Transcript Document

Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp. 32
Wednesday Morning
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Cartoon -. Opening Demo Warm-up problem
Physlet
Topics
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Finish up Mutual inductance
Ferromagnetism
Maxwell equations
Displacement current
Exam
Demos
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
m2  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.
71. Two coils, connected as shown, separately have inductances
L1 and L2. Their mutual inductance is M.
(a) Show that this combination can be replaced by a single coil of
equivalent inductance given by
Leq  L1  L2  2M .
We assume that the current is
changing at (nonzero) rate di/dt and
calculate the total emf across both
coils.
B
First consider coil 1. The magnetic
field due to the current in that coil
points to the left.
The magnetic field due to current in coil 2 also
points to the left. When the current increases, both
fields increase and both changes in flux contribute
emf’s in the same direction.
B
B
B

Thus, the induced emf’s are
di
di
1  ( L1  M )
and  2  ( L2  M ) .
dt
dt
Therefore, the total emf across both coils is
  1   2  ( L1  L2  2 M )
di
dt
which is exactly the emf that would be produced if the coils were replaced
by a single coil with inductance
Leq  L1  L2  2M .
(b) How could the coils in this figure be reconnected to yield an
equivalent inductance of
Leq  L1  L2  2M ?
We imagine reversing the leads of coil 2
so the current enter at the back of the coil
rather than front (as pictured in the
diagram). Then the field produced by coil
2 at the site of coil 1 is opposite to the field
produced by coil 1 itself.
B
B

The fluxes have opposite signs. An increasing current in coil 1 tends to
increase the flux in that coil, but an increasing current in coil 2 tends to
decrease it.
B
B
B
B
The emf across coil 1 is
di
1  ( L1  M )
dt
Similarly, the emf across coil 2 is
 2  ( L2  M )
di
dt
The total emf across both coils is
  ( L1  L2  2 M )
di
dt
This is the same as the emf that would be produced by a single coil with
inductance
Leq  L1  L2  2M
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
 Bwire ldr 

0il
 il
 il
b
dr  0 ln r aa b  0 ln(1 )
2r
2
2
a
Thus,

N N0 l  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 
M
ln1
2 

b 

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 Coulom bs
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.m2
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?
(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)
(a) The number of iron atoms in the iron bar is
7.9 g



cm3 5.0cm 1.0cm2
23
N

4
.
3

10
.
23
55.847g mol 6.02210 mol


Thus, the dipole moment of the bar is
  2.11023 J T 4.31023   8.9 A  m2 .


(b)   B sin 90o  8.9 A  m2 1.57T   13N  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
8.9A.m2

d
BT   dB  0  3
2 z
d 

AL
Adz 

L
.
A = 1 cm2
dz
P
z
B
 0
2 z 3

0 .01 dz 0
2
BT 

(
)(
) Evaluate at z= 0.01

3
2
2L .06 z
2L
z

B

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
3.

dB
E .ds  
dt
Faraday’s Law
B is increasing in
magnitude

Note that induced E
field is in such a
direction that the B field
it produces opposes the
original B field.
Note there is no electric potential
associated with the electric field
induced by Faraday’s Law
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.

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
0
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
E
 q/A

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
2
2R
0id
B
2r
Inside the capacitor
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?
37. A parallel-plate capacitor has
square plates 1.0 m on a side as
shown in the figure. A current of 2.0 A
charges the capacitor, producing a

uniform electric field E between the

plates, with E perpendicular to the
plates.
(a) What is the displacement current id through the region between the
plates?
At any instant the displacement current id in the gap between the plates
equals the conduction current i in the wires. Thus, id = i = 3.0 A.
(b) What is dE/dt in this region?
The rate of change of the electric field is
dE
1  d E  id
2.0 A
11 V





2
.
3

10
 0

m s
dt  0 A 
dt   0 A 8.851012 F m 1.0m2


(c) What is the displacement current through the square, dashed
path between the plates?
The displacement current through the indicated path is
2


0.50m 
 area enclosed by path 

id  id  
  2.0 A 1.0m   0.50A


 area of each plate 


(d) What is
 
 B  ds around this square, dashed path?
The integral field around the indicated path is
 
6
7



B

d
s


i

1
.
26

10
H
m
0
.
50
A

6
.
3

10
T m
0 d



(e) What is the value of B on this path?????
Summary of Maxwell Equations
Integral form
1. E 
2. B 
3.
4.

 E.dA  q
 B.dA  0
enc
/0
dB
E .ds  
dt
dE
 B.ds  00 dt  0ienc
Warm up set 10 Due 8:00 am Tuesday
1.
HRW6 31.TB.02. [120186] Suppose this page is perpendicular to a uniform magnetic field and the
magnetic flux through it is 5 Wb. If the page is turned by 30° around an edge the flux
through it will be:
4.3 Wb
10 Wb
5.8 Wb
2.5 Wb
5 Wb
2. HRW6 31.TB.08. [120192] Faraday's law states that an induced emf is proportional to:
the rate of change of the electric field
the rate of change of the magnetic field
zero
the rate of change of the magnetic flux
the rate of change of the electric flux
3. HRW6 31.TB.09. [120193] The emf that appears in Faraday's law is:
around a conducting circuit
perpendicular to the surface used to compute the magnetic flux
throughout the surface used to compute the magnetic flux
none of these
around the boundary of the surface used to compute the magnetic flux