Physics 1B schedule Winter 2008
Week
Mon
Friday
Wed
Lecture: the Electric field
1:Jan 7
2: Jan 14
Lecture: Intro, 15.1-15.3
Lecture: Electric Flux &
Gauss’s law
Lecture: The Coulomb law,
Lecture: Gauss
law/examples
3: Jan 21
University Holiday
Lecture: Potential
4: Jan 28
Lecture: capacitance
Lecture: capacitor
combinations
5: Feb 4
Lecture: Resistivity, Electric
power
Quiz 2: chapter 16
6: Feb 11
Lecture: Kirchhoff’s rules
Quiz 3: chapter 17
7: Feb 18
University Holiday
Lecture: Magnetism
Lecture: current loop,
solenoid
8: Feb 25
Lecture: Induced EMF
Lecture: Faraday’s law,
Lecture Inductance,
9: March 3
Lenz’s law
Inductors
Lecture: Energy of the
Lecture: AC circuits, EM
10: March 10 magnetic field
wave
HW: Ch15: 1,10,11,13,15,17,20,24,27,28,30,32,36,38,43,46,48
Ch16: 1,3,5,8,12,15,19,22,23,25,29,31,33,35,43,45,47,49,60
Ch17: 1,3,8,9,11,13,16,19,20,23,31,33,39,45,52,60
Ch18: 1,3,5,7,13,17,21,26,31,33,35,
Ch19: 1,3,8,9,11,15,19,22,24,27,29,34,37,38,41,44,47,49,57,61
Ch20: 1,5,8,11,13,16,18,23,25,27,29,31,34,37,39,
Quizzes:
HW problems, problems in class, more…
4 best out of 5
No make-up quizzes for any reason
Quiz 1: chapter 15
Lecture: equi-potential
surfaces
Lecture: Electric current,
Ohm’s law
Lecture: Resistors, series
parallel
Lecture: RC circuits
Lecture: torque on current
loop, Ampere’s law
Quiz 4: chapter 18-19
Quiz 5: chapter 20
Lecture: discussion of the
final exam
Instructor: D.N. Basov
T.A.: Eric Spencer
PB: TH 7-8:50pm LEDDN AUD
Final exam: all material in ch 15-21
no make up final for any reason
Last update:
February 10, 2008
19 Magnetism
Bar Magnet [19-1]
• Bar magnet ... two poles: N and S
Like poles repel; Unlike poles attract.
• Magnetic Field lines: (defined in same way as
electric field lines, direction and density)
S
•
N
Does this remind you of a similar case in electrostatics?
Bar Magnet [19-1]
r
E
Electric Field Lines
of an Electric Dipole
r
B
Magnetic Field Lines
of a bar magnet
S
N
19.3 Magnetic Fields
Magnetic field produces force on moving charges
F
B
q
θ
v
A charge q moving with velocity v in a magnetic
field B experiences a force F.
Force: F = qvB sinθ
Depends on the velocity and magnetic field
Depends on angle between vectors v and B
Is max when vectors v and B are perpendicular
Is perpendicular to the direction of B and v
19.3 Magnetic Fields
F
B
q
θ
v
r
F
+ --
r
F
Force: F = qvB sinθ
Depends on the velocity and magnetic field
Depends on angle between vectors v and B
Is max when vectors v and B are perpendicular
Is perpendicular to the direction of B and v
direction of F given by the right hand rule
r
F
19.3 Magnetic Fields
Magnetic field magnitude defined
F = qvB sinθ
F
B=
qv sinθ
Units of B:
Ns
= Tesla (T)
Cm
Also commonly used gauss (g)
104 gauss=1Tesla
Nicola Tesla
1865-1943
19.3 Magnetic Fields
Magnetic field notations
x x x x x x
x x x x x x
x x x x x x
B field into
page
B field out of the
page
think of arrows
19.3 Magnetic Fields
Magnetic Field:
Electric Field:
Electric Field vs. Magnetic Field
Magnetic Field:
Source
Acts on
Force
Direction
Field Lines
Electric
Magnetic
Charges
Charges
F = Eq
Parallel E
Moving Charges
Moving Charges
F = q v B sin(θ)
Perpendicular to v,B
Magnetic force on a current carrying wire 19.4
length L
in a perpendicular B field
F
B
q
I
A
L
L
F= sum of forces on all charges = ∑qvB
since ∑q=ρLA
then F=ρLAvB
∆q
since I = ∆t = ρAv
Finally
F = BIL
ρ is the charge/volume
Magnetic force on a current carrying wire 19.4
For angle θ between L and B
F
θ
q
B
I
F = BIL sinθ
B parallel to direction of wire, θ=0, F=0
B perpendicular to direction of wire θ=90o, F= BIL
Magnetic force on a current carrying wire 19.4
A transmission line carries a current of 100 A from east
to west. The magnetic field is 0.05 mT in the northward
direction. Find the force on a 100 m section of wire.
What direction is the force?
F = ILB
= 100(100)(0.05 x10 −3 ) = 0.5N
N
Direction
I
W
E
B
S
Looking from above
F is downward
Loudspeaker
Radial
direction
Torque on a current loop and electric motors 19.5
B field is uniform and in the plane of the current loop
Find the forces acting on the wires in the loop.
(a and b are the lengths)
F2= 0
F3= BIb
F1= BIb
into page
out of page
F4= 0
Torque on a current loop and electric motors 19.5
The current loop in a B field generates a torque
around the center proportional to the area of the loop
Side view
The two forces generate
a torque around the center
a
a
τ=F1( )+F2 ( )
2
2
τ=BIba
τ =BIA
F1=F2=BIb
A=axb=area of loop
counterclockwise
Torque on a current loop and electric motors 19.5
Loop makes an angle with B
Normal to the loop plane
τ = BIA sinθ
The torque tilts the loop
so the normal is parallel
to B
Torque on a current loop and electric motors 19.5
Loop with N turns of wire
N turns of wire
total current =NI
τ = NBIA sinθ
Torque increases with N, B, I and A
Torque is maximum when θ=90o,when the loop is
parallel to the field
Torque is zero when θ=0 when loop is perpendicular
to the field
A 3A current wire-loop (with 100 turns) and an area of
0.2 m2 makes an angle of 30o with a magnetic field of
0.3T.
a) Find the torque exerted on the coil.
b) What is the direction of rotation?
c) What happens if the current is reversed in the coil?
30o
B
a)
X
Θ=60o
τ = NBIA sinθ
= 100(0.3)(3.0)(0.2)sin60 = 1.6 x101Nm
b)
c)
counter clockwise direction
the torque will have the same magnitude but in the
opposite (clockwise) direction,
Torque on a current loop and electric motors 19.5
A current loop in a magnetic field produces a torque
Problem
A dc current does not produce complete rotation
τ = BIA sinθ
B
X
>
τ2
X
τ1
X
> τ3 =0
dc current only rotates coil until it is perpendicular to the field
Torque on a current loop and electric motors 19.5
Solution with direct current source
is to use a commutator.
Split-ring commutator reverses the current
direction when τ=0.
Motion of a charged particle in a magnetic field [19.6]
Fig.19-19
• Lorentz force:
F = qvB
x x x x x x x x x x x x
x x x x x x x x x x x vx B
• centripetal acc:
x x x x x x x x x x x x
2
v
v
=
a
F q
F
R
R=?
• Newton's 2nd Law:
F = ma ⇒
⇒
v2
qvB = m
r
mv
r =
qB
Motion of a charged particle in a magnetic field [19.6]
Application:
Mass spectrometer
Molecular ions
At velocity v
mv
r=
qB
Ions separated by mass
Magnetic field of a long, straight wire and Ampere’s law [19.7]
Hans Oersted 1820
No Current
Current turned on
Magnetic field of a long, straight wire and Ampere’s law [19.7]
Magnetic field lines around a current in straight wire - circle
with radius R
B
R
I
µo I
B=
2π R
B decreases with distance
µo = permeability of free space
=4πx10-7T·m/A
Thumb-along I
Fig.19.24
Fingers- around I
point along B
Magnetic field of a long, straight wire and Ampere’s law [19.7]
Connection between current and magnetic filed
Special case: General relation: The law of Andre Marie Ampere
(1775-1836)
straight wire
µo I
B=
2π R
closed path
I
∆L
Bll
B
∆L - line segment
B - magnetic field
Bll – component of B
parallel to ∆L
∑ B ∆L=µ I sum over all segments in the closed loop
ll
o
Ampere's Law: for a closed loop path, the sum of the
length elements times the magnetic field in the direction
of the length element is equal to the permeability times
the electric current enclosed in the loop.
Magnetic field of a long, straight wire and Ampere’s law [19.7]
The magnetic field around a straight wire calculated
from Ampere’s Law B
R
I
∑ ∆L = 2π R
∆L
The B field has a constant value at a constant radius R.
B and ∆L are in the same direction
Therefore, from Ampere’s Law
∑ B ∆L=B∑ ∆L=B2πR=µ I
ll
o
µoI
B=
2πR
from Ampere’s Law
Application of Ampere’s Law [19.7]
A coaxial cable has an inner conductor carrying current in one
direction and an outer conductor carrying an equal current in the
opposite direction. Find the B field due to the currents at a radius R
outside the coaxial cable.
I
I
R
B=0 since the total
current is equal to
zero.
The B fields due to the
two currents cancel
Magnetic force between 2 parallel conductors [19.8]
• Force on length L of wire b due r
F •
to field of wire a?
µI
Ba =
F = ILB
0 a
Ib
2πd
Ia
µ0Ia IbL
Magnitude
F
=
= b
2π d
of F on b
• Calculate force on length L of
wire a due to field of wire b:
µ0 I b
Bb =
2πd
µ0Ia Ib L
Magnitude
F
=
= a
2π d
of F on a
d
Ib
r
F
×
Ia
d
Two parallel wires 1.0 m in length separated by 4.0 cm
each carry a current of 20 A in opposite directions. Find
the force exerted between the two wires.
I=20 A
I=20 A
µoI2L 4π10−7 (20)2 (1)
−3
=
=
2x10
N
F=
2π(0.04)
2πR
Comparison:
Electric Field vs. Magnetic Field
Source
Acts on
Force
Direction
Electric
Magnetic
Charges
Charges
F = Eq
Parallel E
Moving Charges
Moving Charges
F = q v B sin(θ)
Perpendicular to v,B
Charges Attract
Currents Repel
Field Lines
Opposites
19.9 Magnetic field of a current loops and solenoids
Current loop
Solenoid
Electromagnets
19.9 Magnetic field of a current loops and solenoids
Magnetic field in a current loop
B field- out of loop
Straight wire
Current loop:
B field is the sum of fields
19.9 Magnetic field of a current loops and solenoids
B field due to current loop
looks like a magnetic dipole
Side view
B out of loop
N
I
S
B into loop
A current loop creates a magnetic dipole
the magnitude of the magnetic field at the center of a
circular loop with a radius R and carrying current I.
19.9 Magnetic field of a current loops and solenoids
Motivation. To construct electromagnets, i.e. a device
to convert current to magnetic field.
Some elements of design of electromagnets. The
magnetic field due to current through loops or coils of
wire.
single coil
solenoid multiple turns of wire
19.9 Magnetic field of a current loops and solenoids
Solenoid
B field lines
current out
of page
S
N
side view
current into
page
Current in a solenoid produces magnetic dipole
19.9 Magnetic field of a current loops and solenoids
Air Core Solenoid vs. Bar Magnet
19.9 Magnetic field of a current loops and solenoids
B field in solenoid
High field inside solenoid
Lower fields outside
Uniform relatively constant field in central region
19.9 Magnetic field of a current loops and solenoids
B-field in center by Ampere’s Law
L
∑
closed − loop
Bll ∆L = µo ∑ I = µo NI
N=no. of turns in length L
Only segment 1 contributes because
Bll∆L for other segments =zero
1
BL = µo NI
B=
4
2
3
µo NI
L
B = µo nI
N
n=
L
i.e. 2 in
the picture
19.9 Magnetic field of a current loops and solenoids
An electromagnet with has 100 turns of wire wound
around an air core with length of 3.0 cm. If a current
of 20 A is passed through the wire, what is the B field
at center of the magnet.
⎛N ⎞
B = µo nI = µo ⎜ ⎟ I
⎝L⎠
⎛ 100 ⎞
20
B = (4π x10 ) ⎜
⎟
⎝ 0.03 ⎠
B = 0.08T
−7
19.10 Magnetic domains and materials
Magnetic materials owe their properties
to magnetic dipole moments of electrons in atoms.
Applications
• permanent magnets,
• magnetic core electromagnets
• magnetic recording, magnetic tape, computer drives,
• credit cards
19.10 Magnetic domains and materials
An electron acts as a magnetic dipole
Spinning
charge
Classical model for magnetic dipole moment of electron
19.10 Magnetic domains and materials
Magnetic properties of matter
µ/µo
diamagnetic
Carbon
1-2x10-5
slightly less than vacuum
paramagnetic
Iron alum salt
1x10-5
slightly more than vacuum
ferromagnetic
Iron metal
1000-3000
much more than vacuum
19.10 Magnetic domains and materials
Soft magnetic materials
e.g. iron
Easily magnetized but doesn’t retain magnetization for long
Used as core for electromagnets
Hard magnetic materials
e.g. metal alloys Alnico (Aluminum, Nickel, Cobalt)
Hard to magnetize but retains the magnetization for a
long time
Used as permanent magnets.
19.10 Magnetic domains and materials
Magnetic Domains
Magnetism due to magnetic domains.
Each domain has millions of atoms with magnetic
moments coupled
Separated by domain boundaries
Soft magnetic materials-Boundary movement
10-4m
B
domain boundary shifts in B field to give
magnetization along B field direction
Hard magnetic materials
B
Magnetic dipoles reorient in the domains
to give a net magnetic moment.
Harder to do, i.e requires higher B field.
but also harder to reverse.
19.10 Magnetic domains and materials
Magnetization
Soft magnetic materials e.g. Fe nail can be magnetized
by exposure to a strong B field.
non-magnetic
magnetic
19.10 Magnetic domains and materials
Magnetic material
Solenoid
Iron core
Magnetic dipoles in iron are aligned by the
B field to produce a larger B field
19.10 Magnetic domains and materials
Iron core electromagnet
solenoid with many turns
iron-core
I
V
B=µnI
µ
≈ 1000
µo
The B field in the electromagnet is much higher with an iron core
than an air core.
19.10 Magnetic domains and materials
Applications of Iron core electromagnets
Electric motors, loudspeakers, electrical machinery
Iron-core
electromagnet
N
Magnetic
Field
N
S
S
19.10 Magnetic domains and materials
Magnetic recording
alternating current
Electromagnet
Iron core- magnetic
fields confined in the
core
Gap
Fringe B-fields
Magnetic
tape-magnetized
by fringe fields
19.10 Magnetic domains and materials
Magnetic tape
Information coded in the orientation of magnetic particles
Magnetization can be read on playback to generate a
voltage signal
Similar recording for computer hard disks, credit cards.
Information can be erased by magnetic fields.