Voltage drop across inductor
From last time…
Inductors

Constant current

Current changing in time
Flux = (Inductance) X (Current)

Φ = LI

No voltage difference
L
Voltage difference across inductor
ΔVL = Vb − Va = −L
dI
dt
€
Vb
€
Inductors in circuits
Tue. Nov. 9, 2009


Physics 208, Lecture 20
1
Tue. Nov. 9, 2009
€
⇒
dIL
=
dt
dIL
dt
IL
Vbattery
L
IL instantaneously zero, but increasing in time
€
€
A.  VL = 0
Kirchoff’s loop law:
B.  VL= Vbattery
VR + VL = Vbattery
IL(t)
Slope dI / dt = Vbattery / L
R and L in series, IL=0 IR=0, VR=0
C.  VL= Vbattery / R
0
D.  VL= Vbattery / L
0
Physics 208, Lecture 20
Just a little later…
3
Time ( t )
Tue. Nov. 9, 2009
Physics 208, Lecture 20
Vbattery
R
Switch closed at t=0
IL(t)
dt
€
Slope dI / dt = Vbattery / L
Initial slope
0
0
Time ( t )
A short time later ( t=0+Δt ), the current is increasing …
4
V
Later slope dI = battery − IL (t = Δt ) R
IL(t)
0
2
IL ( t = 0) = 0
What is voltage across L just after switch closed?
0
Physics 208, Lecture 20
VL ( t = 0) = −Vbattery = −L
Before switch closed, IL = 0
dI
Current through inductor cannot ‘jump’ VL = −L
dt
Just after switch closed, IL= 0.
Tue. Nov. 9, 2009
Vbatt
R
RL Circuit

Va
I
L
dI Vbattery
=
dt
L
€
Time ( t )
€ inductor in equilibrium,
What is current through
a long time after switch is closed?
A.  More slowly
IL>0, and IR=IL
A.  Zero
B.  More quickly
VR≠0, so VL smaller
B.  Vbattery / L
C.  At the same rate
VL= -LdI/dt, so dI/dt smaller
C.  Vbattery / R
IL
Equilibrium:
currents not changing
dIL / dt =0, so VL=0
VR=Vbattery
IL = IR =Vbattery / R
Tue. Nov. 9, 2009
Physics 208, Lecture 20
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Tue. Nov. 9, 2009
Physics 208, Lecture 20
6
1
RL summary
Question
I ( t ) = I∞ (1− e−t /(L / R ) ) = I∞ (1− e−t / τ )
I(t)
I∞ = Vbattery /R
τ = L /R = time constant
Switch closed at t=0
€
I(t)
What is the current through
R1 immediately after the
switch is closed?
R1
R2
L
A.  Vbattery / L
I∞ = Vbattery /R
€
€
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
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Tue. Nov. 9, 2009
Thinking about electromagnetism
Arise from charges
Capacitor, Q=CV

Arise from time-varying B-field Inductor, Faraday effect
Magnetic Fields
Arise from currents

Inductor, Φ=LI
Arise from time-varying E-field


Many similarities between electricity, magnetism
Some symmetries, particularly in time-dependence
8
Maxwell’s unification

Electric Fields
Physics 208, Lecture 20
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
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Tue. Nov. 9, 2009
Physics 208, Lecture 20
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Physics 208, Lecture 20
12
The EM
Spectrum

• A Transverse wave.

• Electric/magnetic fields perpendicular to
propagation direction
Types are
distinguished by
frequency or
wavelength
Visible light is a
small portion of
the spectrum
• Can travel in empty space
f = v/λ, v = c = 3 x 108 m/s (186,000 miles/second)
Tue. Nov. 9, 2009
Physics 208, Lecture 20
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Tue. Nov. 9, 2009
2
Sizes of EM waves

Quick Quiz
Visible light

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
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
A. kilometers
B. meters
What is its wavelength?
C. centimeters
A.  230 m
D. micrometers
B.  0.044 m
C.  2.3 m
" =c/ f =
3 #10 8 m /s
= 3cm
1010 /s
D.  44m
Tue. Nov. 9, 2009
Physics 208, Lecture 20
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Tue. Nov. 9, 2009
Mathematical description
!
Physics 208, Lecture 20
14
EM Waves from an Antenna
x
y
 
E = E o cos( kz − ω t)
 
B = Bo cos( kz − ω t)
k=
 
E⊥B
z
Bo = E o /c

2π
, ω = 2πf
λ

€
€
 
Propagation direction = E × B
€

€

Tue. Nov. 9, 2009
Physics 208, Lecture 20
15
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
€
Transatlantic signals
Detecting EM waves
FM antenna
AM antenna
Spark gap
Gulgielmo Marconi’s transatlantic transmitter
Oriented vertically for radio waves
Tue. Nov. 9, 2009
Physics 208, Lecture 20
Capacitor
banks
Induction
coils
17
Tue. Nov. 9, 2009
Physics 208, Lecture 20
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3
Transatlantic receiver
Energy and EM Waves
uE = εo E 2 ( r,t ) /2
Energy density in E-field
Energy density in B-field
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.
uB = B 2 ( r,t ) /2µo
2
2
Total uTot = εo E /2€+ B /2µo
2 €
= εo E /2 + E 2 /2c 2µo = εo E 2 ( r,t ) = B 2 ( r,t ) / µo
uTot = εo E 2 = εo E o2 cos2 ( kz − ω t) moves w/ EM wave
at speed c
€
€
Tue. Nov. 9, 2009
Physics 208, Lecture 20
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Tue. Nov. 9, 2009
Energy density uE moves at c

€
20
Example: E-field in laser pointer
Power and intensity in EM waves

Physics 208, Lecture 20
Instantaneous energy flow =
energy per second passing plane
2
= cεo E o2 cos 2 (ω t)
  = cuTot = cε o E
3 mW laser pointer.

Beam diameter at board ~ 2mm

Intensity =

How big is max E-field?
€
W/m2
This is power density

Oscillates in time

2
2
Time average of this is Intensity = cεo E max /2 = cBmax /2µo
2( 318W /m )
(3 ×10 m /s)(8.85 ×10 C
2
E max =
Physics 208, Lecture 20
10−3 W
2
2 = 318W /m
π (0.001m)
2
cεo E max
/2 = 318W /m 2

Tue. Nov. 9, 2009

21
8
−12
Tue. Nov. 9, 2009
2
/N ⋅ m 2 )
= 489N /C = 489V /m
Physics 208, Lecture 20
22
€
€
Question
Spherical waves

Sources often radiate EM wave in all directions





A radio station transmits 50kW of power from its
antanna. What is the amplitude of the electric
field at your radio, 1km away.
Light bulb
The sun
Radio/tv transmission tower
A.  0.1 V/m
Spherical wave, looks like plane wave far away
Intensity decreases with distance

Power spread over larger area

I=
Psource
4π r 2
B.  0.5 V/m
C.  1 V/m
D.  1.7 V/m
Source power
E.  15 V/m
Spread over this
surface area
€
Tue. Nov. 9, 2009
Physics 208, Lecture 20
I=
23
Tue. Nov. 9, 2009
50,000W
4 π (1000m)
2
= 4 ×10−3 W / m2
2
cεo E max
/2 = 4 ×10−3 W /m 2
€
2( 4 ×10 W /m )
(3 ×10 m /s)(8.85 ×10 C
2
−3
E max =
8
−12
2
/N ⋅ m 2 )
= 1.73N /C = 1.73V /m
€ Physics 208, Lecture 20
24
4
Radiation Pressure
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
 1  
S = E × B ≡ Poynting Vector
µo

€ 


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
Saw EM waves carry energy
They also have momentum
When object absorbs energy U from EM wave:

€

Momentum Δp is transferred
Δp = U /c ( Will see this later in QM )
U /Δt
= P /c
Result is a force F = Δp /Δt =
c
Pressure = Force/Area =
prad
Power
Intensity
P /A
=
= I /c
c
Radiation€
pressure
on perfectly absorbing object
25
Tue. Nov. 9, 2009
Physics 208, Lecture 20
26
€
Question
Radiation pressure & force
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?
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
A.  100 m/s
Resulting force = (radiation pressure) x (area)
€
B.  56 m/s
€
C.  17 m/s
prad = 2I /c
D.  3.6 m/s
Frad = prad A = 2IA /c =
Physics 208, Lecture 20
27
3 ×10 8 m /s
= 0.1N
a = Frad /m = 10−3 m /s2
E.  0.7 m/s
Tue. Nov. 9, 2009
2(1300W /m 2 )(1.145 ×10 4 m 2 )
v = at = (10−3 m /s2 )( 3600s) = 3.6m /s
Tue. Nov. 9, 2009
Physics 208, Lecture 20
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€
5