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Sinusoidal Function of Distance
A sinusoidal function of distance is characterized by its: amplitude, wavelength, and phase constant
Y(x) = YP sin (kx - )
Yp = amplitude = wavelength = phasek = wavenumber [k = /]
k
Y(x)
x
Yp
-Yp
Sinusoidal Function of Time
is the angular frequency (angular speed) [radians per second].f is the frequency [cycles per second, or Hertz (Hz)]
f
A sinusoidal function of time is characterized by its: amplitude, frequency, and phase constant
V(t) = VP sin (t - )
Vp = amplitudef = frequency [f = = phaseT = period [T = 1/f = /
/
V(t)
t
Vp
-Vp
V(t) = VP sin (t - )
/
V(t)
t
Vp
-Vp
Use of a rotating vector to generate a sinusoidal function of time.
Plot the y component of the vector,as a function of time, as the vectorrotates with constant angular speed.
•
Alternating Voltages and Currents
When we plug a lamp into a wallsocket, the voltage and currentsupplied vary sinusoidally, with afrequency f of 60 cycles per second
sin
sin
sin
MAX
MAX
MAX
V V t
VVI t
R R
I I t
f = 60 sec-1 = 60 Hz = 2f
Root Mean Square (rms) Values
2 2 2
sin
sin
MAX
MAX
I I t
I I t
The average value of I2 is ½ I2
MAX
2 21
2 MAXAVI I
The root mean square (rms)value of the current isdefined as :
2
:
1
2
rms AV
rms MAX
I I
then
I I
Alternating Voltages and Currents
The power dissipated in the resistor is:2 2 sinMAXP I R I R t
The average power dissipated in theresistor is:
2
2 2
2 2
1using sin 2
sin
1
2
AV
AV MAX AV
AV MAX rms
t
P I R t
P I R I R
P = I2R instantaneous powerP = I2
rmsR average power
The AC generator has a maximum voltage of VMAX = 24 Vand frequency f = 60 Hz. The resistor has R = 265 .Find:a) The rms voltageb) The rms currentc) The average dissipated powerd) The maximum instantaneous value of the dissipated power
Transformers
A transformer is used to change the voltage in anelectric circuit
max
max
sin
sinP P
S S
V V t
V V t
An alternating current in the primary circuit creates an alternating magnetic flux, that is concentrated in the iron core.
The alternating magnetic flux induces an alternating emf,and hence an alternating current, in the secondary circuit
Transformers
Primary coil
Secondary coil
PP P
SS S
Nt
Nt
Since P = S P P
S S
N
N
But Ɛ V, then:
P
Transformer equation
V
VP
S S
N
N
Transformers
P
Transformer equation
V
VP
S S
N
N
Since energy is conserved, the power in the primary circuit
equals the power in the secondary circuit
P P S SI V I V
Bug Zapper
If the zapper operates at 4000 V, and the primary coilplugs to a standard 120 V outlet, and has 27 turns:
How many turns does the secondary coil have?
What is the ratio of the current in the primary, to the current in the secondary circuit? (IP/IS)
Plane Electromagnetic Waves
E and B are perpendicularto each other, and to the direction of propagationof the wave.
The direction of propagationis given by the right hand rule: Curl the fingers from Eto B, then the thumb pointsin the direction of propagation.
Electromagnetic wavespropagate in vacuumwith speed c, the speed of light.
Plane Electromagnetic Waves
Waves are in phase,but fields oriented at 900
Speed of wave is c
At all times E = c B
c m s 1 3 100 08/ /
E(x, t) = EP sin (kx-t)
B(x, t) = BP sin (kx-t) z
j
x
Ey
Bz
c
Moving waveF(x, t) = FP sin (kx - t ) = 2 f = angular frequencyf = frequencyk = 2 k = wavenumber = wavelengthv = / k = f
F(x)
x
v
c m s 1 3 100 08/ /
For electromagnetic wavesthe speed of propagation is:
and c = f
The distance between Earth and the Sun is 1.50x1011 m.How long does it take for the light to cover this distance?
Visible Region of the Electromagnetic Spectrum
Find the frequency of red light with wavelength 700 nm.Find the wavelength of light with frequency 7.5x1014 Hz
Plane Electromagnetic Wave
x
Ey
Bz
c
The figure represents green-blue lightwith = 500 nm
a
b
P
a) What is the distance from a to b?b) How long does it take for the wavefront to move from a to b?c) What does an electric field sensor detect at point P
as a function of time?
The Electromagnetic Spectrum
Radio Waves: 106 – 109 HzRadio/TV, from antennas.
Microwaves: 109 – 1012 HzRadar, telephone, cooking.
Infrared: 1012 – 1014 HzHeat, remote controls.
Visible: 4.3x1014 – 7.5x1014 HzDetected by our eyes.
Ultraviolet: 7.5x1014 – 1017 HzDamaging to skin/organisms.
X-Rays: 1017 – 1020 HzDamaging. Radiography.
Gamma Rays: 1020 HzDamaging. Used in therapyand sterilization.
c = f
Energy in Electromagnetic Waves
• Electric and magnetic fields contain energy. Potential energy stored in the field: uE and uB
uE: ½ 0 E2 electric field energy densityuB: (1/0) B2 magnetic field energy density
• The energy is put into the oscillating fields by the sources that generate them.
• This energy can then propagate to locations far away, at the velocity of light.
Energy in Electromagnetic Waves
The energy density stored in electric and magnetic fields
uE: ½ 0 E2 electric field energy densityuB: (1/0) B2 magnetic field energy density
was calculated for static fields (Econstant, Bconstant)
Electromagnetic fields are constantly changingand therefore it is more appropriate to treatthe energy in the field using average values.
2 2
2 20
0
1Using:2
we have, for the total energy density
in the electromagnetic field:
1 12 2
rms MAXAV
AV rms rms
A A A
u E B
Intensity of an Electromagnetic Wave
The amount of energy that a wave deliversper unit area, per unit time,
is referred to as the Intensity of the wave
In the figure, a beam of light of cross-sectional area A, shines on a surface.All the light energy contained in the volume V = A (ct) strikes thesurface in the time t.
The energy in the volume V is: U = u V. Then,
the intensity of the wave is:
2 20
0
2 20
0
1 12 2
and
1 12 2AV rms rms
U uAc tI uc
A t A t
I c E B
I c E cB
Intensity of an Electromagnetic Wave
The average intensity of an electromagnetic wave is:
Using E = c B we have:
Using Erms = (1/2) EMAX and Brms = (1/2) BMAX
2 20
0
1 12 2AV rms rmsI c E cB
2 20
0AV rms rms
cI c E B
2 20
0
1 12 2AV MAX MAXI c E cB
At a given instant in time, the electric field in a beam of lighthas a magnitude of 510 N/C. What is the magnitude of the magnetic field at that same time?
Since Intensity is energy per unit area per unit timeIntensity is power per unit area
A garage is illuminated by a light bulb dangling from a wire.The bulb radiates uniformly in all directions and consumes 50 W.Calculate:a) The average intensity of light 1 m from the bulbb) The rms values of E and B, 1 m from the bulb(assume 5% of the dissipated power is converted into light)
and
AVAV
U PI
A t A
PI
A
A small laser emits a cylindrical beam of light 1 mm in diameterwith and average power of 5 mW.a) Calculate the average intensity of the beamb) Compare with the 50 W bulb of the previous problem(Hint: use a 1 m separation to compare)
Wave Momentum and Radiation Pressure
An electromagnetic wave not only carries energy Ubut also carries momentum p.
The momentum of a wave is related to its energy by:
p = U / c
For an electromagnetic wave absorbed by an area A in time t, the total energy received is:
U = uAV A c t
Then, the momentum received by the surface is:
p = U/c = uAV A t = IAV A t / c
Wave Momentum and Radiation Pressure
The momentum received by the surface is:
p = IAV A t / c
Then, the average force exerted by the light on the surface is:
FAV = p / t = IAV A / c
And the average pressure on the surface is:
AVRadiation Pressure AVI
c
One a sunny day the average intensity of sunlight on earth isabout 1.0x103 W/m2. Find:a) The average radiation pressure due to sunlightb) The average force exerted on a beach towel that is
1 m by 2.5 m in size (assume the towel absorbs the light)