Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2 for Electrical Engineering
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
Grader: Gady Afek
Week 10. Maxwell’s equations – Introduction • The problem with
Ampère’s law • Maxwell’s fix •   E • Stokes’s theorem
Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38.
Introduction
The set of four fundamental equations for E and B,

E  dA 
q
(Gauss’s law)
0
 B  dA  0

d
E  dr    B
dt
 B  dr   0 I
,
(Faraday’s law)
(Ampère’s law)
together with the Lorentz force law FEM = q (E + v × B), sum
up everything we have learned so far about electromagnetism!
Introduction
This lecture and the next describe one of the great advances in
the history of science, namely, how J. C. Maxwell discovered a
problem with Ampère’s law, how he solved the problem, and
how he was led to the amazing prediction of electromagnetic
waves.
These two lectures are is also more mathematical than the other
lectures. We need the mathematics in order to analyze
Maxwell’s work efficiently. We need two operators – the
“curl”   E and the “divergence”   E – and corresponding
theorems: “Stokes’s theorem” and the “divergence theorem”.
Introduction
New notation: If S denotes a surface (which may be curved),
then ∂S denotes the boundary of the surface S:
z
S
Rutgers
x
∂S
y
The problem with Ampère’s law
We have seen that Ampère, building on Oersted’s discovery
and on the law of Biot and Savart, postulated the law named
after him:
 B(r)  ds  0 I .
I
B
I=0
ds
The problem with Ampère’s law
Now let’s compare Ampère’s law with Faraday’s law:
 B  dr   0 I

(Ampère’s law)

d
E  dr  
B  dA .
dt
S
(Faraday’s law)
S
With the new notation, we can write the left side of Faraday’s
law as an integral over the closed loop ∂S, and the right side as
an integral over any surface S bounded by the closed loop ∂S.
The problem with Ampère’s law
Now let’s compare Ampère’s law with Faraday’s law:
 B  dr   0 I

(Ampère’s law)

d
E  dr  
B  dA .
dt
S
(Faraday’s law)
S
With the new notation, we can write the left side of Faraday’s
law as an integral over the closed loop ∂S, and the right side as
an integral over any surface S bounded by the closed loop ∂S.
How do we know that the surface S on the right side can be
any surface?
The problem with Ampère’s law
Now let’s compare Ampère’s law with Faraday’s law:
 B  dr   0 I

(Ampère’s law)

d
E  dr  
B  dA .
dt
S
(Faraday’s law)
S
With the new notation, we can write the left side of Faraday’s
law as an integral over the closed loop ∂S, and the right side as
an integral over any surface S bounded by the closed loop ∂S.
Can we write Ampère’s law this way?
The problem with Ampère’s law
Yes, we can!
 B  dr  0  J  dA
S
.
(Ampère’s law)
S
With the new notation, we can write the left side of Ampère’s
law as an integral over the closed loop ∂S, and the right side as
an integral over any surface S bounded by the closed loop ∂S.
Here J is the current density.
The problem with Ampère’s law
This is indeed what Ampère intended with his law,
 B  dr  0  J  dA
S
,
S
namely that the right side is the current flowing through any
surface S bounded by the closed loop ∂S on the left side.
However, there is a problem with Ampère’s law when we
consider an electrical circuit with a capacitor in it.
The problem with Ampère’s law
How do we apply Ampère’s law,
 B  dr  0  J  dA, to this
S
electrical circuit?
S
–q
∂S
q
Surface S2
I
Surface S1
The problem with Ampère’s law
How do we apply Ampère’s law,
S
electrical circuit?
Let’s suppose the
capacitor in this
figure is charging
(I = dq/dt > 0) and
apply Ampère’s
law to Surface S1:
 B  dr  0  J  dA, to this
S
–q
∂S
B
q
r
Surface S2
(2πr)B = μ0I, so
B = μ0I/2πr .
I
Surface S1
The problem with Ampère’s law
How do we apply Ampère’s law,
 B  dr  0  J  dA, to this
S
electrical circuit?
Let’s suppose the
capacitor in this
figure is charging
(I = dq/dt > 0) and
apply Ampère’s
law to Surface S2:
S
–q
∂S
q
Surface S2
(2πr)B = 0, so
B=0.
I
Surface S1
The problem with Ampère’s law
How do we apply Ampère’s law,
S
electrical circuit?
Current flows
through Surface S1.
No current flows
through Surface S2.
 B  dr  0  J  dA, to this
S
–q
∂S
B
q
r
Hence we have a
μ0I/2πr = B = 0:
CONTRADICTION!
Surface S2
I
Surface S1
The problem with Ampère’s law
How do we apply Ampère’s law,
 B  dr  0  J  dA, to this
S
electrical circuit?
S
–q
∂S
B
q
r
Surface S2
CAN WE FIX THIS
CONTRADICTION?
I
Surface S1
The problem with Ampère’s law
You may say, “This is silly. Ampère’s law is not like Faraday’s
law, no matter what Ampère thought. Forget about S1 and S2!
Just consider how much current flows through the closed loop
∂S .”
Yes, but suppose the closed loop straddles the capacitor:
q
–q
I
∂S
How much current is flowing through the loop ∂S now?
Maxwell’s fix
J. C. Maxwell noticed this problem with Ampère’s law.
And there is another problem: In all of physics, interactions
are mutual. Newton’s third law states that to every action there
is an equal and opposite reaction. In electrostatics, a charge
creates an electric field and an electric field accelerates a
charge. In magnetostatics, a moving charge
creates a magnetic field and a magnetic field
accelerates a moving charge.
Maxwell’s fix
J. C. Maxwell noticed this problem with Ampère’s law.
And there is another problem: In all of physics, interactions
are mutual. Newton’s third law states that to every action there
is an equal and opposite reaction. In electrostatics, a charge
creates an electric field and an electric field accelerates a
charge. In magnetostatics, a moving charge
creates a magnetic field and a magnetic field
accelerates a moving charge.
But according to Faraday’s law, a changing
magnetic field creates an electric field….
Is this interaction mutual?
Maxwell’s fix
J. C. Maxwell noticed this problem with Ampère’s law.
And there is another problem: In all of physics, interactions
are mutual. Newton’s third law states that to every action there
is an equal and opposite reaction. In electrostatics, a charge
creates an electric field and an electric field accelerates a
charge. In magnetostatics, a moving charge
creates a magnetic field and a magnetic field
accelerates a moving charge.
Maxwell speculated that a changing electric
field could create a magnetic field.
Maxwell’s fix
Maxwell’s speculation could save Ampère’s law!
The capacitor in this
figure is charging.
If we consider
Surface S1, there is
a magnetic field B
because of the
current I through
S1. If we consider
Surface S2, there is
a magnetic field B
because of the
changing electric
field in the capacitor!
–q
∂S
B
q
r
Surface S2
I
Surface S1
Maxwell’s fix
This set of four fundamental equations for E and B,

E  dA 
q
0
(Gauss’s law)
 B  dA  0

d
E  dr    B
dt

d
B  dr   0 I   0  0  E ,
dt
is called Maxwell’s equations!
(Faraday’s law)
(Ampère’s law
as modified by
Maxwell)
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,

d
B  dr   0 I   0 0  E ,
dt
fixes the problem, for
the case of a parallelplate capacitor.
–q
∂S
B
q
r
If we consider
Surface S1, we have
(2πr)B = μ0I .
Surface S2
I
Surface S1
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,

d
B  dr   0 I   0 0  E ,
dt
fixes the problem, for
the case of a parallelplate capacitor.
If we consider
Surface S2, we have
(2πr)B = μ0ε0 dΦE/dt.
For a parallel-plate
capacitor, it still
equals μ0I!
–q
∂S
B
q
r
Surface S2
I
Surface S1
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,

d
B  dr   0 I   0 0  E ,
dt
fixes the problem, for the case of a parallel-plate capacitor.
If we consider Surface S1, we have (2πr)B = μ0I .
If we consider Surface S2, we have (2πr)B = μ0ε0 dΦE/dt.
For a parallel-plate capacitor of area A and plate separation d,
ΦE = AE = AV/d = Aq/Cd, and C = ε0 A/d, so ΦE = qA/Cd = q/ε0.
We get dΦE/dt = I/ε0 so (2πr)B = μ0ε0 dΦE/dt = μ0I!
Note I = ε0 dΦE/dt is often called the displacement current.
Maxwell’s fix
Without going into the details, we can already anticipate
electromagnetic radiation: A changing electric flux will
generate a transient magnetic field around it, which will
generate a transient electric field, etc. etc. while the wave
spreads through space.
 E
We will now develop the mathematical tools we need for
analyzing Maxwell’s equations. We begin with the curl
  E and Stokes’s theorem.
 E
The curl of a vector field E(x,y,z), written   E , is the
following vector field:
 






  E   E z  E y , E x  E z , E y  E x  .
z
z
x
x
y
 y

We can write it as a determinant:
i
j
k
  E   / x  / y  / z
Ex
Ey
Ez
.
 E
Example 1: Show that   E  0 if E(r)  V (r) for some
function V(r).
 E
Example 1: Show that   E  0 if E(r)  V (r) for some
function V(r).
Answer: We assume E(r)  V (r), which means that
Ex  V / x, E y  V / y, Ez  V / z. Substituting
these values into   E , we obtain
  V  V  V  V  V  V 

  E  

,

,

 y z z y z x x z x y y x 
 (0,0,0)  0 .
We write   V  0: the curl of a gradient vanishes.
 E
Example 2: If r = (x,y,z), what is   r ?
 E
Example 2: If r = (x,y,z), what is   r ?
Answer: r  r 2 / 2  ( x 2  y 2  z 2 ) / 2, i.e. r is a gradient,
so   r  0.
 E
Example 3: If B(r) = –yi+xj, what is   B ?
 E
Example 3: If B(r) = –yi+xj, what is   B ?
i
j
k
Answer:   B   / x  / y  / z  2k .
y
x
0
z
B
y
x
Stokes’s theorem
Examples of tessellated surfaces:
Stokes’s theorem
Examples of tessellated surfaces:
Stokes’s theorem
There is even computer-aided basket weaving:
Stokes’s theorem
Stokes’s theorem tells us how to compute an integral of the
form
 E  dr around a closed loop ∂S as an integral over a
S
surface S which has ∂S as its boundary.
For the proof of Stokes’s theorem, we assume that any surface
S can be approximated arbitrarily well by triangulation, i.e. by
reducing S to arbitrarily small triangles.
Stokes’s theorem
Stokes’s theorem tells us how to compute an integral of the
form
 E  dr around a closed loop ∂S as an integral over a
S
surface S which has ∂S as its boundary.
Here is an example of a triangulated surface:
S
∂S
Stokes’s theorem

Stokes’s theorem tells us that E  dr around the closed loop ∂S
S
 nE  dr over all the triangles in the
is equal to the sum
n
surface S that has ∂S as its boundary.
n-th triangle
∂S
S
Stokes’s theorem
To see why Stokes’s theorem is true, all we have to do is look

at two neighboring triangles and at E  dr for each one:
Stokes’s theorem
To see why Stokes’s theorem is true, all we have to do is look

at two neighboring triangles and at E  dr for each one:
The integrals on their common side sum to zero, because the
directions of integration are opposite!
Stokes’s theorem
When we put all the triangles together, the only integrals that
contribute (that are not paired with opposing integrals) are the
integrals along ∂S !
S
∂S
Stokes’s theorem
When we put all the triangles together, the only integrals that
contribute (that are not paired with opposing integrals) are the
integrals along ∂S !

Stokes’s theorem thus tells us that E  dr 
S
 nE  dr , i.e.
n
that the integral along ∂S equals the sum of the contribution of
each triangle.
We will now compute the contribution of an infinitesimal
triangle. In the limit of infinitely many triangles, the sum over
triangles will become an integral over the surface S.
Stokes’s theorem
1. We will assume that each triangle is a right-angled triangle,
since any triangle can be cut into two right-angled triangles.
2. We will place one corner of the triangle at a point (u,v) on a
uv-plane, and assume E depends linearly on Δu and Δv (since
the triangle is infinitesimally small).
3. We will use a lemma: If f(s) depends linearly on s, then the
integral of f(s) on the interval from s1 to s2 equals
s2

1
f( s ) ds  f( s2 )  f( s1 )s2  s1  ,
2
s1
i.e. the average of f(s1) and f(s2) times the separation s2 – s1.
Stokes’s theorem
1. We will assume that each triangle is a right-angled triangle,
since any triangle can be cut into two right-angled triangles.
2. We will place one corner of the triangle at a point (u,v) on a
uv-plane, and assume E depends linearly on Δu and Δv (since
the triangle is infinitesimally small).
1
3. The shaded area is f( s2 )  f( s1 )s2  s1 .
2
f(s2)
f(s)
f(s1)
s1
s2
s
Stokes’s theorem
So here is our right-angled triangle:
(v) Ev (u, v  v)  (u ) Eu (u, v  v)
(v) 2  (u ) 2
Ev(u,v+Δv)
(v) Ev (u  u, v)  (u) Eu (u  u, v)
(v) 2  (u) 2
Ev(u,v)
Eu(u,v)
Eu(u+Δu,v)
Stokes’s theorem

If we apply our lemma to compute E  dr for this triangle, we
obtain

1
E  dr  (u )[ Eu (u, v)  Eu (u  u , v)]
2
1
 [( v) Ev (u  u , v)  (u ) Eu (u  u , v)
2
 (v) Ev (u , v  v)  (u ) Eu (u , v  v)]
1
 (v)[ Ev (u , v  v)  Ev (u , v)] .
2
Stokes’s theorem

If we apply our lemma to compute E  dr for this triangle, we
obtain

1
E  dr  (u )[ Eu (u, v)  Eu (u  u , v)]
2
1
 [( v) Ev (u  u , v)  (u ) Eu (u  u , v)
2
 (v) Ev (u , v  v)  (u ) Eu (u , v  v)]
1
 (v)[ Ev (u , v  v)  Ev (u , v)] .
2
Stokes’s theorem

If we apply our lemma to compute E  dr for this triangle, we
obtain

1
E  dr  (u )[ Eu (u, v)  Eu (u  u , v)]
2
1
 [( v) Ev (u  u , v)  (u ) Eu (u  u , v)
2
 (v) Ev (u , v  v)  (u ) Eu (u , v  v)]
1
 (v)[ Ev (u , v  v)  Ev (u , v)] .
2
Stokes’s theorem

If we apply our lemma to compute E  dr for this triangle, we
obtain

1
E  dr  (u )[ Eu (u , v)  Eu (u , v  v)]
2
1
 [( v) Ev (u  u , v)  Ev (u, v)]
2
1
 Ev Eu 
 (u ) (v) 

.

2
v 
 u
Stokes’s theorem

If we apply our lemma to compute E  dr for this triangle, we
obtain

1
E  dr  (u )[ Eu (u , v)  Eu (u , v  v)]
2
1
 [( v) Ev (u  u , v)  Ev (u, v)]
2
1
 Ev Eu 
 (u ) (v) 

.

2
v 
 u
Stokes’s theorem

If we apply our lemma to compute E  dr for this triangle, we
obtain (in the limit of infinitesimal Δu and Δv):

1
 Ev Eu 
E  dr  (u ) (v) 


2

u

v


   E  dA .
This is the scalar product of the curl of E and the oriented area
element dA!
Stokes’s theorem
So here, at last, is Stokes’s theorem in all its glory:


E  dr  (  E)  dA
S
S
Stokes’s theorem
Faraday’s law says


d
d
E  dr    B  
B  dA .
dt
dt
S
But Stokes’s theorem says
S
 E  dr   (  E)  dA ,
S
and therefore

S
S
d
  E dA   B  dA .
dt

S
B
.
This is true for any S, hence   E  
t
Stokes’s theorem
Stokes’s theorem applied to the Ampère-Maxwell law:
 (  B)  dA  
d
B  dr   0 I   0 0  E
dt
S
S


d
  0 J  dA   0 0
E  dA .
dt
S
S
E 

and therefore   B   dA    0 J   0 0
  dA .
t 


S

S
E
.
This is true for any S, hence   B   0 J   0 0
t
Stokes’s theorem
With the aid of Stokes’s theorem, we have converted two of
Maxwell’s equations to differential form:
B
E  
t
E
  B   0 J   0 0
.
t
To convert the other two of Maxwell’s equations, we will need
the “divergence theorem”.
Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 2:
In 1929, M. R. Van Cauwenberghe was the first to measure
“displacement current”, on a round parallel-plate capacitor of
capacitance C = 100 pF and effective radius R = 40.0 cm. The
potential across the capacitor, which alternated with frequency
f = 50.0 Hz, reached V0 = 174 kV. What was the maximum
field B measureable at the edge of the capacitor? (Note: p =
pico = 10–12.)
Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 2:
In 1929, M. R. Van Cauwenberghe was the first to measure
“displacement current”, on a round parallel-plate capacitor of
capacitance C = 100 pF and effective radius R = 40.0 cm. The
potential across the capacitor, which alternated with frequency
f = 50.0 Hz, reached V0 = 174 kV. What was the maximum
field B measureable at the edge of the capacitor? (Note: p =
pico = 10–12.)
Answer: We can write (2πR)B = μ0I, where I = dq/dt =CdV/dt
is the charging rate of the capacitor. Let V = V0 sin(ωt) where
ω = 2πf, then dV/dt = V0 2πf cos(ωt). The maximum field B
satisfies 2πRB = μ0CV0 2πf, hence we obtain B = μ0CV0 f/R
= (4π×10–7 T·m/A)(10–10 F)(1.74×105 V)(50.0 Hz)/(0.0400 m)
= 2.73×10–8 T.