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Our Goal:
Maxwell’s Equations
Gauss's law ( electric ) :
q
E
A
d
⋅
=
S∫
εo
Gauss's law in (magnetic):
0
∫ B ⋅ dA =
S
Faraday's law:
d ΦB
−
∫ E ⋅ ds =
dt
Ampere-Maxwell law
∫ B ⋅ ds = μo I + μo
d ΦE
dt
Electric Force and Field
q1q2
F12 = k 2 rˆ
r
dq
=
dE k ∫ 2 rˆ
E ∫=
r
  1
Φ E= 
∫ E ⋅ dA =
ε0
∫ ρ dV=
qin
ε0
Magnetic Force and Field
F
= qv × B
 
Φ B= 
∫ B ⋅ dA = 0
µo I ds × rˆ
B=
2
∫
r
4π
∫ B ⋅ ds=
μo ∫ J ⋅ dA= μo I
Continuous Charge Distribution
Note: For spheres, cylinders, infinite planes, rods, etc with high symmetry, the
electric field is found using Gauss’s Law which we’ll do next chapter!
dq
=
dE k ∫ 2 rˆ
E ∫=
r
Q
λ≡
linear charge density
L
Q
σ≡
surface charge density
A
Q
ρ≡
volume charge density
V
dq = λ dx
dq = σ dA
dq = ρ dV
Gauss’s Law
The net electric flux through ANY closed
surface is due to the net charge contained
inside that surface (divided by ε0)!
  qin
Φ E= 
∫ E ⋅ dA =
ε0
Biot-Savart Law
• Empirical observations are
summarized in the
mathematical equation called
the Biot-Savart law:
µo I ds × rˆ
B=
4π ∫ r 2
• The magnetic field described
by the law is the field due to
the current-carrying conductor
– Don’t confuse this field with
a field external to the
conductor

 μ I ds × ˆr
dB = o
4π r 2
Ampere’s Law
Ampere’s law states that the line integral of B . ds
around any closed path equals µoI where I is the total
steady current passing through any surface bounded by
the closed path.
∫ B ⋅ ds=
μo ∫ J ⋅ dA= μo I
s
I
B
⋅
d
=
μ
o
∫
Gauss’ Law in Magnetism
•
•
•
•
The magnetic flux associated
with a magnetic field is defined
in a way similar to electric flux
The unit of magnetic flux is
T.m2 = Wb is a weber
Magnetic fields do not begin or
end at any point
– The number of lines
entering a surface equals
the number of lines leaving
the surface
Gauss’ law in magnetism says
the magnetic flux through any
closed surface is always zero:
 
Φ B = ∫ B ⋅ dA
 
Φ B= 
∫ B ⋅ dA = 0
Maxwell’s Equations
Gauss's law ( electric ) :
q
E
A
d
⋅
=
S∫
εo
Gauss's law in (magnetic):
0
∫ B ⋅ dA =
S
Faraday's law:
d ΦB
−
∫ E ⋅ ds =
dt
Ampere-Maxwell law
∫ B ⋅ ds = μo I + μo
d ΦE
dt
Currents Produce Magnetic Fields
Smallest Magnet:
Electron
How to Find the B Field?
If there is high symmetry: Use Ampere’s Law
μ I
∫ B ⋅ ds =
o
Otherwise use the Biot-Savart Law:
µo I ds × rˆ
B=
2
∫
r
4π
Ampere’s Law
Ampere’s law states that the line integral of B . ds
around any closed path equals µoI where I is the total
steady current passing through any surface bounded by
the closed path.
⋅
=
I
B
d
s
μ
o
∫
∫ B ⋅ ds =
μo I = μo ∫ J ⋅ dA
Field Due to a Long Straight Wire
The current is uniformly distributed through the
cross section of the wire
• Outside of the wire, r > R
⋅ ds
∫ B =
B=
B( 2πr
=
) μo I
μo I
2πr
• Inside the wire, r < R
s
∫ B ⋅ d=
B( 2πr=
) μo I ' →
 μo I 
B=
r
2 
 2πR 
r2
I=
'
I
2
R
Magnetic Fields due to Currents
Magnitude of the Field:
µ0 I
B=
2π r
µ0
4π x10 T ⋅ m / A
−7
“mew not” : The permeability of free space.
Permeability is a measure of a materials ability to be
permeated by magnetic fields – ferromagnetic materials
have a higher permeability.
Currents Produce Magnetic Fields
Smallest Magnet:
Electron
B for a Long, Straight Conductor
• The magnetic field lines
are circles concentric with
the wire
• The field lines lie in
planes perpendicular to to
wire
• The magnitude of B is
constant on any circle of
radius a
• The right-hand rule for
determining the direction
of B is shown
Direction for a loop
Right Hand Rules
What is the direction of the current?
Electromagnets
What’s the direction of the B field?
A cross-sectional view of a coaxial cable is shown.
The center conductor is surrounded by a rubber
layer, which is surrounded by an outer conductor,
which is surrounded by another rubber layer. In a
particular application, the current in the inner
conductor is 1.00 A out of the page and the current
in the outer conductor is 3.00 A into the page.
Determine the magnitude and direction of the
magnetic field at points a and b.
Adding B Fields
The wires carry a current of 8.0A.
Find the magnitude of the net B field at A and B.
First, define positive directions. Up out of plane is positive.
µ 0 I1 µ 0 I 2 µ 0 I  1 1 
−=
B=
− 
netA

2π r1 2π r2
2π  r1 r2 
4π x10−7 Tm / A(8 A) 1
1
(
)
−
2π
.03m .15m
−5
BnetA = 4.27 x10 T
Is there any place along the line
where the fields cancel?
I1
I2
Current Affair
An electric current is flowing through
two parallel wires in the same
direction. Do the wires tend to
a) repel each other
b) Attract each other
c) Exert no force on each other
d) Twist at right angles to each other
e) Spin
F
Current Affair
F
F
Magnetic Force between two Current Carrying
Wires ON Each Other:
Opposite CURRENTS repel!
Like CURRENTS attract!
Magnetic Force between two Current Carrying
Wires ON Each Other:
F
= IL × B
F = ILB sin θ
Fon 2 = I 2 LB1at 2 sin θ
µ 0 I1

= I 2 L(
) sin 90
2π r
Fon 2
µ 0 I1 I 2 L
=
= − Fon1
2π r
Problem
The wires carry a current of 8.0A.
What is the force per meter on each wire? What direction?
Fon 2
µ 0 I1 I 2 L
=
= − Fon1
2π r
N
T=
A⋅ m
4π x10−7 Tm / A(8 A) 2
Fon 2 µ 0 I1 I 2
=
=
2π (.12m)
L
2π r
= 1.07 x10−4 N / m
I1
They repel each other.
I2
Questions
Is the net B field greater, less or the same at A and B?
Do the wires exert forces on each other?
Do the wires exert torques on each other?
If so, what directions?
Question
Is the wire attracted or repelled by the coil?
Compare E & B
Electric Force and Field
q1q2
F12 = k 2 rˆ
r
dq
=
dE k ∫ 2 rˆ
E ∫=
r
  1
Φ E= 
∫ E ⋅ dA =
ε0
∫ ρ dV=
qin
ε0
Magnetic Force and Field
F
= qv × B
 
Φ B= 
∫ B ⋅ dA = 0
µo I ds × rˆ
B=
2
∫
r
4π
∫ B ⋅ ds=
μo ∫ J ⋅ dA= μo I
Compare E & B
Symmetry
Use Gauss’s Law
qin
E
dA
⋅
=
∫
ε0
λl
E ( 2πrl ) =
ε0
λ
E=
2πε0 r
Use Ampere’s Law
μ I
∫ B ⋅ ds =
o
B( 2πr ) = μo I
μo I
B=
2πr
Compare E & B:
Infinite Current Sheet
Use Ampere’s Law
J
B = µ0
2
Infinite Charged Plane
Use Gauss’s Law
σ
E=
2ε 0
Circular Charge Distribution
Ering =
(x
kxQ
2
+a
)
2 3/ 2
Bx =
μo I R 2
2(x + R
2
2
)
Far from ring: Dipole Field
Ering
kxQ
= 3
a
μo I R 2
Bx =
2x 3
3
2
Compare Finite Rods
λ
E=
2πε 0 y
μo I
B=
2πa
Compare Gauss’s Law
Flux for Dipoles
 
Φ B= 
B
⋅
d
A
=
0
∫
  qin
Φ E= 
∫ E ⋅ dA =
ε0
Compare Dipoles
p ≡ 2aq, τ =
p × E, U =
-p ⋅ E
μ × E, U =
µ ≡ IA, τ =
-μ ⋅ B
Compare Dipoles
E0 & B0: External Field
Enet
= E0 − Eind
Reduces External Field
Bnet= B0 + Bind
Increases External Field
(ferromagnetic)
How to Find the B Field?
If there is high symmetry: Use Ampere’s Law
μ I
∫ B ⋅ ds =
o
Otherwise use the Biot-Savart Law:
µo I ds × rˆ
B=
2
∫
r
4π
Ampere’s Law
Ampere’s law states that the line integral of B . ds
around any closed path equals µoI where I is the total
steady current passing through any surface bounded by
the closed path.
∫ B ⋅ ds=
μo ∫ J ⋅ dA= μo I
s
I
B
⋅
d
=
μ
o
∫
A long cylindrical conductor of radius R carries a current I as
shown. The current density J, however, is not uniform over the
cross section of the conductor but is a function of the radius
according to J = br, where b is a constant. Find an expression
for the magnetic field B (a) at a distance r1 < R and (b) at a
distance r2 > R, measured from the axis.
Biot-Savart Law
• Empirical observations are
summarized in the
mathematical equation called
the Biot-Savart law:
µo I ds × rˆ
B=
4π ∫ r 2
• The magnetic field described
by the law is the field due to
the current-carrying conductor
– Don’t confuse this field with
a field external to the
conductor

 μ I ds × ˆr
dB = o
4π r 2
B for a Long, Straight Conductor
 μ I ds × ˆr
dB = o
4π r 2
• The thin, straight wire is
carrying a constant current
ds × ˆr= dxˆi × (cos θˆi + sin θˆj )= ( dx sin θ ) kˆ
a
r=
, x = r cos θ = a cot θ ,dx = −a csc 2 θdθ
sin θ

2
sin
ds × ˆr
θ
1
2
ˆ
=
−
sin
csc
=
sin θdθkˆ
θk
a
θdθ
2
2
r
a
a
μo I θ2
B=
sin θ dθ
∫
θ
4πa 1
μo I
( cos θ1 − cos θ2 )
4πa
•
If the conductor is an infinitely long,
straight wire, θ1 = 0 and θ2 = π
μo I
B=
2πa
Same result with Ampere’s Law
which is way easier
• Outside of the wire, r > R
⋅ ds
∫ B =
B=
B( 2πr
=
) μo I
μo I
2πr
• Inside the wire, r < R
s
∫ B ⋅ d=
B( 2πr=
) μo I ' →
 μo I 
B=
r
2 
 2πR 
r2
I=
'
I
2
R
The segment of wire carries a current of I = 5.00 A, where the radius of
the circular arc is R = 3.00 cm. Determine the magnitude and direction of
the magnetic field at the origin.
B for a Circular Current Loop
• The loop has a radius
of R and carries a
steady current of I
• Find B at point P
Bx =
μo I R 2
2(x + R
2
2
)
3
2
• At x = 0:
Bx
μo I R 2
μo I
=
3
2
2
2 ( x + R ) 2 2R
 μ I ds × ˆr
dB = o
4π r 2
Magnetic Field of a Coil
(Not a Solenoid)
Magnetic Field at the center of the loop:
B=
µ0 I
2R
N
Magnetic Field of a Solenoid
• A solenoid is a long
wire wound in the form
of a helix
• The field distribution is
similar to that of a bar
magnet
• As the length of the
solenoid increases
– the interior field becomes
more uniform
– the exterior field becomes
weaker
The Field Outside is nearly zero
because the fields from different
loops cancel each others
Ideal Solenoid
The field inside is
uniform the field
outside is ~zero.
∫ B ⋅ ds = ∫ B ⋅ ds = B ∫ ds = B
path1
path1
∫ B ⋅ ds = B = µ oNI
N
=
B μ=
I μo n I
o

What current is required in the windings
of a long solenoid that has 1 000 turns
uniformly distributed over a length of
0.400 m, to produce at the center of the
solenoid a magnetic field of magnitude
1.00 × 10–4 T?
N
=
B μ=
I μo n I
o

Magnetic Field of a Toroid
• Find the field at a
point at distance r
from the center of the
toroid
• The toroid has N turns
of wire
 
) μo N I
ds B( 2πr
=
∫ B ⋅=
μo N I
B=
2πr
A magnetic field of 1.30 T is to be set up in an iron-core
toroid. The toroid has a mean radius of 10.0 cm, and
magnetic permeability of 5 000 μ0. What current is required
if the winding has 470 turns of wire? The thickness of the
iron ring is small compared to 10 cm, so the field in the
material is nearly uniform.
μo N I
B=
2πr
Coil, Solenoid, Toroid
Coil:
B=
µ0 I
2R
N
N
B μ=
I μo n I
Solenoid:=
o

Toroid:
μo N I
B=
2πr
Magnetic Dipoles
The magnetic dipole moment
of a current loop enclosing an
area A is defined as
The SI units of the magnetic
dipole moment are A m2. The
on-axis field of a magnetic
dipole is
Smallest Magnet:
Electron
• The electrons move in
circular orbits
• The orbiting electron
constitutes a tiny current
loop
• The magnetic moment of
the electron is associated
with this orbital motion

• L is the angular
momentum

• µ is magnetic moment
1
=
μ I=
A
evr
2
Classical Atom
q e
e
=
I
= =
∆t T 2πr / v
In Bohr’s 1913 model of the hydrogen atom, the electron is in
a circular orbit of radius 5.29 × 10–11 m and its speed is 2.19 ×
106 m/s. (a) What is the magnitude of the magnetic moment
due to the electron’s motion? (b) If the electron moves in a
horizontal circle, counterclockwise as seen from above, what is
the direction of this magnetic moment vector?
Domains
• All ferromagnetic materials are made up of
microscopic regions called domains
– The domain is an area within which all
magnetic moments are aligned
• The boundaries between various domains
having different orientations are called
domain walls
Domains, Unmagnetized Material
• The magnetic
moments in the
domains are
randomly aligned
• The net magnetic
moment is zero
Domains, External Field Applied
• A sample is placed in
an external magnetic
field
• The size of the
domains with
magnetic moments
aligned with the field
grows
• The sample is
magnetized
Domains, External Field Applied
• The material is placed
in a stronger field
• The domains not
aligned with the field
become very small
• When the external
field is removed, the
material may retain a
net magnetization in
the direction of the
original field
Ferromagnetism
• Some substances exhibit strong magnetic effects
called ferromagnetism
• Some examples of ferromagnetic materials are:
–
–
–
–
–
iron
cobalt
nickel
gadolinium
dysprosium
• They contain permanent atomic magnetic
moments that tend to align parallel to each other
even in a weak external magnetic field
Paramagnetism
• Paramagnetic substances have small but
positive magnetism
• It results from the presence of atoms that
have permanent magnetic moments
– These moments interact weakly with each other
• When placed in an external magnetic field,
its atomic moments tend to line up with the
field
– The alignment process competes with thermal
motion which randomizes the moment
orientations
Diamagnetism
• When an external magnetic field is applied
to a diamagnetic substance, a weak
magnetic moment is induced in the
direction opposite the applied field
• Diamagnetic substances are weakly repelled
by a magnet
– Weak, so only present when ferromagnetism or
paramagnetism do not exist
Curie Temperature
•
•
The Curie temperature is
the critical temperature
above which a
ferromagnetic material loses
its residual magnetism
– The material will
become paramagnetic
Above the Curie
temperature, the thermal
agitation is great enough to
cause a random orientation
of the moments
Meissner Effect
• Certain types of
superconductors also
exhibit perfect
diamagnetism in the
superconducting state
– This is called the
Meissner effect
• If a permanent magnet is
brought near a
superconductor, the two
objects repel each other
Earth’s Magnetic Field
• The Earth’s magnetic field
resembles that achieved by
burying a huge bar magnet
deep in the Earth’s interior
• The Earth’s south
magnetic pole is located
near the north geographic
pole
• The Earth’s north
magnetic pole is located
near the south geographic
pole
Earth’s interior consists of a rocky
mantle and an iron rich core
The Earth’s magnetic field is caused by
dynamo movements in Earth’s core
Earth’s magnetic field flips
poles. Consecutive reversals
are spaced 5 thousand years
to 50 million years apart. The
last reversal happened
740,000 years ago. Some
researchers think our planet is
overdue for another one, but
nobody knows exactly when
the next reversal might occur.
The Sun's north magnetic pole is
pointing through the Sun's
southern hemisphere, until the
year 2012 when they will reverse.
This transition happens, as far as
we know, at the peak of every 11year sunspot cycle -- like
clockwork.
Babcock’s magnetic dynamo is one possible explanation of the
sunspot cycle where magnetic field lines become complexly
entangled after many solar rotations
The Sun’s magnetic fields create sunspots
Zeeman effect - spectral
lines split in regions of high
magnetic fields
Magnetic field lines connect sunspots on the Sun’s photosphere
Solar magnetic fields also create
other atmospheric phenomena
• plages
• filaments
Solar magnetic fields also create
other atmospheric phenomena
• plages
• filaments
• prominences
Solar magnetic fields also create
other atmospheric phenomena
•
•
•
•
plages
filaments
prominences
solar flares
Solar magnetic fields also create
other atmospheric phenomena
•
•
•
•
•
plages
filaments
prominences
solar flares
coronal holes
Solar magnetic fields also create
other atmospheric phenomena
•
•
•
•
plages
filaments
prominences
solar flares
• coronal holes
• coronal mass
ejections
(CMEs)
Reversals of the Earth’s
Magnetic Field
• The direction of the Earth’s magnetic field
reverses every few million years
– Evidence of these reversals are found in basalts
resulting from volcanic activity
– The rocks provide a timeline for the periodic
reversals of the field
• The rocks are dated by other means to determine the
timeline
This Week’s Lab
Using a Tangent Galvanometer to
Measure the Earth’s Magnetic Field
Magnetic Field at the center of the loop:
B=
µ0 I
2r
N
Horizontal
Earth Field
Total Field
1/--pages
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