other method)
E(r) = −∇φ and B(r) = ∇ × A.
A quasi-static system means a system for
which
(i).. the sources and fields change slowly;
(ii) .. and one or more of the red terms in
the field equations can be neglected.
Zangwill lists several examples of quasistatic approximations.
If the system radiates significant energy
as electromagnetic waves, then the quasistatic approximation does not apply.
Example. Suppose ρ(r,t) changes slowly in
time. Then we should be able to use what
we know about static systems to analyze
the fields. In particular, electromagnetic
induction may be negligible.
This is called quasi-electrostatics.
(1/L)EC ~ ρ /ε0 ;
i.e.,
and
∇ ∙ E = ρ /ε0
∇ × E = − ∂B /∂t is negligible
∇∙B=0
∇ × B = μ0 j + μ0ε0 ∂E /∂t
∇ ∙ j = − ∂ρ /∂t
Dimensional analysis. What do we mean
by “changes slowly”? How slowly?
Consider a system with charge density ρ
(r,t) , for which the length constant is L
and the time constant is T; also, define
ω = 1/T (“frequency”).
There are two sources of E: Faraday’s law
and Coulomb’s Law. The characteristic
sizes of EF and EC will be...
EC ~ L ρ /ε0
(1/L) EF ~ ω B
where (1/L) B ~ μ0 j , and (1/L) j ~ ω ρ ;
i.e., EF ~ L ω L μ0 L ω ρ = ω2 L3 μ0ρ
EF
ω 2 L 3 μ 0ρ
So,
=
= ω 2 L 2 μ0 ε 0
EC
L ρ /ε0
We can neglect the electromagnetic
induction if ω2 L2 μ0 ε0 is << 1; i.e.,
ω L << 1/ (μ0 ε0 )½ = 3 × 108 m/s.
(The speed of light in vacuum is
c = 1/ (μ0 ε0 )½ = 3 × 108 m/s .)
Then, approximately,
1
E(r,t) = −∇ 4πε
0
∫
ρ(r’,t) d3r’
| r − r’ |
Example: Charging a capacitor
As the capacitor
charge increases,
what is the
magnetic field B(r,
t)?
Note: We used quasi-electrostatics to calculate
Q(t). We used ∇ ∙ E = ρ /ε0 ;
and we assumed that ∇ × E = 0 so that we could
write E = V/d; and thus Q = CV with
C = ε0 A/d. What about ∇ × E = - ∂B/∂t?
Another example: quasi-magnetostatics
If j(r,t) changes slowly in time, then there
is an electric field because of
electromagnetic induction. So, is the
displacement current important?
Dimensional analysis:
(1/L) EF ~ ω B and (1/L) B ~ μ0 j
Compare current and displacement
current:
jD ε 0 ω L ω B
=
j
B / ( μ0 L)
= ω 2 L 2 μ0 ε 0 ;
so if ω L << c, then we can neglect the
displacement current.
Example.