Physics 122B
Electricity and Magnetism
Lecture 26
Review
Martin Savage
Lecture 26 Announcements
The Final Exam is Tuesday June 5 at 2.30 – 4.20 pm
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Physics 122B - Lecture 26
2
Charges and Forces
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Dipoles and Forces (2)
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Coulomb’s Law
Like charges repel.
Charles Augustine
de Coulomb
(1736-1806).
Opposite charges attract.
Coulomb’s Law:
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F1 on 2  F2 on 1  K
Physics 122B - Lecture 26
q1 q2
r
2
( Magnitude of force )
5
Using Coulomb’s Law
1. Coulomb’s Law applies only to point charges.
(This is particularly important because charge are
free to move around on conductors.)
1. Strictly speaking, Coulomb’s Law applies only to
electrostatics (non-moving charges).
(However, it is usually OK provided v<<c).
1. Electrostatic forces can be superposed.
Linear superposition !!!!
Fnet  F1 on j  F2 on j  F3 on j 
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Fields
Fields:
1. The electric field permeates all of space
2. Charges interact with the electric field
inducing a non-zero value of E (subject to
boundary conditions).
3. A charge in an electric field experiences a
force F exerted by the field.
E ( x, y , z ) 
Fon q at (x,y,z)
q
3. This relation assigns a field vector to every
point in space.
4. If q is positive, the electric field vector points
in the same direction as the force vector.
5. Does E depend on q (the charge that detects
it)? No, because the force is proportional to q.
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F  qE
7
Unit Vector Notation
Positive q
Negative q
1
q
E
rˆ
2
4 0 r
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Two Positive Charges
Field Map
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Field Lines
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Types of Symmetry
Translational symmetry:
translation along a line
does not change object.
Full rotational symmetry:
any rotation about any
axis does not change
object: Sphere
Cylindrical symmetry:
any rotation about one
axis does not change
object: Cylinder
Reflectional symmetry:
mirror image reflection
is same as object.
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A
(yes)
A
E
(no)
E
Partial rotational symmetry:
720 rotation about one axis
does not change object: Star
720
10
The E-Field of a Charged Ring
A thin uniformly charged ring of radius R has a
total charge Q. Find the electric field on the axis
of the ring .. the z-axis …(perpendicular to the
page).
The linear charge density of the ring is l =
Q/(2R). The system has cylindrical symmetry for
rotations about the axis, so along the z-axis Ex=Ey=0
and we need only to find Ez. Consider a small
segment of the circumference of the ring of width
dl = R df. The contribution to the electric field on
the z-axis is :
1 dQ
1 l Rdf
z
dEz 
cos


4 0 ri 2
4 0 R 2  z 2 R 2  z 2
Ez 

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1
l Rz
4 0  R 2  z 2 3/ 2
1
2

0
df 
1
l Rz
2 0  R 2  z 2 3/ 2
Qz
4 0  R 2  z 2 3/ 2
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11
An Infinite Charged Plane
An infinite plane of charge is the limiting case
of a disk of charge when the disk radius is allow
to become very large, i.e., R. As we have seen,
in that case:
Ez ( R  ) 

 constant
2 0
The E field must change direction across the
plane, so:
Eplane
 
 2

0

 

 2 0
kˆ if z>0
(independent of distance)
kˆ if z<0
Note that the system has translational symmetry
along any direction in the plane and rotational
symmetry about any perpendicular. Therefore, E
cannot depend on (x,y) position and Ex=Ey=0.
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Charge Conjugation Symmetry
Charge Conjugation
There is an additional symmetry associated with electric charges.
If we change all positive charges to negative and all negative charges to
positive without changing the magnitudes of the charges, the field must
reverse direction, but must not change in any other way.
Charge Conjugation
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Flux Surface Integral
For closed Gaussian surfaces (i.e. encloses a non-zero volume)
we write the flux integral as:
 e   E  dA
Here, the “loop” on the integral sign indicates a surface
integral over a closed surface. Since a simple non-intersecting
closed surface has a definite inside and outside, we take the
vector direction of dA as always pointing out of the volume (and
normal to the surface of course).
This integral is related to the amount of charge enclosed by
the surface !!
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Gauss’s Law
The previous arguments lead to the conclusion that
the correspondence between the charge Qin enclosed
by a surface and the net flux e through that surface
is a general result. It is called Gauss’s Law, and is
usually written as:
 e   E  dA 
Qin
0
Johann Carl Friedrich Gauss
(1777 – 1855)
Gauss’s Law is the first of four master equations, collectively called
Maxwell’s Equations, that together constitute a “unified field theory” of
electromagnetism. In essence, Gauss’s Law says that diverging field lines
from a point indicate the presence of an electric charge at that point,
and that this charge can be “detected” by surrounding the point with a
surface and observing the flux through the surface.
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Summary: The E-Field of
a Solid Sphere of Charge
xr
out
x
rin
R
Q
Eoutside  (1/ 4 0 )Q / r 2
Einside  (1/ 4 0 )Qr / R 3
Esurface  (1/ 4 0 )Q / R 2
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The Surface Field of a Conductor
Suppose that the electric field at the surface
of a conductor had a component tangent to the
surface. Then mobile electric charges at the
surface would feel a force and would move in
response to the field, until equilibrium was
established and the tangential field vanished.
Therefore, the field at the surface of a conductor
must be perpendicular to the surface, with no
tangential components.
Suppose that a charged conductor has local
surface charge density . Put a perpendicular
cylinder of end-cap area A through this region of
the surface. The field on the inner end cap is
zero, and the field is perpendicular to the area
vectors on the curved wall of the cylinder, so e =
EsurfA  Qin/0  A/0. Therefore, Esurf = /0. (Note
that this matches the surface field of a sphere.)
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Current and Drift Velocity
If the electrons have an average drift
speed vd, then on the average in a time
interval Dt they would travel a distance Dx in the wire,
where Dx = vd Dt. If the wire has cross sectional area
A and there are n electrons per unit volume in the wire,
then the number of electrons moving through the cross
sectional area in time Dt is Ne = n A Dx = n A vd Dt = i Dt .
Therefore,
This table gives n
for various metals.
i  nAvd
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Establishing the
Electric Field in a Wire (2)
The figure shows the region of the wire near the neutral midpoint. The
surface charge rings become more positive to the left and more negative
to the right.
In Chapter 26, we found that a ring of charge makes an on-axis E field that:
1. Points away from a positive ring and toward a negative ring;
2. Is proportional to the net charge of the ring;
3. Decreases with distance from the ring.
The non-uniform surface charge distribution creates an E field
inside the wire. This pushes the electron current through the
wire
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A Model of Conduction (1)
Suppose E  0 : K  12 mv 2  32 kT
gives v  105 m/s. However, v  0.
Now turn on an E field. The straight-line
trajectories become parabolic, and because of the
curvature, the electrons begin to drift in the
direction opposite E, i.e., “downhill”.
ax=F/m=eE/m so
vx=vix+ axDt = vix+ Dt eE/m
This acceleration increases an electrons
kinetic energy until the next collision, a “friction”
that heats the wire….energy is imparted to the
atoms of the lattice.
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Kirchhoff’s Junction Law
 I  I
in
I
i
out
 0; summed over all the currents to any "junction".
i
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Conductivity and Resistivity
The current density J = nevd is directly proportional to the electron
drift speed vd. Our microscopic conduction model gives vd = eE/m, where
 is the mean time between collisions. Therefore:
 e E  n e 
J  nevd  n e 
E

m
 m 
2
The quantity ne2/m depends only on the properties of the conducting
material, and is independent of how much current density J is flowing.
This suggests a definition:
2
ne 
conductivity:  
so J  J
 E=
m
E
This result is fundamental and tells us three things:
(1) Current is caused by an E-field exerting forces on charge carriers;
(2) Current density J and current I=JA depends linearly on E;
(3) Current density J also depends linearly on . Different materials
have different  values because n and  vary with material type.
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Resistors and Resistance
Conducting material that carries current
along its length can form a resistor,
a circuit element characterized by an
electrical resistance R:
R ≡ rL/A
where L is the length of the conductor and A is
its cross sectional area.
R has units of ohms ( W ).
Multiple resistors may be combined in
series, where resistances add, or in parallel,
where inverse resistances add.
I
Rnet
Rnet
For identical resistors can
Series
Connection
[L]:
simply
add
the lengths
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Rnet  R1  R2  R3
For identical resistors can
simply Connection
add the areas
Parallel
[(1/A)]:
Physics 122B - Lecture 26
1
1
1
1
 

Rnet R1 R2 R3
23
The Potential Energy of
Like and Unlike Charge Pairs
U elec 
Kq1q2
1 q1q2

r
4 0 r
This approach can be applied to pairs of electrically charged particles,
whether they have the same or opposite charges. However, for like-sign
particles (a) the system energy is positive and decreases with separation,
while for opposite-sign particles (b) the system is typically “bound”, so that
the net energy is negative and increases (closer to zero) with increasing
separation.
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The Electric Force as
a Conservative Force
The electrical force is a “conservative
force”, in that the amount of energy involved in
moving from point i to point f is independent of
the path taken.
This can be demonstrated in the field of a
single point charge by observing that tangential
paths involve no change in energy (because r is
constant). Therefore, an arbitrary path can be
approximated by a succession of radial and
tangential segments, and the tangential
segments eliminated.
What remains is a straight line path from
the initial to the final position of the moving
charge, indicating a net work that will be the
same for all possible paths.
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Multiple Point Charges
We have established that both energy and electrical forces
obey the principle of superposition, i.e., they can be added linearly
without “cross terms”. Therefore, for multiple point charges:
N
Kqi q j
i<j
ri j
U elec  
Here, “i<j” means that for summing over N particles, the sum
over i runs from 1 to N, and the sum over j runs from i+1 to N for
each value of i. This it a mathematical trick to avoid counting pairs
of point charges twice or having i=j terms, which would give a zero
in the denominator.
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The Electric Potential
In Chapter 25 we introduced
the concept of an electric field E,
which can be though of as a
normalized force, i.e., E = F/q,
the field E that would produce a
force F on some test charge q.
We can similarly define the
electric potential V as a chargenormalized potential energy, i.e.,
V=Uelec/q, the electric potential V
that would give a test charge q an
electric potential energy Uelec
because it is in the field of some
other source charges.
We define the unit of electric potential as the volt: 1 volt = 1 V =
1 J/C = 1 Nm/C. Other units are: kV=103 V, mV=10-3 V, and mV=10-6 V.
Example: A D-cell battery has a potential of 1.5 V between its terminals.
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The Electric Potential Inside
a Parallel Plate Capacitor
Consider a parallel-plate capacitor with
d  3.0 mm,  =4.42 10-9 C/m2


E   ,  to -    500 N/C, to right 
 0

U elec  U q+sources
 qEs for a charge q,
located at s from the - plate
(with U0=0)
DVC  V  V  Ed  1.5 V
DV
d
E  C ; DVC  Ed 
d
0
V  U elect / q  Es (V inside a parallel-plate
capaitor with V=0 @ - plate)
Note that 1 N/C = 1 V/m.
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Rules for Equipotentials
1. Equipotentials never intersect
other equipotentials. (Why?)
2. The surface of any static
conductor is an equipotential
surface. The conductor volume
is all at the same potential.
3. Field line cross equipotential
surfaces at right angles. (Why?)
4. Dense equipotentials indicate a
strong electric field. The potential V decreases in the direction
in which the electric field E points, i.e., energetically “downhill”
for a + charge
5. For any system with a net charge, the equipotential surfaces
become spheres at large distances.
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The Electric Potential
of Many Charges
The principle of superposition allows us to
calculate the potentials created by many point
charges and then add the up. Since the potential V
is a scalar quantity, the superposition of potentials
is simpler than the superposition of fields.
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qi
V 
i 4 0 ri
1
30
Finding E from V
DV 
DU q+sources
q
 Es  
Ex  
W

  Es Ds
q
DV
 dV 
 lim  

Ds Ds 0  ds 
dV
dV
dV
; Ey  
; Ez  
dx
dy
dz
 dV ˆ dV ˆ dV ˆ  In other words, the E field components are
E  
i
j
k  determined by how much the potential V
dy
dz  changes in the three coordinate directions.
 dx
dV
d  1 q
1 q
For a point charge: E  Er  
 

dr
dr  4 0 r  4 0 r 2
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Kirchhoff’s Loop Law
Since the electric field is
conservative, any path between
points 1 and 2 finds the same potential
difference. Any path can be approximated
by segments parallel and perpendicular to
equipotential surfaces, and the
perpendicular segments must cross the
same equipotentials.
Since a closed loop starts and ends at
the same point, the potential around the
loop must be zero. This is Kirchhoff’s Loop
Law, which we will use later.
DVloop    DV i  0
i
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A Conductor in
Electrostatic Equilibrium
A conductor is in electrostatic
equilibrium if all charges are at rest and
no currents are flowing. In that case,
Einside=0. Therefore, all of it is at a
single potential: Vinside=constant.
Rules for conductor.
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Forming a Capacitor
Any two conductors can form a
capacitor, regardless of their shape.
Q
C
DVC
The capacitance depends only on the geometry of the conductors,
not on their present charge or potential difference.
(In fact, one of the conductors can be moved to infinity, so the
capacitance of a single conductor is a meaningful concept.)
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Combining Capacitors
Parallel: Same DV, but different Qs.
Cparallel 
Q  Q2  Q3 
Q
 1
DVC
DVC
 C1  C2  C3 
Series: Same Q, but different DVs.
Cseries 
Q
Q

DVC DV1  DV2  DV3 

1
 DV1 / Q    DV2 / Q    DV3 / Q  

1
1/ C1  1/ C2  1/ C3 
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 C1 || C2 || C3 ||
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Energy Stored in a Capacitor
1
DU  dqDV  qdq
C
Q
2
1
1 Q
U C   qdq  2
C0
C
2
Q
U C  12
 12 C DVC 2
C
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Energy in the Electric Field
Volume of E-field
U C  C DV 
1
2
2
1
2
0 A
d
 Ed 
2

0
2
 Ad  E 2
energy stored U C  0 2
uE 

 E
storage volume Ad 2
Example: d=1.0 mm, DVC=500 V
E
DVC
500 V
5


5.0

10
V/m
d
1.0 10-3 m
uE 
0
2
E 
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2
1
2
 5.0 10
5
V/m  /  4  9.0 109 Vm/C   1.1 J/m 3
2
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37
Dielectric Materials*
There is a class of polarizable dielectric
materials that have an important application in
the construction of capacitors. In an electric
field their dipoles line up, reducing the E field
and potential difference and therefore
increasing the capacitance:
E off
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 0 A
Q
C

DVC
d
E on
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38
Electric Fields and Dielectrics
In an external field EO, neutral molecules can polarize. The induced
electric field E’ produced by the dipoles will be in the opposite direction
from the external field EO. Therefore, in the interior of the slab the
resulting field is E = EO-E’.
The polarization of the material has the net effect of producing a sheet
of positive charge on the right surface and a sheet of negative charge on
the left surface, with E’ being the field made by these sheets of charge.
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Resistors and Ohm’s Law
J E 
I  JA 
R
rL
A
E
r
E
r
A
DV / L
r
I
A
DV
;
R
DV
 r L / A
DV  IR
Ohm’s Law
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Applying Kirchhoff’s
Loop Law to Many Loops*
1. Define a minimum set of current
loops. Label all elements.
R1=
2. Write a loop equation for each
loop. (Battery or 0 = DV).
3. Solve equations for currents
4. Calculate other variables of
interest.
i1 – i3
R2=
i1
Vbat=
R4=
R3=
Loop equations:
R5=
i2
Loop 1: Vbat  R1i1  R2 (i1  i3 )  R3 (i1  i2 )
Loop 2: 0  R3 (i2  i1 )  R5 (i2  i3 )
Loop 3: 0  R2 (i3  i1 )  R4i3  R5 (i3  i2 )
i3
i1 – i2
Loop L:
 (i
L
 ii ,other ) Ri  0
i
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Energy and Power (1)
P  IE = I 2 R  E 2 / R
Example: A 90 W load
resistance is connected
across a 120 V battery.
How much power is delivered
by the battery?
P
dU
 power  rate of energy transfer
dt
DU  qDVbat  qE
Pbat 
E (120 V)

 1.33 A
R (90 W)
P  IE  (1.33 A)(120 V)  160 W
dU bat d
dq
 qDVbat 
DVbat  I DVbat
dt
dt
dt
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I
P  E 2 / R  (120 V)2 /(90 W)  160 W
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42
RC Circuits
I = - dQ/dt
Q
Q
dQ
DVC  DVR   IR   R
0
C
C
dt
Qf
t
ln

Qi
RC
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dQ
1

dt
Q
RC
 t 
 exp  
Qi
 RC 
Qf
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Qf

Qi
t
dQ
1

dt

Q
RC 0
Q f  Qi et / RC
Exponential decay!
43
Magnetic Field Lines
The magnetic field can be graphically
represented as magnetic field lines, with
the tangent to a given field line at any
point indicating the local field direction
and the spacing of field lines indicating the
local field strength. The field line
direction indicates the direction of force
on an isolated north magnetic pole.
B-field lines
never cross.
B-field line spacing
indicates field
strength
Field Map
Field Lines
weak
strong
B-field lines always
form closed loops.
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Cross Product Form
of the Biot-Savart Law
The Biot-Savart Law can be represented
more compactly using a vector cross
product. This automatically gives a B field
that is perpendicular to the plane of the
charge velocity and radius vector to the
point at which the field is being evaluated.
0 q  v  rˆ 
B
4
r2
Note that the r in the numerator
is ^
r (unit vector), not r (vector)!
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Magnetic Force
A current consists of moving charges. Ampere’s experiment
implies that a magnetic field exerts a force on a moving charge.
This is true, although the exact form of the force relation was not
discovered until later in the 19th century. The force depends on the
relative directions of the magnetic field and the velocity of the
moving charge, and is perpendicular to both..

F  q vB
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
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46
Cyclotron Motion
Consider a positive charged particle
with mass m and charge q moving at
velocity v perpendicular to a uniform
magnetic field B. The particle will
move in a circular path of radius rcyc
because of the force F on the particle,
which is:
mv 2
F  qvB 
rcyc
f cyc 
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rcyc 
mv
qB
v
q B

(independent of rcyc and v)
2 rcyc m 2
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The Hall Effect
When a charged particle moves in
a vacuum, it experiences a force
that is perpendicular to its
velocity in a magnetic field. In
1879, Edwin Hall, a graduate
student at Johns Hopkins Univ.,
discovered that the same
behavior is true of charged
particles moving in a conductor.
DVH
Fm  evd B  Fe  eE  e
w
DVH  wvd B
vd 
J I/A
I


ne
ne
wtne
Edwin Herbert Hall
(1855 – 1938)
DVH  wB
I
IB

wtne tne
The sign of the mobile charges matters !!!
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48
The Magnetic Field of a Current
Consider the charge
DQ moving at speed v
through a short
segment of wire Ds.
DQ v = I Dt v = I Dt
Ds
= I Ds
Dt
Therefore, we can use the Biot-Savart law
to find the magnetic field B produced by the
wire segment:
0 q v  rˆ 0 I Ds  rˆ
B

2
4 r
4 r 2
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Physics 122B - Lecture 26
0 ds  rˆ
dB 
I
4
r2
The Biot-Savart Law
for current elements
49
Ampere’s Law
From Biot-Savart Law :

r r
B.dl = 0 I
Current penetrating the surface enclosed
by closed loop integration
Magnetic analogue of Gauss’s Law
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50
The Magnetic Field
of a Solenoid (2)
We can use Ampere’s Law to calculate the
field of an ideal long solenoid by choosing the
integration path carefully. We choose a
rectangular LxW loop, with one horizontal side
outside the solenoid and the vertical sides passing
through.
If the loop encloses N wires, then Ithrough = NI.
Therefore, Ampere’s Law says that:

O Bds

 B  ds  B L  B W  B L  B W

3
4
W
2
1
r Br  ds  0 NI
1
2
3
4
The first side in inside and parallel to B, so B1=B.
Sides 2 and 4 are perpendicular to B (no radial
B), so B2=B4=0. Side 3 is outside the solenoid,
so B3=0. Therefore, B = 0NI/L
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Physics 122B - Lecture 26
If n = N/L is the
number of turns per
unit length, then:
Bsolenoid  0 nI
51
The Force between
Two Parallel Wires
Bwire
0 2 I

4 d
Fparallel wires  I1 LB2
 0 2 I 2 
 I1 L 

4

d


0 2 L I1 I 2

4 d
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Parallel wires carrying current in the
same direction attract each other.
Parallel wires carrying current in
opposite directions repel each other.
Physics 122B - Lecture 26
52
Ferromagnetism (2)
Although iron is a magnetic material, a typical piece
of iron is not a strong permanent magnet. It turns out,
as shown in the figure on the right, that a piece of iron
is divided into small regions called magnetic domains. A
typical domain size is roughly 0.1 mm. The magnetic
moments of all of the iron atoms within each domain are
perfectly aligned, so that each individual domain is a
strong magnet.
The picture shows a photograph of domains in iron.
Each domain is magnetized in a different direction.
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53
Hysteresis*
Some ferromagnetic
materials can be permanently
magnetized, and “remember”
their history of magnetization.
The “hysteresis curve” shows
the response of a ferromagnetic
material to an external applied
field. As the external field is
applied, the material at first
has increased magnetization,
but then reaches a limit at (a)
and saturates. When the
external field drops to zero at
(b), the material retains about
60% of its maximum
magnetization.
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Unmagnetized
Physics 122B - Lecture 26
Partially
magnetized
Saturated
54
Motional EMF
Consider a length l of conductor moving to the right in a magnetic field
that is into the diagram. Positive charges in the conductor will experience
an upward force and negative charges a downward force. The net result is
that charges will “pile up” at the two ends of the conductor and create an
electric field E. When the force produced by E becomes large enough to
balance the magnetic force, the movement of charges will stop and the
system will be in equilibrium.
FB  qvB
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FE  qE
FB  FE
Physics 122B - Lecture 26
 E  vB
55
This is also true ``locally’’
Example: Potential Difference
along a Rotating Bar
A metal bar of length l rotates with
angular velocity w about a pivot at one
end. A uniform magnetic field B is
perpendicular to the plane of rotation.
What is the potential difference
between the ends of the bar?
E  Bv  Bw r
v wr
l
DV  Vtip  Vpivot    Er dr
0
l
l
0
0
   ( Bw r )dr  Bw  rdr  12 Bwl 2
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Eddy Currents (3)
Now consider a sheet of conductor
pulled through a magnetic field. There
will be induced current, just as with the
wire, but there are now no well-defined
current paths.
As a consequence, two “whirlpools” of
current will circulate in the conductor.
These are called eddy currents.
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A magnetic braking system.
Physics 122B - Lecture 26
57
Magnetic Flux
in a Nonuniform Field
So far, we have assumed that
the loop is in a uniform field. What
if that is not the case?
The solution is to break up the
area into infinitesimal pieces, each
so small that the field within it is
essentially constant. Then:
d m  B  dA
m 

B  dA
area of loop
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58
Lenz’s Law (1)
Heinrich Friedrich Emil Lenz
(1804-1865)
In 1834, Heinrich Lenz announced a rule for determining the direction
of an induced current, which has come to be known as Lenz’s Law.
Here is the statement of Lenz’s Law:
There is an induced current in a closed conducting loop if and only if
the magnetic flux through the loop is changing. The direction of the
induced current is such that the induced magnetic field opposes the
change in the flux.
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59
Faraday’s Law
Consider the loop shown:
 m   B  dA  BA  Bl x
d m d
dx
 Bl x  Bl
dt
dt
dt
E  Blv  Bl
dx
dt
d m
Therefore, E 
dt
This is Faraday’s Law. It can be stated as follows:
An emf E is induced in a conducting loop if the magnetic flux m
through the loop changes with time, so that E = |dm/dt| for the
loop. The emf will be in the direction that will drive the induced
current to oppose the flux change, as given by Lenz’s Law.
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60
What does Faraday’s
Law Tell Us?
Faraday’s Law tells us that all induced currents are the associated
with a changing magnetic flux. There are two fundamentally
different ways to change the magnetic flux through a loop:
(1) The loop can move, change size, or rotate, creating motional emf;
(2) The magnetic field can change in magnitude or direction.
We can write:
E
d m
d ( B  A)
dA

 B

dt
dt
dt
motional
emf
A
dB
dt
new
physics
The second term says that an emf can be created simply by changing
a magnetic field, even if nothing is moving.
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61
Generators
The figure shows a coil with
N turns rotating in a magnetic
field, with the coil connected to
an external circuit by slip rings
that transmit current
independent of rotation. The
flux through the coil is:
 m  A  B  AB cos 
 AB cos wt
Ecoil  N
d m
d
 ABN  cos wt   w ABN sin wt
dt
dt
Therefore, the device produces emf and current that will vary
sinusoidally, alternately positive and negative. This is called an alternating
current generator, producing what we call AC voltage.
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62
Transformers
When a coil wound around an
iron core is driven by an AC voltage
V1cos wt, it produces an oscillating
magnetic field that will induce an
emf V2cos wt in a secondary coil
wound on the same core. This is
called a transformer.
The input emf V1 induces a
current I1 in the primary coil that
is proportional to 1/N1. The flux in
the iron is proportional to this, and
it induces an emf V2 in the secondary
coil that is proportional to N2. Therefore, V2 = V1(N2/N1). From
conservation of energy, assuming no losses in the core, V1I1 = V2I2.
Therefore, the currents in the primary and secondary are related by the
relation I1 = I2(N2/N1).
A transformer with N2>>N1 is called a step-up transformer, which
boosts the secondary voltage. A transformer with N2<<N1 is called a stepdown transformer, and it drops the secondary voltage.
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63
(Self-) Inductance
We define the inductance L of a coil of wire producing flux m as:
m
L
I
The unit of inductance is the henry: 1 henry = 1 H = 1 T m2/A = 1 Wb/A
The circuit diagram symbol used to represent inductance is:
Example: The inductance of a long solenoid with N turns of
cross sectional area A and length l is:
0 NI
 per turn  BA
B
 m  N  per turn  NBA 
7/19/2016
0 N A
l
2
l
I
Physics 122B - Lecture 26
Lsolenoid
 m 0 N 2 A


I
l
64
Potential Across an Inductor (2)
Ecoil  L
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dI
dt
DVL   L
Physics 122B - Lecture 26
dI
dt
65
Energy Density in the Fields
This is the magnetic analog of the energy stored in an electric field
UE =
0
2
1 2
uB 
B
2 0
E
2
Magnetic
Electric
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66
The LR Circuit
DVR  DVL  0
dI
R
  dt
I
L
 RI  L
I
dI
0
dt
t
dI
R


I I L 0 dt
0
 I 
t
ln    
L/R
 I0 
I (t )  I 0e  t /( L / R )
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67
Inductors: Early and Late
Initially, when a switch closes an inductor appears to have an infinite
resistance and has a maximum potential drop across it. Ultimately, the
inductor reaches a steady current flow with no potential drop across
it. Therefore, at t=0 the inductor behaves like at open circuit (R=∞),
and at t=∞ the inductor behaves like a short circuit (R=0). This behavior
is opposite that of a conductor.
Example:
Circuit
at t=0
at t=∞
Calculate initial potentials. Calculate final currents.
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The Oscillation Cycle
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Capacitor AC Circuits (1)
Consider an AC current iC through a capacitor as shown. The capacitor
voltage vC = E = E0cos wt = VCcos wt. The charge on the capacitor will
be q = CvC = CVCcos wt.
iC 
dq d
  CVC cos wt   wCVC sin wt
dt dt
iC  wCVC cos(wt   / 2)
The AC current through a capacitor leads the capacitor voltage
by /2 rad or 900.
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70
RC Filter Circuits
Now consider a circuit that
includes both a resistor and a
capacitor. Because the
capacitor voltage VC and the
resistor voltage VR are 900
apart in the phasor diagram,
they must be added like the
sides of a right triangle:
E0 2  VC 2  VR 2  ( IR) 2  ( IX C ) 2
 ( R 2  X C 2 ) I 2   R 2   wC   I 2


E0
2
I
R 2   wC 
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2
VR  IR 
Vc  IX C 
Physics 122B - Lecture 26
E0 R
R 2   wC 
2
E0 / wC
R 2   wC 
2
71
AC Inductor Circuits
Consider an AC current iR through an inductor. The changing
current produces an instantaneous inductor voltage vL.
di
vL  L L
dt
If the inductor is connected in an AC
circuit as shown, then Kirschoff’s loop
law tells us that:
vL
VL
di

dt

cos wtdt
DVsoruce  DVL = E  vL  0 E(t )  E0 cos wt  vL
L
L
L
V
V
V




iL  L  cos wtdt  L sin wt  L cos  wt    I L cos  wt  
L
wL
wL
2
2


In the phasor diagram, the
inductor current iL lags the
voltage vL by 900, so that iL
peaks T/4 later than vL.
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The Series RLC Circuit
The figure shows a resistor, inductor,
and capacitor connected in series. The
same current i passes through all of the
elements in the loop. From Kirchhoff’s
loop law, E = vR + vL + vC.
Because of the capacitive and inductive
elements in the circuit, the current i will
not in general be in phase with E, so we will
have i = I cos(wt-f) where f is the phase
angle between current and voltage. If
VL>VC then the current i will lag E and f>0.
E02  VR2  (VL  VC ) 2   R 2  ( X L  X C ) 2  I 2
E0
E0
I

2
2
R  (X L  XC )
R 2  (w L  1/ wC ) 2
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Resonance
I
E0
R 2  (w L  1/ wC ) 2
The current I will be a maximum when wL=1/wC.
This defines the resonant frequency of the system w0:
w0 
1
LC
7/19/2016
I
E0
2
2 
 w0  
2
R   Lw  1    
  w  
2
Physics 122B - Lecture 26
74
Displacement Current
Q
Q
 e  EA 
A
0 A
0
d  e 1 dQ I


dt
 0 dt  0
I disp   0
d e
dt

d e 

 B  ds  0  I through  I disp   0  I through   0 dt 
d e
 B  ds  0 I through  0 0 dt

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Maxwell’s Equations
 E  dA  Q
in
/ 0
 B  dA  0
(magnetic monopole
charge goes here)
Gauss’s Law
Gauss’s Law for magnetism
d m
 E  ds   dt
d e
 B  ds  0 I through  0 0 dt

Faraday’s Law
(magnetic monopole
current goes here)

Ampère-Maxwell Law
F  q( E  v  B)
Lorentz Force Law
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76
A Prelude to Waves
Maxwell’s formulation of electricity and
magnetism has an interesting consequence.
The equations can be manipulated to give a
wave equations for E and B of the form:
d 2E
d 2E
  0 0 2
2
dx
dt
This can be recognized as describing
an electromagnetic wave traveling
through space with a velocity of:
vEM wave 
1
 0 0
(4  9.0 109 Nm 2 /C 2 )

(4 107 N/A 2 )
 3.0 108 m/s
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Lecture 26 Announcements
The Final Exam is Tuesday June 5 at 2.30 – 4.20 pm
Good Luck
7/19/2016
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78