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Ch – 34 Electromagentic Fields
and Waves
Reading Quiz – Ch. 34
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1. Experimenter A creates a magnetic field in the
laboratory. Experimenter B moves relative to A.
Experimenter B sees:
a. the same magnetic field.
b. a magnetic field of different strength.
d. an electric field.
e. both a magnetic and an electric field.
2. Which of the following was not discussed in the
sections 1-4 of this chapter:
a. Gauss’s Law
b. The Lorentz Force
c. Electrical Conduction in Gases
d. Galilean Transformations
Learning Objectives – Ch 34
• To understand that electric and magnetic fields
are interdependent. There’s just a single
electromagnetic field that presents different
faces, in terms of and to different observers.
• Electromagnetic fields obey four general laws,
called Maxwell’s equations.
• Electromagnetic fields can exist without source
charges or currents in the form of a selfsustaining electromagnetic wave.
• Maxwell’s equations predict that all
electromagnetic waves travel at the same
speed.
Electromagnetic fields

E
1
q
1 q
ˆ
r

2
2
40 r
40 r
 0 qv  rˆ 0 qv sin 
B

2
4 r
4 r 2
away from q
direction given by right hand rule
Field vectors for E and B
Field Lines
The Lorentz Force
A charge moves
through a region
where there is both
an external electric
and an external
magnetic field.


FE  qE

 
FB  qv  B
The
Lorentz
Force


FE  qE

 
FB  qv  B
The net force on the
charge is due to both
these forces.
This is known as the
Lorentz Force law:

  
F  q( E  v  B)
Case 1: Moving or Stationary?
• Bill says there will be a B field
generated by the moving charge,
in addition to the E field
generated by the charge
• But Sharon claims there will be
an E field, but not a B field,
because she “sees” the charge
as being stationary relative to
her*.
*This story is a bit foolish until you imagine
Sharon being on the Moon, or at least in a
space station, so she can really believe the
charge in her hand is stationary.
Case 2: What causes the force?
• Bill turns on a constant,
external magnetic field.
Bill claims there should
be a upward force due to
B for the moving charge.
• Sharon feels the force on
the charge, but claims it
must be from something
else because stationary
charges can’t experience
a force in a constant B
field.
Galilean Relativity – Review of ch.
6-4
• Reference frame S’ (think
of a space station) moves
with constant velocity V
relative to reference
frame S (think of Earth).
• An observer at rest in S
sees the origin of S’ whiz
by at V. Of course the
guy in the S’ sees S go
by at –V.
Galilean Relativity
• Observers in both
reference frames see
a particle whiz by.
• Observer in S would
say its velocity is v.
Observers in S’ would
say the velocity is v’
where:
• v’ = v-V and
• v = v’ + V
Galilean Relativity
v’ = v-V and
v=v+V
How does the particle’s
acceleration compare in the
two frames?
dv’/dt = dv/dt – dV/dt
since V is constant:
a’ = a and F’ = F
Observers in all inertial reference
frames agree about the force.
What differs is how they
perceive the B and E fields that
cause the force.
Case 1: Moving or Stationary?
• Bill measures the E field
and the B field due to
the moving charge in
reference frame S.
• Shouldn’t Sharon, in
frame S’, measure the
same fields? After all,
aren’t those fields just
“there”?
Let’s experiment
• Establish a region
where Eext is known to
be 0, but there is a
non-zero Bext
• Send a moving
charge through B.

  
F  q( E  v  B)


Since
FE there
qE is no Eext

 
FB  qv  B
Case 2: B causes the force for an
observer in frame S
• Bill can measure an
upward force and
 the magnetic

from
FE  qE
field:

 
FB  qv  B
and v = V:

 
FB  qV  B
How about Sharon, in S’?
• She observes the
same magnitude
upward force, one
that is proportional to
q. She concludes
that, whether there is
a Bext or not, there
must be an Eext,
because from her
point of view, the
charge is at rest.
How about Sharon, in S’?

  
F  q ( E ' v 'B ' )
• And since there is no
v’ = 0….
Case 2: E’ causes the force for an
observer in frame S’
• Sharon can measure
an upward force and
from the external E
field, even though Bill
swears there is none.


FE '  qE '
How do the fields relate

 
FB  qV  B


FE '  qE '
• But both observers
agree on the
magnitude of the
force, therefore:
• E’ = V X B
The General Case
• An experimenter in S
creates external
fields, B and E.
• A charge moving with
velocity v relative to S
experiences a Lorentz
force of:

  
F  q( E  v  B)
The General Case
• The charge is at rest
in frame S’ so the
force in S’ can be due
only to the E field:


F '  qE '
• Since both observers
agree about F:
• E’ = E + V x B
Transformation for E’
E’ = E + V x B
This expression
transforms electric
and magnetic fields in
S into the electric field
measured in S’.
Transformation for B’
The charge at rest in
frame S produces an
E field but no B field:

E
1
q
1 q
rˆ  B=0 2
2
40 r
40 r
What are the fields
measured by Sharon
in frame S’? She
sees a charge moving
with velocity v’ = -V.
Transformation for B’
Since B = 0:
E’ =

E
1
q
1 q
rˆ 
2
40 r
40 r 2
But Sharon also
measures a B’ field
since in her frame the
charge is moving:
1 2q
1 q
E
B’ = - (u0/4π)qV/r
X2 rˆ 
40 r
40 r 2
Transformation for B’
Since B = 0:

E
q
1
rˆ 
2
40 r
40
But Sharon also
measures a B’ field
since in her frame the
charge is
 moving:
12 q
1
E
B’ = - (u0/4π)qV/r
X2 rˆ 
40 r
40
1
= -ε0μ0 VX 1 q ˆ
E
r
2
40 r
40
E’ =
1
q
r2
q
r2
q
r2
Transformation for B’
Since B = 0:

E
q
1
rˆ 
2
40 r
40
But Sharon also
measures a B’ field
since in her frame the
charge is
 moving:
12 q
1
E
B’ = - (u0/4π)qV/r
X2 rˆ 
40 r
40
1
= -ε0μ0 VX 1 q ˆ
E
r
2
40 r
40
E’ =
1
q
r2
q
r2
q
r2
Transformation for B’
B’ = -ε0μ0 V X E.
If observers in frame S
create a magnetic
field B in addition to E
field, then:
B’ = B - ε0μ0 V X E.
Galilean Field Transformations
Which diagram shows the fields in frame S’
Answer is b
A conductive loop in frame S
In frame S, a conductive
loop is moving with
velocity v, into a
“stationary” magnetic
field. The B field causes
an force on the moving
charges. The resulting is
charge separation and an
emf. Note there is no
pre-existing electric field.

 
FB  qv  B
ε = vLB
E = dε/dL = vB
Same loop in frame S’
In frame S’, the loop is
stationary, but B
travels to the left with
velocity -v. An
observer in this frame
agrees that there is a
force on the charge,
but since the loop is
stationary, the force
must be due to an
electric field, E’.
The same loop in frame S’
Using the field
transformations
(recall E in frame S
equals 0):
     
E'  E  v  B  v  B
  1   
B'  B  2 v  E  B
c
The same loop in frame S’
The observer in frame
S’ sees the same B
field as the observer
in S, but in addition
he sees an E field as
well. The B field does
not exert a force in S’.
     
E'  E  v  B  v  B
  1   
B'  B  2 v  E  B
c
The same loop in frame S’
The electric force, E’ in
frame S’ exerts the
same magnitude
force and generates
the same emf as the
B field in frame S.
     
E'  E  v  B  v  B
  1   
B'  B  2 v  E  B
c
E’ is the induced electric field of
Faraday’s Law
In frame S, this electric
field is the result of the
charge separation
caused by the
magnetic force. In
frame S’ it is the
cause:
E = vB
  
E'  v  B
E’ is a non-Coulomb Field
It is not created by static
electric charges, but
rather by a changing
magnetic field.
  
E'  v  B
Calculating the emf
Recall W = -∆U and
∆V = ∆U/q
ε = W/q
 
   Eds
This is another
example of a line
integral. Two
cases are of
interest.
Calculating the emf
 
   Eds
For a closed curve of
length, L:
1. If E is everywhere
perpendicular to the
integration path, ε = 0.
2. If E is everywhere
tangent to the
integration path, ε =
EL.
Sign Convention
Use right hand rule with
thumb showing
positive direction of
flux and fingers
showing positive
direction for emf
(current direction)
Faraday’s Law, revisited
Using this convention, ε and dΦ/dt will
always have opposite signs.
Therefore:
ε = - dΦm/dt
A full statement of Faraday’s Law:
 
d m
d
   Eds  

dt
dt


 
 B  dA
Displacement Current
Ampere’s law relates
the integral of B
around curve C to the
current passing
through surface S.
This is for the case of
a flat surface.
 
 Bds  u0 I through
Displacement Current
It still holds if the
surface is not flat. In
both cases shown,
both S1 and S2 are
bounded by curve C.
In both cases,
Ithrough-1 = Ithrough-2
Displacement Current
• If we use this logic for
the plates of a
capacitor, it fails us
since Ithrough-1 does not
equal Ithrough-2.
• There is no current
through S2 but there
is a changing electric
flux through S2 as
capacitor charges
Displacement Current
 
 Bds  u0 I through
•
•
•
•
•
(Ithrough S1)
Φe = EA for capacitor
E = Q/(ε0A)
Φe = Q/ε0, I = dQ/dt
d Φe /dt = I / ε0
ε0 d Φe /dt is called
the displacement
current idisp
Ampere-Maxwell Law
 
B
d
s

u
(
I


d

/
dt
)
0
through
0
e

Induced B and E fields
• “Ordinary” electric fields are created by
charge but we can create an electric field
by having a changing magnetic field.
• “Ordinary” magnetic fields are created with
a current, but we can create a magnetic
field by having a changing electric field.
Faraday’s Law Problem(#12)
• The graph shows the
current as a function
of time for a 20-cm
long, 4.0 cm diameter
solenoid with 400
turns Draw a graph of
the induced E field as
a function of time at a
point 1.0 cm from the
axis of the solenoid.
Faraday’s Law Problem(#12)
• For 0<t<0.1 s, E = -6.28 x 10-4 V/m
• For 0.1<t<0.2 s, E = +3.14 x 10-4 V/m
Displacement Currents (#18)
• A 10-cm diameter plate capacitor has a
1.0 mm spacing. The E field increases at
the rate of 1.0 x 106 V/m-s. What is the
magnetic field strength:
• A: on the axis
• B: 3.0 cm from the axis
• C. 7.0 cm from the axis
Displacement Currents (#18)
• A 10-cm diameter plate capacitor has a
1.0 mm spacing. The E field increases at
the rate of 1.0 x 106 V/m-s. What is the
magnetic field strength:
• A: 0 T
• B: 1.67 x 10-13 T
• C: 1.98 x 10-13 T
Sign Convention
Recall Faraday’s Law:
ε = |dΦm/dt|, direction
given by Lenz’ Law.
Use of Lenz’ Law
means we can’t use
Cartesian
conventions of
positive and negative
to establish sign of
Φm, or ε.
So make an arbitrary
choice for Φ and a
convention for ε:
Sign Convention
•
•
Curl right hand around
surface in most painfree direction.
The direction in which
your thumb points is
positive. The area
vector A, always points
in the positive direction.
Φ is the vector dot
product of B and A; it
will be positive or
negative, depending on
the direction of B.
Sign Convention
•
•
Don’t forget that its
dΦm/dt that is
important, so adjust
sign according to
whether Φm increases
or decreases with time.
A positive emf creates
an induced current in
the direction of your
fingers; a negative emf
creates a current in the
opposite direction.