Dr. Alain Brizard
Electromagnetic Theory I (PY 302)
Faraday’s Law & Maxwell’s Equations
(Griffiths – Chapter 7: Sections 2-3)
• Electromagnetic Induction
The flux rule states that a changing magnetic flux ΦB =
induces an emf E on the boundary ∂S:
dΦB
E = −
= −
dt
Z
S
∂B
· da =
∂t
R
S
I
B · da through a surface S
E · dl.
(1)
∂S
The flux rule is also known as Faraday’s Law (in integral form). The minus sign in Faraday’s
Law is associated with Lenz’s Law: the induced current on the boundary ∂S generated by
the induced emf flows in a direction which opposes the change in magnetic flux.
It is worth mentioning, here, that there is an infinite number of open surfaces S with
R
boundary ∂S but that the magnetic flux S B · da through surface S (with boundary ∂S)
R
is equal to the magnetic flux S 0 B · da through surface S 0 (with boundary ∂S 0) if the
boundaries are identical ∂S 0 = ∂S. This statement is proved by considering the volume V
enclosed by the two surfaces S and S 0 (such that ∂V = S − S 0 ):
Z
B · da −
Z
B · da =
S0
S
I
B · da =
∂V
Z
∇ · B dτ = 0,
V
which follows from the Divergence Theorem and Gauss’s Law for magnetic fields.
Returning to Faraday’s Law (1), we use Stoke’s Theorem on the last integral to obtain
−
Z
S
I
Z
∂B
· da =
E · dl =
∇ × E · da,
∂t
∂S
S
which becomes Faraday’s Law in differential form:
∇×E = −
∂B
.
∂t
(2)
Hence, the presence of a time-dependent magnetic field induces an electric field whose curl
is not zero. Consequently, this induced electric field cannot be expressed in terms of the
gradient of a scalar field (since ∇ × E 6= 0). The clue to the nature of this new electric
field is provided by the representation of the magnetic field B = ∇ × A in terms of the
vector potential A:
∂B
∂A
= ∇×
= − ∇×E
∂t
∂t
→
1
∂A
∇× E +
∂t
!
= 0.
From the Helmholtz Decomposition Theorem for vector fields, the electric and magnetic
fields are, therefore, expressed in terms of the scalar potential Φ and vector potential A as
E = − ∇Φ −
∂A
and B = ∇ × A,
∂t
(3)
where the new term in the electric field is called the inductive electric field. Note that
these definitions immediately imply Faraday’s Law and Gauss’s Law for magnetic fields:
∇ × E = − ∂B/∂t and ∇ · B = 0, respectively.
For example, when a uniform magnetic field B(t) is time dependent, it induces a timedependent nonuniform electric field E(r, t). To show this, we invoke the formula
A(r, t) = −
r
× B(t)
2
for the vector potential of a uniform magnetic field; note, here, that
∇×A = −
B(t)
1
3
B(t)
· ∇r +
∇ · r = − B(t) + B(t) = B(t).
2
2
2
2
The inductive electric field is, thus, given as
E(r, t) = −
r dB(t)
∂A
= ×
,
∂t
2
dt
and that this field is divergenceless
∇·E =
1 dB(t)
· ∇ × r = 0.
2 dt
• Inductance
Suppose now we are interested in calculating the magnetic flux ΦB 2 through a wire loop
(labeled 2) due to the magnetic field B1 produced by another wire loop (labeled 1):
ΦB 2 =
Z
S2
B1 · da2 = M1→2 I1,
where the magnetic flux ΦB 2 must clearly be proportional to the current I1 flowing in loop
1 and the constant of proportionality M1→2, known as the mutual inductance, is a purely
geometric quantity which depends on the relative orientation and positions of the loops as
well as their shapes. A similar argument leads to the formula
ΦB 1 =
Z
S1
B2 · da1 = M2→1 I2.
If we now replace the magnetic field Bj = ∇ × Aj with the vector potential
Aj =
µ0
Ij
4π
I
∂Sj
2
dlj
,
|r − rj |
we find
ΦB k =
I
∂Sk
Aj · dlk =
µ0
Ij
4π
I
I
∂Sj
∂Sk
dlj · dlk
= Mj→k Ij .
|rj − rk |
Hence, the mutual inductance is symmetric, M1→2 = M12 = M2→1 , and is expressed as
M12 =
µ0
4π
I
I
∂S
∂S 0
µ0
dl1 · dl02
=
0
|r − r |
4π I1I2
Z
dτ
V
Z
V0
dτ 0
J1 (r) · J2 (r0 )
.
|r − r0|
For example, we consider the mutual inductance of two current loops (of radii a and b)
separated by a distance Z; furthermore, the two loops are parallel and share the same axis
(see Figure below).
Here, the current densities are
J1 (r) = I1 δ(r − a)δ(z − Z) θb1 (θ) and J2 (r0 ) = I2 δ(r0 − b)δ(z 0) θb2(θ0 ),
where θb1 · θb2 = cos(θ − θ0 ), so that
Z 2π
Z 2π
µ0
cos(θ − θ0 )
M =
ab
dθ
dθ0 q
4π
0
0
Z 2 + a2 + b2 − 2 ab cos(θ − θ0 )
Z π q
Z 2π
q
d
cos χ dχ
µ0 q
√
= − µ0 abq
abq
1 − 2q cos χ dχ ,
=
2
1 − 2 q cos χ
dq 0
0
where we used the substitution (θ, θ0) → (θ, χ = θ − θ0), and integrated over θ, and
introduced
the dimensionless quantity q = ab/(Z 2 +a2 +b2). Note that the Taylor expansion
√
of 1 − 2q cos χ yields
q
1 − 2q cos χ = 1 − q cos χ −
q2
q3
5 q4
cos2 χ −
cos3 χ −
cos4 χ − · · · ,
2
2
8
3
so that
Z
π q
d
πq
15 2
q + ··· ,
1 − 2q cos χ dχ =
1 +
dq 0
2
8
and the mutual inductance between the two circular loops is
−
!
µ0 π q 3
15 2
µ0
q + ··· '
M =
abq 1 +
2
8
2π
π a2 · π b2
,
Z3
which is valid when Z a, b (with q ' ab/Z 2 1).
A single wire loop also experiences a self-inductance, labeled L, which opposes a change
in the current flowing through it:
dI
dΦB
= −L .
dt
dt
Here, we note that both self-inductance L and mutual inductance M are measured in
henries (SI units), abbreviated H, and defined as H = V· s. For example, the self-inductance
of a rectangular toroidal coil (height h, inner radius a, and outer radius b) with N turns is
ΦB = L I
Z
→
E = −
Z
Z
h
N
µ0 2
N b
µ0 NI
L =
=
N h ln(b/a).
B · da =
dr
dz
I
I a
2π r
2π
0
The self-inductance, thus, plays the role of electromagnetic inertia, i.e., the large selfinductance of an electrical circuit makes it difficult to either initiate current flow in the
circuit or vary the current flowing through the circuit.
• Magnetic Energy in a Current Distribution
The rate of work done in allowing a current I to flow through a circuit with selfinductance L is
!
dW
dI
d
I2
= − E I = LI
=
L
,
dt
dt
dt
2
and, thus, the work done in setting a current distribution is
W =
I
1
ΦB I =
2
2
I
A · dl =
1
2
Z
A · J dτ.
If we now make use of Ampère’s Law, J = µ−1
0 ∇ × B, and integrate by parts, we find
Z
Z
1
1
A · ∇ × B dτ =
|B|2 dτ −
W =
2µ0 V
2µ0
V
Z
I
1
2
=
|B| dτ −
(A × B) · da .
2µ0
V
∂V
Z
∇ · (A × B) dτ
V
If the integration volume V is |R3, for which the magnetic field B must vanish on ∂V , the
magnetic energy stored in a current distribution is
W =
1
2
Z
3
|R
4
2
µ−1
0 |B| dτ.
(4)
• Maxwell’s Equation
Following up on Faraday’s Law, which states that a time-dependent magnetic field
generates a nonuniform electric field, we naturally ask the question: Will a time-dependent
electric field generate a nonuniform magnetic field? In order to provide an answer, we begin
with the following puzzle involving Ampère’s Law:
0 = ∇ · ∇ × B = µ0 ∇ · J 6= 0,
where ∇ · J does not vanish for time-dependent charge and currents distributions. Hence,
Ampère’s Law is incomplete for time-dependent fields.
Fortunately, from the charge conservation law and Gauss’s Law for the electric field, we
find
∂ρ
∂E
∇·J = −
= − 0 ∇ ·
,
∂t
∂t
and, therefore, the solution proposed by Maxwell is to modify Ampère’s Law and write
∇ × B = µ0 J + µ0 0
∂E
,
∂t
(5)
which could be called the Ampère-Maxwell Law. It states that, even in the absence of a
current density J, a time-dependent electric field will produce a nonuniform magnetic field.
In summary, Maxwell’s equations are
∇ · E = −1
0 ρ
∇·B = 0
(Gauss’s Law for E)
(Gauss’s Law for B)
∂B
∂t
∂E
∇ × B = µ0 J + µ0 0
∂t
∇×E = −
(Faraday’s Law)
(Ampère-Maxwell’s Law)
An important note must be made concerning the charge conservation law used above.
On the one hand, the charge density ρ is known to be the sum of the free-charge density
ρf and the bound-charge density ρb = − ∇ · P (expressed in terms of the polarization P),
so that
∂ρ
∂ρf
∂P
=
− ∇·
.
∂t
∂t
∂t
On the other hand, the current density must now be the sum of the free-current density Jf ,
the bound-current density Jb = ∇ × M (a divergenceless current expressed in terms of the
5
magnetization M), and a polarization current Jp associated with the fact the polarization
P is time dependent. Hence,
∇ · J = ∇ · Jf + ∇ · Jp = −
∂ρf
∂ρb
∂ρ
= −
−
,
∂t
∂t
∂t
and, since the free-charge and free-current distributions must satisfy the charge conservation
law separately (i.e., ∇ · Jf = − ∂ρf /∂t), we find
∇ · Jp = −
∂P
∂ρb
= ∇·
∂t
∂t
Jp =
→
∂P
.
∂t
From these definitions, we may now express the Ampère-Maxwell Law solely in terms of
the free-current density as
∂D
∇ × H = Jf +
,
∂t
−1
where D = 0 E + P = E and H = µ−1
0 B − M = µ B.
• Energy-Momentum Transport
When we combine the electric and magnetic energies found in a disbtribution of charges
and currents, we find
1
W =
2
Z
|R3
2
0 |E| +
µ−1
0
|B|
2
1
dτ =
2
!
Z
|R3
D · E + H · B dτ,
where the last expression takes into account the permitivity and permeability µ of the
medium. If we introduce the electromagnetic energy density
E =
1 2
0 |E|2 + µ−1
,
0 |B|
2
we find, by using Maxwell’s equations, that
∂E
∂E
∂B
−1
= 0E ·
+ µ−1
=
E
·
B
·
∇
×
µ
B
−
J
− µ−1
0
0
0 B · ∇ × E,
∂t
∂t
∂t
which can be re-arranged as
∂E
+ ∇ · S = − E · J,
∂t
(6)
(Poynting) electromagnetic-energy flux. Hence, the
where S = E × µ−1
0 B represents the
R
total electromagnetic energy E = V E dτ is dissipated at a rate − E · J.
Since the electromagnetic field carries energy, it also carries momentum, with momentum density defined as Π = 0 E × B. Hence, using Maxwell’s equations once again, we
find
∂B
∂E
∂Π
= 0
× B + 0 E ×
= µ−1
∇
×
B
−
J
× B − 0 E × ∇ × E,
0
∂t
∂t
∂t
6
which can be re-arranged as
∂Π
+ ∇ · T = − ρ E − J × B,
∂t
where the momentum-stress tensor for the electromagnetic field is defined as
T = EI −
(7)
0 EE + µ−1
0 BB .
• Electromagnetic Waves in Free Space
An important limit of Maxwell’s equations is provided by the case of electromagnetic
fields in free space (defined by the absence of charge and current densities). In free space,
Maxwell’s equations are thus
∇·E
= 0 =
∇·B
∂B
1 ∂E
and
∇×B = 2
,
∂t
c ∂t
where we note that µ0 0 = 1/c2 (c denotes the speed of light). By taking the curl of
Faraday’s Law and using the Ampère-Maxwell Law, we find
∇×E = −
∇×∇×E = − ∇×
which becomes
1 ∂2
∇ − 2 2
c ∂t
2
1 ∂ 2E
∂B
= − 2
,
∂t
c ∂t2
!
E = 0,
after using the vector identity
∇ × ∇ × E = ∇(∇ · E) − ∇2E = − ∇2E,
where the second equality follows from Gauss’s Law for E in free space. Note that the
magnetic field satisfies a similar equation
1 ∂2
∇2 − 2 2
c ∂t
!
B = 0.
The plane-wave solutions for the electric and magnetic fields are



E(x, t)
B(x, t)





= 
e
E
e
B





ei (k · x−ωt) + 
e∗
E
e∗
B



e−i (k · x−ωt) ,
where ∗ denotes the complex conjugate and the electromagnetic waves satisfy the dispersion
relation ω 2 = (|k|c)2, i.e., electromagnetic waves are light waves. Lastly, we find that the
time-averaged energy density and Poynting flux are
0
|k|2c2
E =
1 +
2
ω2
!
e 2 = |E|
e 2 and S =
|E|
0
7
kc2
E = c2 Π.
ω