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Lecture 16-1
Magnetic field of a solenoid
L
• A constant magnetic field could be
produced by an infinite sheet of current.
In practice, however, it is easier and
more convenient to use a solenoid.
• A solenoid is defined by a current I
flowing through a wire that is wrapped n
turns per unit length on a cylinder of
radius R and length L.
Stretched-out solenoid
R
Lecture 16-2
Dipole Moments in Applied Fields
Magnetic Dipole
Electric dipole

p
 External
fields tend to
align dipoles.
B app
E app
p
Ep

 E decreases at center
B
 B increases at center
Lecture 16-3
Potential Energy of Dipole
• Work must be done to change the
orientation of a dipole (current loop)
in the presence of a magnetic field.
• Define a potential energy U (with
zero at position of max torque)
corresponding to this work.
B
x
F
q
F

θ
U   τdθ

U = +μ B cos θ
90
Therefore,
U   μB cos θ
U  μB cos θ 90
θ
.
Lecture 16-4
Reading Quiz 1
Three currents I1, I2, and I3 are directed perpendicular to
the plane of this page as shown. The value of the Ampere’s
Law line integral of B∙ dl counterclockwise around the
circular path is μo3I1 What is the magnitude of the currents
in I2 and I3?
a.
b.
c.
d.
e.
I2=0, I3 can be any value
I2=0, I3 can only be zero
I2=I1, I3 can be any value
I2=2I1, I3 can be any value
I2=0.5I1, I3 can be any value
I1
I3
I2
Lecture 16-5
Ampere’s Law in Magnetostatics
Biot-Savart’s Law can be used to derive another relation: Ampere’s Law
The path integral of the dot product of magnetic field and unit
vector along a closed loop, Amperian loop, is proportional to
the net current encircled by the loop,

C
Bt dl 
0 (i1  i2 )

C
B d l  0 I C
• Choosing a direction of
integration.
• A current is positive if it
flows along the RHR normal
direction of the Amperian
loop, as defined by the
direction of integration.
Lecture 16-6
Magnetization and “Bound” Current in Matter
• Strong externally applied field Bapp aligns the
magnetic moments in matter. Magnetization
M 

V

d 
M 

dV


• Ampere: Aligned magnetic moments in magnetized
matter arise due to microscopic current loops inside
the material. A Bound current
d   Adi
magnetic moment dµ due
to Amperian current di
d  Adi di
M 


Adl Adl dl
current /length
Equivalent to a solenoid of nI=M
 Bm  0nI  0 M
Lecture 16-7
MAGNETIC MATERIAL TYPES
Magnetic materials are placed inside a solenoid with a magnetic field Bs.
The following kinds of materials change the resultant B inside
the solenoid.
A| Stainless steel (paramagnetic)
~ +1% increase
B| Iron
(ferromagnetic)
~ +10-100 times increase
C| Copper
(diamagnetic)
~ -10-4 decrease
Lecture 16-8
Magnetic Susceptibility
• Magnetic susceptibility Χm
m  0
m  0
M  B app
m
 M
B app
0
 Bm  m Bapp
paramagnet
diamagnet
B  Bapp  0 M  (1  m ) Bapp
Relative permeability Km
B  K m B app
 B app 
  0 K m  


 0 
  Km 0
permeability
Lecture 16-9
Magnetism Exhibited by Materials
• Diamagnetism: (small) magnetic moment opposite
to the external magnetic field Bapp is induced
Any material – but
shows only if nonparamagnetic
repelled from region of large B
• Paramagnetism: magnetic moment of individual
atoms become aligned parallel to the applied
magnetic field Bapp
attracted toward region of large B
• Ferromagnetism: magnetic moment of individual
atoms are already (partially) aligned in some direction
even if Bapp=0
• Antiferromagnetism: like ferromagnetism except that
alternating moments are (partially) aligned opposite to each
other (when B=0)
transition
element, rare
earth, ...
Fe, Ni,...
Mn, Cr,...
Lecture 16-10
Paramagnetism and diamagnetism E-77
Lecture 16-11
Hysteresis for a Ferromagnet
Lack of retraceability shown is called hysteresis.
 Memory in magnetic disk and tape
 Alignment of magnetic domains
retained in rock (cf. lodestones)
Area enclosed in hysteresis loop

Energy loss per unit volume
 hard magnet: broad hysteresis loop
(hard to demagnetize, large energy loss,
higher memory)
 soft magnet: narrow hysteresis loop
(easy to demagnetize,…)
Lecture 16-12
INDUCTION
• Bar magnet approaches coil
S
N
v
Current induced in coil
• Reverse poles of magnet
N
S
v
Current in opposite
direction
N
• Bar magnet stationary
S
No induced current
v
• Coil moving around bar magnet
Same currents
induced in coil
S
N
What’s in common?: Change of Magnetic flux = EMF!
Lecture 16-13
Induction: Coil and B Fields 6D04
Lecture 16-14
Magnetic Flux
B
 B  BA2 cosq  B nA
1 Wb = 1 T m2
B 
B
ndA
S
Bi
Gauss’s Law
for Magnetism

S
B ndA  0
over closed surface
 B  NBA cosq
(N turns)
Lecture 16-15
Faraday’s Law of Induction
The magnitude of the induced EMF in conducting loop is
equal to the rate at which the magnetic flux through the
surface spanned by the loop changes with time.
d B

dt
where
 B   B ndA
N
S
Minus sign indicates the sense of EMF: Lenz’s Law
• Decide on which way n goes
Fixes sign of ΔϕB
• RHR determines the
positive direction for EMF
N
Lecture 16-16
Warm-up quiz 2
The magnetic field is decreasing, what’s the direction of the
induced currents in the closed rectangular loop?
A. Clockwise
B. Counterclockwise
C. No induced currents.
Lecture 16-17
Induced Electric Field from Faraday’s Law
• EMF is work done per unit charge:
ε W /q
• If work is done on charge q, electric field E must be present:
ε
E
nc
W   q Enc ds
ds
Rewrite Faraday’s Law in
terms of induced electric field:
r
d B
v
—
 Enc gds   dt
This form relates E and B!
B
• Note that  E  ds  0for E fields generated by charges at rest
(electrostatics) since this would correspond to the potential difference
between a point and itself. => Static E is conservative.
• The induced E by magnetic flux changes is non-conservative.
Lecture 16-18
Conducting Loop in a Changing Magnetic Field
Induced EMF has a direction such that it opposes
the change in magnetic flux that produced it.
approaching
 Magnetic moment μ
created by induced currrent
I repels the bar magnet.
Force on ring is repulsive.
moving away
 Magnetic moment μ
created by induced currrent
I attracts the bar magnet.
Force on ring is attractive.
Lecture 16-19
Faraday Induction and i2R Losses
6D14
Lecture 16-20
Faraday’s and Lenz’s Laws
 At 1, 3, and 5, ϕB is not changing.
So there is no induced emf.
 At 2, ϕB is increasing into page. So
emf is induced to produce a
counterclockwise current.
 At 4, ϕB in decreasing
into page. So current is
clockwise.
Lecture 16-21
Ways to Change Magnetic Flux
 B  BA cos q
• Changing the magnitude of the field within a conducting loop (or coil).
• Changing the area of the loop (or coil) that lies within the magnetic field.
• Changing the relative orientation of the field and the loop.
motor
generator
Lecture 16-22
Other Examples of Induction
+
-
Switch has been
open for some time:
Switch is just closed:
Nothing happening
EMF induced in Coil 2
+
-
Switch is just opened:
EMF is induced again
Switch is just closed:
EMF is induced in coil
-
+
Back emf
(counter emf)
Lecture 16-23
10:30 Quiz 3 October 20, 2011
A current directed toward the top of the page and a rectangular
loop of wire lie in the plane of the page. Both are held in place
by an external force. If the current I is decreasing, what is the
direction of the magnetic force on the left edge of the loop?
a. Toward the right
b. Toward the left
c. Toward top of page
d. Toward bottom of page
e. No force acts on it.
I
Lecture 16-24
11:30 Quiz 3 October 20, 2011
A current directed toward the top of the page and a
rectangular loop of wire lie in the plane of the page. If the
current I is increasing, what happens to the loop?
a. The loop is pulled toward the
top of the page
b. The loop is pulled toward the
current
c. A clockwise current is induced
in the loop.
d. A counterclockwise current is
induced in the loop.
e. Nothing happens to the loop
I
Lecture 16-25
Quiz B
Three currents I, 2I, and 3I are directed perpendicular to
the plane of this page as shown. What is the value of the
Ampere’s Law line integral of B∙ dl counterclockwise
around the circular path shown?
a.
b.
c.
d.
e.
4μ0I
-2μ0I
2μ0I
6μ0I
zero
2I
3I
I
Lecture 16-26
Quiz C
Three currents I, 2I, and 3I are directed perpendicular to
the plane of this page as shown. What is the magnitude of
the Ampere’s Law line integral of B∙ dl clockwise around
the circular path shown?
a. 5μ0I
b. 3μ0I
c. μ0I
d. 6μ0I
e. zero
3I
I
2I