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Announcements
Change of plans for today:
Demos on light and selected review for today
Faraday’s Law
10 
5T
10 m
What is the current induced in this circuit?
A) 30A
B) 3A
C) 10A
D) 6A
3 m/s
2m
Faraday’s Law
10 
5T
10 m
As the bar moves a current is induced!
3 m/s
2m
dB
E 
dt
There are no batteries anywhere, so we say that a
current is induced, by an induced emf.
Hence, an electric current can be induced in a circuit by a changing
magnetic field, in the opposite direction to the change in flux.
Comparision of Induction
dE
 B  ds  0 I  0 0 dt
dB
 E  ds   dt
•No magnetic monopole, hence no magnetic current
•Electric fields and magnetic fields induce in opposite fashions
Faraday’s Law and Electric Fields
dB
 E  ds   dt
. A cylindrical region of radius R = 3.0 cm contains a uniform magnetic field parallel to its
axis. The field is 0 outside the cylinder. If the field is changing at the rate 0.60 T/s,
the electric field induced at a point 2R from the cylinder axis is:
Using Faraday’s law: 2p (2R)E =-p(R2) dB/dt,
so E= (-(R2) /4) dB/dt=0.0045 V/m
Maxwell’s Equations
Integral Form
qin
 E  dA  
S
 B  dA  0
S
0
dE
 B  ds  0 I  0 0 dt
dB
 E  ds   dt
Gauss’s laws, Ampere’s law and Faraday’s law all combined!
They are nearly symmetric with respect to magnetism and electricity.
The lack of magnetic monopoles is the main reason
why they are not completely symmetric.
Quiz
The
Thediagrams
diagramsshow
showthree
threecircuits
circuitswith
withidentical
identicalbatteries,
batteries,identical
identicalinductors,
inductors,
and
after
the
switch
isisclosed
which
the
andidentical
identicalresistors.
resistors.Just
A
Just
very
after
long
the
time
switch
later,
which
closedhas
which
thehas
has
least
theleast
greatest
current
currentthrough
throughthe
thebattery?
battery?
RL - Circuits
I
•What happens when the
switch S is closed at t = 0?
•Let I be the current in
the circuit
R
E
+
L
–
•Use Kirchoffs rule for loops on the circuit
dI E R
dI
  I
0  E  RI  L
dt L L
dt
E
E
 Rt / L
t /
I  t )  1  e
 1  e )
)
R
R
  L/ R
S
RC–Circuits vs RL-Circuits
   t RC 
I  e

R

•At t=0, ordinary wire
•As t-> infinity, broken wire
 
 tR 
I  1  e L 
R

•At t=0, broken wire, little current
for small t
•As t-> infinity, ordinary wire
•In terms of current control, an inductor can
often be considered as the opposite of a
capacitor
LC – Circuits and Energy
+
V  E cos t )
–
I  CE sin t )
C
L
S1
  1/ LC
S2
At an arbitrary time t, where is the energy stored in this circuit?
A) In the capacitor
B) In the inductor
C) Alternately in the capacitor or the inductor
D) What energy?
U C  C  V )  CE cos t )
2
1
2
1
2
2
2
U tot  C E
1
2
2
U L  LI  12 LC 2E 2 2 sin 2 t )  12 CE 2 sin 2 t )
1
2
2
E
I
C
–
•Switch S1 is closed, then
opened.
•At t = 0, switch S2 is
closed.
•What happens?
+
LC - Circuits
L
S1
S2
V  t  0)  E
dQ
I 
Q  C V
dt
2
2
d  V )
d Q
dI
V  L   L 2  CL
dt
dt
dt 2
LC – Circuits and
Harmonic Oscillators
V  CL
d
2
 V )
dt
2
V  E cos t )
 m d 2x
x
2
k dt
x  A cos(t )
There are many correspondances between
These equations
electrical and mechanical systems!
RLC circuits in Series II
R
L
Q 2 LI 2
U 

2C
2
S
dU Q dQ
dI

 LI
dt C dt
dt
dU
 I 2R
dt
Do some algebra, and use
2
d Q
dQ Q
L 2 R
 0
dt
dt C
dQ
I
dt
C
RLC circuits and Harmonic
Oscillators
R
L
C
2
d Q
dQ Q
L 2 R
 0
dt
dt C
2
d x
dx
m 2 b
 kx  0
dt
dt
S
A damped harmonic oscillator!
Hence, the charge oscillations are the same asdU
the motion
2 of a
 I R
damped harmonic oscillator.
dt
Quiz
A.
B.
C.
D.
Electromagnetic Waves
  ck
kE0   B0
Ey  E0 cos  kx  t )
E0  cB0
E y  cBz
Bz  B0 cos  kx  t )
Electric
Magnetic Field
Field
Direction of Motion
Using Maxwell’s Equations
E y
Ex
Bz


x
y
t
By
By
Ex Ez


z
x
t
E y
Bx Bz

 0 0
z
x
t
Bx
Ez E y


y
z
t
Ex
Bz By

 0 0
y
z
t
Bx
Ez

 0 0
x
y
t
Electromagnetic Waves
E y
Bz

x
t
E y
Bz

  0 0
x
t
•These equations look like sin functions will solve them.
Ey  E0 cos  kx  t )
Bz  B0 cos  kx  t )
kE0 sin  kx   t )   B0 sin  kx   t )
kB0 sin  kx   t )  0 0 E0 sin  kx   t )
kE0   B0
kB0  0 0 E0
Electromagnetic Waves
kE0   B0
kB0  0 0 E0
•These equations imply
k E0 B0  0 0 B0 E0
2
2

k

1
 0 0
2
k
2

1
0 0
 c  2.998 108 m/s
•The speed of light (in vacuum)