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Inductors
Flux = (Inductance) X (Current)
  LI

Inductors in circuits
Tue. Nov. 9, 2009
Physics 208, Lecture 20
1
Voltage drop across inductor

Constant current


No voltage difference
Current changing in time

L
Voltage difference across inductor
dI
VL  Vb Va  L
dt

Vb
Va
I
Vbatt
R
Tue. Nov. 9, 2009
Physics 208, Lecture 20
2
RL Circuit



Before switch closed, IL = 0
dI
Current through inductor cannot ‘jump’ VL  L
dt
Just after switch closed, IL= 0.
What is voltage across L just after switch closed?

A. VL = 0
Kirchoff’s loop law:
B. VL= Vbattery
VR + VL = Vbattery
C. VL= Vbattery / R
R and L in series, IL=0 IR=0, VR=0
D. VL= Vbattery / L
Tue. Nov. 9, 2009
Physics 208, Lecture 20
3
IL t  0  0
VL t  0  Vbattery  L
dIL
dt
IL
dIL Vbattery


dt
L
IL instantaneously zero, but increasing in time
IL(t)
Slope dI / dt = Vbattery / L
0
0
Tue. Nov. 9, 2009
Time ( t )
Physics 208, Lecture 20
4
Just a little later…
Switch closed at t=0
IL(t)
Slope dI / dt = Vbattery / L
0
0
Time ( t )
A short time later ( t=0+Δt ), the current is increasing
…
A. More slowly
IL>0, and IR=IL
B. More quickly
VR≠0, so VL smaller
C. At the same rate
VL= -LdI/dt, so dI/dt smaller
Tue. Nov. 9, 2009
Physics 208, Lecture 20
5
Vbattery
R
Later slope dI  Vbattery  IL t  t R
dt
IL(t)
0
0
L
dI Vbattery
Initial slope

dt
L

Time ( t )
 inductor in equilibrium,
What is current through
a long time after switch is closed?
A. Zero
IL
B. Vbattery / L
C. Vbattery / R
Equilibrium:
currents not changing
dIL / dt =0, so VL=0
VR=Vbattery
IL = IR =Vbattery / R
Tue. Nov. 9, 2009
Physics 208, Lecture 20
6
RL summary
It  I 1 et /(L / R )  I 1 et /  
I(t)
I  Vbattery /R
Switch closed at t=0
  L /R
= time constant

I(t)
I  Vbattery /R


Tue. Nov. 9, 2009
Physics 208, Lecture 20
7
Question
What is the current through
R1 immediately after the
switch is closed?
R1
R2
L
A. Vbattery / L
B. Vbattery / R1
IL cannot ‘jump’. IL=0 just after
closing switch.
C. Vbattery / R2
All current flows through resistors.
D. Vbattery / (R1+R2)
Resistor current can jump.
E. 0
Tue. Nov. 9, 2009
Physics 208, Lecture 20
8
Thinking about electromagnetism
Electric Fields
Arise from charges
Capacitor, Q=CV
Arise from time-varying B-field Inductor, Faraday effect
Magnetic Fields
Inductor, Φ=LI
Arise from currents
Arise from time-varying E-field


Many similarities between electricity, magnetism
Some symmetries, particularly in time-dependence
Tue. Nov. 9, 2009
Physics 208, Lecture 20
9
Maxwell’s unification

Intimate connection
between electricity and magnetism

Time-varying magnetic field
induces an electric field (Faraday’s Law)

Time-varying electric field generates a
magnetic field
1 B
 E  
c t
In vacuum:
1 E
 B 
c t
This is the basis of Maxwell’s unification of electricity
and magnetism into Electromagnetism

Tue. Nov. 9, 2009
Physics 208, Lecture 20
10
• A Transverse wave.
• Electric/magnetic fields perpendicular to
propagation direction
• Can travel in empty space
f = v/, v = c = 3 x 108 m/s (186,000 miles/second)
Tue. Nov. 9, 2009
Physics 208, Lecture 20
11
The EM
Spectrum


Types are
distinguished by
frequency or
wavelength
Visible light is a
small portion of
the spectrum
Tue. Nov. 9, 2009
Physics 208, Lecture 20
12
Sizes of EM waves

Visible light

typical wavelength of 500 nm = = 0.5 x 10-6 m
= 0.5 microns (µm)
AM 1310, your badger radio network,
has a vibration frequency of 1310 KHz = 1.31x106 Hz
What is its wavelength?
A. 230 m
B. 0.044
m
C. 2.3 m
D. 44m
Tue. Nov. 9, 2009
Physics 208, Lecture 20
13
Quick Quiz
A microwave oven irradiates food with electromagnetic
radiation that has a frequency of about 1010 Hz. The
wavelengths of these microwaves are on the order of
A. kilometers
B. meters
C. centimeters
D. micrometers
Tue. Nov. 9, 2009
Physics 208, Lecture 20
14
Mathematical description
x
y
E  E o coskz   t
B  Bo coskz   t
k
2

EB
z
Bo  E o /c
,   2f


r r
Propagation direction = E  B

Tue. Nov. 9, 2009
Physics 208, Lecture 20
15
EM Waves from an Antenna




Two rods are connected to an ac source, charges oscillate
between the rods (a)
As oscillations continue, the rods become less charged, the field
near the charges decreases and the field produced at t = 0
moves away from the rod (b)
The charges and field reverse (c)
The oscillations continue (d)
Tue. Nov. 9, 2009
Physics 208, Lecture 20
16
Detecting EM waves
FM antenna
AM antenna
Oriented vertically for radio waves
Tue. Nov. 9, 2009
Physics 208, Lecture 20
17
Transatlantic signals
Capacitor
banks
Induction
coils
Spark gap
Gulgielmo Marconi’s transatlantic transmitter
Tue. Nov. 9, 2009
Physics 208, Lecture 20
18
Transatlantic receiver
Left to right: Kemp, Marconi, and Paget pose in front of a kite that was
used to keep aloft the receiving aerial wire used in the transatlantic
radio experiment.
Tue. Nov. 9, 2009
Physics 208, Lecture 20
19
Energy and EM Waves
Energy density in E-field
Energy density in B-field
uE  o E r,t /2
uB  B2 r,t /2o
2
2
2

u


E
/2

B
/2o
Total Tot
o
2 
2
2
2
2
 o E /2  E /2c o  o E r,t   B r,t / o
uTot  o E 2  o E o2 cos2 kz  t moves w/ EM wave

Tue. Nov. 9, 2009
at speed c
Physics 208, Lecture 20
20

Power and intensity in EM waves
Energy density uE moves at c


Instantaneous energy flow =
energy per second passing plane
2
2
2
 = cu
Tot  co E  co E o cos  t

This is power density W/m2

Oscillates in time

2
2
Time average of this is Intensity = co E max
/2  cBmax
/2o
Tue. Nov. 9, 2009
Physics 208, Lecture 20
21
Example: E-field in laser pointer

3 mW laser pointer.

Beam diameter at board ~ 2mm

103 W
2

318W
/m
Intensity =
2
 0.001m

How big is max E-field?

2
co E max
/2  318W /m 2
E max 
Tue. Nov. 9, 2009
2318W /m 2 
3 10 m /s8.85 10
8
12
C /N  m
2
Physics 208, Lecture 20
2

 489N /C  489V /m
22
Spherical waves

Sources often radiate EM wave in all directions





Light bulb
The sun
Radio/tv transmission tower
Spherical wave, looks like plane wave far away
Intensity decreases with distance

Power spread over larger area

I
Psource
4 r 2
Source power
Spread over this
surface area
 Tue. Nov. 9, 2009
Physics 208, Lecture 20
23
Question
A radio station transmits 50kW of power from its
antanna. What is the amplitude of the electric
field at your radio, 1km away.
A. 0.1 V/m
B. 0.5 V/m
I
50,000W
4  1000m
2
 4 103W / m2
C. 1 V/m
2
co E max
/2  4 103W /m 2
D. 1.7 V/m
E. 15 V/m
Tue. Nov. 9, 2009

E max 
24 103 W /m 2 
3 10 m /s8.85 10
8
12
C 2 /N  m 2 
 1.73N /C  1.73V /m
 Physics 208, Lecture 20
24
The Poynting Vector

Rate at which energy flows through a unit area perpendicular
to direction of wave propagation

Instantaneous power per unit area (J/s.m2 = W/m2) is also
S


1
o
E  B  Poynting Vector
Its direction is the direction of propagation of
the EM wave
This is time dependent
 Its magnitude varies in time
 Its magnitude reaches a maximum at the
same instant as E and B
Tue. Nov. 9, 2009
Physics 208, Lecture 20
25

Radiation Pressure



Saw EM waves carry energy
They also have momentum
When object absorbs energy U from EM wave:




Momentum p is transferred
Power
p  U /c ( Will see this later in QM )
U /t
Result is a force F  p /t 
 P /c
c
Pressure = Force/Area =
P/A
prad 
 I /c
c
Radiation
pressure
Intensity
on perfectly absorbing object
Tue. Nov. 9, 2009
Physics 208, Lecture 20
26
Radiation pressure & force
EM wave incident on surface exerts a radiation pressure
prad (force/area) proportional to intensity I.
Perfectly absorbing (black) surface: prad  I /c
Perfectly reflecting (mirror) surface: prad  2I /c
Resulting force = (radiation pressure) x (area)


Tue. Nov. 9, 2009
Physics 208, Lecture 20
27
Question
A perfectly reflecting square solar sail is 107m X 107m. It has
a mass of 100kg. It starts from rest near the Earth’s orbit,
where the sun’s EM radiation has an intensity of 1300 W/m2.
How fast is it moving after 1 hour?
A. 100 m/s
B. 56 m/s
C. 17 m/s
D. 3.6 m/s
E. 0.7 m/s
Tue. Nov. 9, 2009
prad  2I /c
Frad  prad A  2IA /c 
21300W /m 2 1.145 10 4 m 2 
3 10 m /s
8
 0.1N
a  Frad /m  103 m /s2
v  at  103 m /s2 3600s  3.6m /s
Physics 208, Lecture 20
28