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Lecture 12 Magnetism of Matter: Maxwell’s Equations Ch. 32
• Cartoon
• Opening Demo
• Topics
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Finish up Mutual inductance
Ferromagnetism
Maxwell equations
Displacement current
Demos
Elmo
Polling
What is the magnetic energy stored in a solenoid or
coil
dU B
di
 Li
dt
dt

dUB  Lidi
2
For an inductor L
UB  12 Li

UB
0
dU B 
i

Lidi
UB 
0

Lidi  12 Li 2
0
Now define the energy per unit volume
U
uB  B
Al

Li 2 L i 2
uB 

Al
l 2A
1
2

Area A
i

l

The energy density formula
is valid in general

Li 2 1
uB 
 2  0 n 2i 2
Al

2
B
uB 
20

1
2
L
 0 n 2 A
l
B  0ni
E2
uE 
20
What is Mutual Inductance? M
When two circuits are near one another and both have currents
changing, they can induce emfs in each other.
1
2
I2
I1
m1  L1I1  M 21I 2
m 2  L2 I 2  M12 I1
M12  M 21  M
On circuit boards you have to be careful you do not put circuits near
each other that have large mutual inductance.
They have to be oriented carefully and even shielded.
75. A rectangular loop of N closely
packed turns is positioned near a
long, straight wire as shown in the
figure.
(a) What is the mutual inductance M
for the loop-wire combination?
(b) Evaluate M for N = 100, a = 1.0 cm,
b = 8.0 cm, and l = 30 cm.
(a) The flux over the loop cross section due to the current i in the wire is
given by

a b
a b
a
a
 Bwireldr 

0il
 il
 il
b
dr  0 ln r aa b  0 ln(1 )
2r
2
2
a
M

N
i
N N0l  b 
M

ln1 
i
2  a 
(b) Evaluate M for N = 100, a = 1.0 cm,
b = 8.0 cm, and l = 30 cm.
N0 l  b 
M
ln1 
2  a 

(b) From the formula for M obtained,


100 4 107 H m  0.30m  8.0 
M
ln 1 

2
1
.
0


 1.3 10 5 H
Ferromagnetism
Iron, cobalt, nickel, and rare earth alloys exhibit ferromagnetism.
The so called exchange coupling causes electron magnetic moments
of one atom to align with electrons of other atoms. This alignment produces
magnetism. Whole groups of atoms align and form domains.
(See Figure 32-12 on page 756)
A material becomes a magnet when the domains line up adding all the
magnetic moments.You can actually hear the domains shifting by bringing
up an magnet and hear the induced currents in the coil. Barkhausen Effect
Two other types of magnetic behavior are paramagnetism or
diamagnetism.
What is the atomic origin of magnetism?
Electron spinning on its axis
Electron orbiting around the nucleus
Spin Magnetic Dipole Moment of the Electron
e
S
m
e  1.6 109 Coulombs
 
m  9.111031 kg
S is the angular momentum due to the electron’s spin. It
has units kg.m2/s. has units of A.m2 - current times area
Recall for a current loop, the magnetic dipole moment =
current times area of loop
In the quantum field theory of the electron, S can not be measured.
Only it’s component along the z axis can be measured. In quantum
physics, there are only two values of the z component of the electron spin.
Therefore, only the z component of can be
measured.Its two possible values are:
eh
z  
4m
Corresponding to the two values of
the electron spin quantum number +1/2
and -1/2
The above quantity is called the Bohr magneton and is equal to:
eh
B  z 
 9.27 1024 A.m 2
4m
The magnetic moment of the electron is the prime origin of
ferromagnetism in materials.
22. The dipole moment associated with an atom of iron in an iron bar
is 2.1x10-23 J/T. Assume that all the atoms in the bar, which is 5.0 cm
long and has a cross-sectional area of 1.0 cm2, have their dipole
moments aligned.
(a) What is the dipole moment of the bar?
The number of iron atoms in the iron bar is
7.9 g cm 5.0cm 1.0cm 

N
 4.3  10
55.847 g mol  6.022  10 mol 
3
2
23
Thus, the dipole moment of the bar is
  2.1  1023 J T 4.3  1023  9.03A  m2 .
23
.
(b) What torque must be exerted to hold this magnet perpendicular
to an external field of 1.5 T? (The density of iron is 7.9 g/cm3)


(b)    Bsin90o  9.03A  m2 1.5T   13.5N  m
(C) Use the dipole formula to find the magnitude and direction of the
magnetic field 1cm from the end of the bar magnet on its central axis at P.
5 cm
.
A = 1 cm2
8.9A.m2
BT 
d 
 dB 

AL
0  .01 dz

0
2
Adz 


L
d
z3
dz
0 
2
BT 
(
)( 2
3

2 L .06 z
2 L
z
B

P
z
B
0
2z 3

0 
2
)=(
)(
)
2
2 L
(0.01)
0.06
0.01
4 107 AN2 8.9A.m 2
  0.05m(.01m) 2
4 1078.9N / A.m
B
5 106
B  0.71 T
BigBite is a 50 ton
electromagnet with
a 25 cm by 100 cm
gap
B = 1 Tesla
Maxwells Equations:
In 1873 he wrote down 4 equations which govern all
classical electromagnetic phenomena.
You already know two of them.
1. E 
 E .dA  q
enc
/0

2. B 
 B.dA  0
A magnetic field changing with time can produce an
electric field: Faraday’s law
3.

dB
E .ds  
dt
Line integral of the electric field
around the wire equals the change of
Magnetic flux through the area
Bounded by the loop
Electric lines curl around
changing magnetic field lines
Example
New Question: Can a changing electric field with
time produce an magnetic field?
Yes it can and it is called
Maxwell’s law of induction
.
dE
 B.ds  00 dt
Maxwell’s law of
induction
dE
 B.ds  00 dt
Consider the charging of our circular
plate capacitor
B field also
induced at
point 2.
When capacitor stops charging
B field disappears.
current ever
actually flows
through the
capacitor

Find the expression for the induced magnetic field B
that circulates around the electric field lines of a
charging circular parallel plate capacitor
r<R
Flux within
the loop of
radius r
dE
 B.ds  00 dt
E
 B.ds  (B)(2r) since B parallel ds
B
r
R

2 dE
(B)(2r)  00r
 dt
dE
d(AE)
dE
00
 00
 00 A
dt
dt
dt
r< R
r>R
B
00r dE
2
dt
B
00 R 2 dE
2r
dt
Ampere-Maxwell’s Law
dE
 B.ds  00 dt
This term has units of current

4.
 B.ds   i
0
Maxwell combined the above
two equations to form one
equation
dE
 B.ds  00 dt  0ienc
How do we interpret this equation?

What is the displacement current?
dE
 B.ds  00 dt  0ienc
dE
0
 id
dt
 B.ds   i
0 d
This is called the
displacement current id
 0ienc

The term is really is a transfer of electric and magnetic energy from one
plate to the other while the plates are being charged or discharged.
When charging stops, this term goes to zero. Note it is time dependent.

Show that the displacement current in the gap of the two
capacitor plates is equal to the real current outside the gap
dE
0
 id
dt
dE
 B.ds  00 dt
Can I detect the magnetic field associated with displacement current?

Calculation of id
First find the real current i
 q/A
E 
0
0
q  0 AE
i
dq
dE
 0 A
dt
dt
For the field inside a parallel plate capacitor
Solving for q
This is the real current i charging the capacitor.
Next find the displacement current
dE
id  0
dt
d(AE)
dE
id  0
 0 A
dt
dt
displacement current =
real current. No charge
actually moves across
the gap.
Calculate Magnetic field due to displacement current
Current is uniformly spread over the circular plates of the capacitor.
Imagine it to be just a large wire of diameter R. Then use the
formula for the magnetic field inside a wire.
0id
B(
)r Inside the capacitor
2
2R
0id
B
2r
Outside the capacitor
Question 11: A circular capacitor of radius R is being
charged through a wire of radius R0. Which of the
points a, b, c, and d correspond to points 1, 2, and 3
on the graph
Where is the radius R0 and R on the graph?
Summary of Maxwell Equations
Integral form
1. E 
2. B 
 E .dA  q
 B.dA  0
dB
E .ds  
dt
enc
/0
3.

4.
 
d E
 B.ds  0 0 dt  0ienc
Chapter 32 Problem 56
Suppose that  4 are the limits to the values of mc for an
electron in an atom.
(a) How many different values of the z component µorb,z of the
electron’s orbital magnetic dipole moment are possible?
(b) What is the greatest magnitude of those possible values?
Next suppose that the atom is in a magnetic field of magnitude
0.250T, in the positive direction of the z axis.
(c)What are the maximum potential energy
(d)What are the minimum potential energy associated with
those possible values of µorb,z?
Chapter 32 Problem 61
In Figure 32-39, the capacitor with circular plates of radius R = 16.0
cm is connected to a source of emf script e = script em sin ωt, where
script em = 220 V and ω = 120 rad/s. The maximum value of the
displacement current is id = 7.60 µA. Neglect fringing of the electric
field at the edges of the plates.
Fig. 32-39
(a) What is the maximum value of the current i in the circuit?
(b) What is the maximum value of dE/dt, where E is the electric flux
through the region between the plates?
(c) What is the separation d between the plates?
(d) Find the maximum value of the magnitude of B between the
plates at a distance r = 11.0 cm from the center.