Physics 2 for Electrical Engineering
Ben Gurion University of the Negevwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
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 • • Stokes’s theorem Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38.
E
Introduction
The set of four fundamental equations for E and B,
together with the Lorentz force law FEM = q (E + v × B), sum up everything we have learned so far about electromagnetism!
,
0
0
0
Id
dtdd
d
qd
B
rB
rE
AB
AE
(Ampère’s law)
(Gauss’s law)
(Faraday’s law)
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” and the “divergence” – and corresponding theorems: “Stokes’s theorem” and the “divergence theorem”.
E E
Introduction
New notation: If S denotes a surface (which may be curved), then ∂S denotes the boundary of the surface S:
Rutgers
S ∂S
x y
z
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:
I = 0
I
Bds
.)( 0Id srB
The problem with Ampère’s law
Now let’s compare Ampère’s law with Faraday’s law:
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.
.
0
ABrE
rB
ddtdd
Id
SS
(Faraday’s law)
(Ampère’s law)
The problem with Ampère’s law
Now let’s compare Ampère’s law with Faraday’s law:
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?
.
0
ABrE
rB
ddtdd
Id
SS
(Faraday’s law)
(Ampère’s law)
The problem with Ampère’s law
Now let’s compare Ampère’s law with Faraday’s law:
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?
.
0
ABrE
rB
ddtdd
Id
SS
(Faraday’s law)
(Ampère’s law)
The problem with Ampère’s law
Yes, we can!
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.
. 0 AJrB dd SS
(Ampère’s law)
The problem with Ampère’s law
This is indeed what Ampère intended with his law,
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.
, 0 AJrB dd SS
Surface S1
Surface S2
I
–q
q
∂S
The problem with Ampère’s law
How do we apply Ampère’s law, to this
electrical circuit?
,0 AJrB dd SS
Surface S1
Surface S2
I
–q
q
r
B∂S
The problem with Ampère’s law
How do we apply Ampère’s law, to this
electrical circuit?
Let’s suppose thecapacitor in this figure is charging(I = dq/dt > 0) and apply Ampère’s law to Surface S1:
(2πr)B = μ0I, so B = μ0I/2πr .
,0 AJrB dd SS
Surface S1
Surface S2
I
–q
q
∂S
The problem with Ampère’s law
How do we apply Ampère’s law, to this
electrical circuit?
Let’s suppose thecapacitor in this figure is charging(I = dq/dt > 0) and apply Ampère’s law to Surface S2:
(2πr)B = 0, so B = 0 .
,0 AJrB dd SS
Surface S1
Surface S2
I
–q
q
r
B∂S
The problem with Ampère’s law
How do we apply Ampère’s law, to this
electrical circuit?
Current flows through Surface S1.No current flows through Surface S2.
Hence we have aμ0I/2πr = B = 0:
CONTRADICTION!
,0 AJrB dd SS
Surface S1
Surface S2
I
–q
q
r
B∂S
The problem with Ampère’s law
How do we apply Ampère’s law, to this
electrical circuit?
CAN WE FIX THIS
CONTRADICTION?
,0 AJrB dd SS
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:
How much current is flowing through the loop ∂S now?
I
∂S
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 changingmagnetic 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.
Surface S1
Surface S2
I
–q
q
r
B∂S
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!
Maxwell’s fix
This set of four fundamental equations for E and B,
is called Maxwell’s equations!
,
0
000
0
E
B
dtdId
dtdd
d
qd
rB
rE
AB
AE
(Ampère’s law as modified by Maxwell)
(Gauss’s law)
(Faraday’s law)
Surface S1
Surface S2
I
–q
q
r
B∂S
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,
fixes the problem, for the case of a parallel-plate capacitor.
If we considerSurface S1, we have (2πr)B = μ0I .
, 000 EdtdId rB
Surface S1
Surface S2
I
–q
q
r
B∂S
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,
fixes the problem, for the case of a parallel-plate capacitor.
If we considerSurface S2, we have (2πr)B = μ0ε0 dΦE/dt.For a parallel-plate capacitor, it stillequals μ0I!
, 000 EdtdId rB
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,
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.
, 000 EdtdId rB
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 and Stokes’s theorem.
E
E
The curl of a vector field E(x,y,z), written , is the following vector field:
We can write it as a determinant:
E
. , ,
xyzxyz Ey
Ex
Ex
Ez
Ez
Ey
E
. ///
zyx EEEzyx
kjiE
E
Example 1: Show that if for some function V(r).
0E )()( rrE V
E
Example 1: Show that if for some function V(r).
Answer: We assume , which means that
Substituting
these values into , we obtain
We write : the curl of a gradient vanishes.
)()( rrE V
)()( rrE V
./ ,/ ,/ zVEyVExVE zyx
E
. (0,0,0)
, ,
0
E
xV
yyV
xzV
xxV
zyV
zzV
y
0E
0 V
E
Example 2: If r = (x,y,z), what is ?r
E
Example 2: If r = (x,y,z), what is
Answer: i.e. r is a gradient, so
?r
.0r ,2/)(2/ 2222 zyxr r
E
Example 3: If B(r) = –yi+xj, what is ?B
E
Example 3: If B(r) = –yi+xj, what is
Answer: . 20/// kkji
B
xy
zyx
x
y
z
B
?B
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 around a closed loop ∂S as an integral over a
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.
S
rE d
Stokes’s theorem
Stokes’s theorem tells us how to compute an integral of the
form around a closed loop ∂S as an integral over a
surface S which has ∂S as its boundary.
Here is an example of a triangulated surface:
S
rE d
∂S
S
Stokes’s theorem
Stokes’s theorem tells us that around the closed loop ∂S
is equal to the sum over all the triangles in the
surface S that has ∂S as its boundary.
S
rE d
n
nd rE
∂S
n-th triangleS
Stokes’s theorem
To see why Stokes’s theorem is true, all we have to do is look
at two neighboring triangles and at for each one: rE d
Stokes’s theorem
To see why Stokes’s theorem is true, all we have to do is look
at two neighboring triangles and at for each one:
The integrals on their common side sum to zero, because the directions of integration are opposite!
rE d
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 i.e.
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.
, n
ndd rErE
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
i.e. the average of f(s1) and f(s2) times the separation s2 – s1.
, )f()f(21 )f( 1212
2
1
ssssdsss
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. The shaded area is . )f()f(21
1212 ssss
s
f(s)
s1 s2
f(s1)
f(s2)
Stokes’s theorem
So here is our right-angled triangle:
Eu(u,v) Eu(u+Δu,v)Ev(u,v)
Ev(u,v+Δv)
22 )()(
),()(),()(
uv
vuuEuvuuEv uv
22 )()(
),()(),()(
uv
vvuEuvvuEv uv
Stokes’s theorem
If we apply our lemma to compute for this triangle, we
obtain
rE d
. )],(),([)(21
)],()(),()(
),()(),()[(21
)],(),()[(21
vuEvvuEv
vvuEuvvuEv
vuuEuvuuEv
vuuEvuEud
vv
uv
uv
uu
rE
Stokes’s theorem
If we apply our lemma to compute for this triangle, we
obtain
rE d
. )],(),([)(21
)],()(),()(
),()(),()[(21
)],(),()[(21
vuEvvuEv
vvuEuvvuEv
vuuEuvuuEv
vuuEvuEud
vv
uv
uv
uu
rE
Stokes’s theorem
If we apply our lemma to compute for this triangle, we
obtain
rE d
. )],(),([)(21
)],()(),()(
),()(),()[(21
)],(),()[(21
vuEvvuEv
vvuEuvvuEv
vuuEuvuuEv
vuuEvuEud
vv
uv
uv
uu
rE
Stokes’s theorem
If we apply our lemma to compute for this triangle, we
obtain
rE d
. )( )(21
)],(),()[(21
)],(),()[(21
vE
uE
vu
vuEvuuEv
vvuEvuEud
uv
vv
uurE
Stokes’s theorem
If we apply our lemma to compute for this triangle, we
obtain
rE d
. )( )(21
)],(),()[(21
)],(),()[(21
vE
uE
vu
vuEvuuEv
vvuEvuEud
uv
vv
uurE
Stokes’s theorem
If we apply our lemma to compute for this triangle, we
obtain (in the limit of infinitesimal Δu and Δv):
This is the scalar product of the curl of E and the oriented area element dA!
rE d
.
)( )(21
AE
rE
d
vE
uE
vud uv
Stokes’s theorem
So here, at last, is Stokes’s theorem in all its glory:
SS
AErE dd )(
Stokes’s theorem
Faraday’s law says
But Stokes’s theorem says
and therefore
This is true for any S, hence
, )( SS
AErE dd
. SS
ABrE ddtd
dtdd B
. SS
ABAE ddtdd
.t
BE
Stokes’s theorem
Stokes’s theorem applied to the Ampère-Maxwell law:
and therefore
This is true for any S, hence
. 000
SS
AEJAB dt
d
.000 t
EJB
.
)(
000
000
SS
SS
AEAJ
rBAB
ddtdd
dtdIdd E
Stokes’s theorem
With the aid of Stokes’s theorem, we have converted two of Maxwell’s equations to differential form:
To convert the other two of Maxwell’s equations, we will need the “divergence theorem”.
. 000 t
t
EJB
BE
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.
