Alternating Current
Electromagnetic Waves
Sinusoidal Function of Distance
A sinusoidal function of distance is characterized
by its: amplitude, wavelength, and phase constant
Y(x)
Y(x) = YP sin (kx - f )
Yp
x
f/k
-Yp

Yp = amplitude
 = wavelength
f = phase
k = wavenumber [k = 2/]
Sinusoidal Function of Time
A sinusoidal function of time is characterized
by its: amplitude, frequency, and phase constant
V(t)
V(t) = VP sin (wt - f )
Vp
f/w
-Vp
t
Vp = amplitude
f = frequency [f = w/2]
f = phase
T = period [T = 1/f = 2/w]
w is the angular frequency (angular speed) [radians per second].
f is the frequency [cycles per second, or Hertz (Hz)]
w = 2 f
w
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 vector
rotates with constant angular speed.
V(t)
Vp

f/w
-Vp
V(t) = VP sin (wt - f )
t
Alternating Voltages and Currents
When we plug a lamp into a wall
socket, the voltage and current
supplied vary sinusoidally, with a
frequency f of 60 cycles per second
f = 60 sec-1 = 60 Hz
w = 2f
V = VMAX sin wt
V VMAX
I= =
sin wt
R
R
I = I MAX sin wt
Root Mean Square (rms) Values
I = I MAX sin wt
The average value
of I2 is ½ I2MAX
I 
2
I =I
2
2
MAX
sin wt
2
AV
1 2
= I MAX
2
The root mean square (rms)
value of the current is
defined as :
I rms =
I 
2
AV
then :
I rms
1
=
I MAX
2
Alternating Voltages and Currents
The power dissipated in the resistor is:
2
P = I 2 R = I MAX
R sin wt
The average power dissipated in the
resistor is:
using  sin 2wt 
AV
=1
2
2
PAV = I MAX
R  sin 2 wt 
AV
1 2
2
PAV = I MAX R = I rms
R
2
P = I2R instantaneous power
P = I2rmsR average power
The AC generator has a maximum voltage of VMAX = 24 V
and frequency f = 60 Hz. The resistor has R = 265 .
Find:
a) The rms voltage
b) The rms current
c) The average dissipated power
d) The maximum instantaneous value of the dissipated power
Transformers
A transformer is used to
change the voltage in an
electric circuit
VP = VP max sin wt
VS = VS max sin wt
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
 P
Primary coil  P =  N P
t
 S
Secondary coil  S =  N S
t
Since P = S 
 P NP
=
 S NS
But Ɛ  V, then:
Transformer equation
VP N P
=
VS N S
Transformers
Transformer equation
VP N P
=
VS N S
Since energy is conserved,
the power in the primary circuit
equals the power in the secondary circuit
I PVP = I SVS
Bug Zapper
If the zapper operates at 4000 V, and the primary coil
plugs 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
Ey
Bz
c
x
Plane Electromagnetic Waves
Plane Electromagnetic Waves
E and B are perpendicular
to each other, and to the
direction of propagation
of the wave.
The direction of propagation
is given by the right hand
rule: Curl the fingers from E
to B, then the thumb points
in the direction of propagation.
Electromagnetic waves
propagate in vacuum
with speed c, the speed
of light.
Plane Electromagnetic Waves
Ey
Bz
E(x, t) = EP sin (kx-wt) jˆ
B(x, t) = BP sin (kx-wt) zˆ
Waves are in phase,
but fields oriented at 900
Speed of wave is c
c = 1 / 00 = 3 108 m / s
At all times E = c B
c
x
Moving wave
F(x)

v
x
F(x, t) = FP sin (kx - wt )
w = 2 f
w = angular frequency
f = frequency
k = 2 / 
k = wavenumber
 = wavelength
v=w/k=f
For electromagnetic waves
the speed of propagation is:
c = 1 / 00 = 3 108 m / s
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
Ey
The figure represents
green-blue light
with  = 500 nm
Bz
a
P
b
c
x
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
The Electromagnetic Spectrum
Radio Waves: 106 – 109 Hz
Radio/TV, from antennas.
Ultraviolet: 7.5x1014 – 1017 Hz
Damaging to skin/organisms.
Microwaves: 109 – 1012 Hz
Radar, telephone, cooking.
X-Rays: 1017 – 1020 Hz
Damaging. Radiography.
Infrared: 1012 – 1014 Hz
Heat, remote controls.
Gamma Rays:  1020 Hz
Damaging. Used in therapy
and sterilization.
Visible: 4.3x1014 – 7.5x1014 Hz
Detected by our eyes.
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 density
uB: (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 density
uB: (1/0) B2 magnetic field energy density
was calculated for static fields (Econstant, Bconstant)
Electromagnetic fields are constantly changing
and therefore it is more appropriate to treat
the energy in the field using average values.
Using: Arms =
A 
2
= 1
2
AMAX
2
we have, for the total energy density
AV
in the electromagnetic field:
2
2
u AV = 1  0 Erms
 1
Brms
2
2 0
Intensity of an Electromagnetic Wave
The amount of energy that a wave delivers
per 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 (ct) strikes the
surface in the time t.
The energy in the volume V
is: U = u V. Then,
the intensity of the wave is:
U uAct
I=
=
= uc
At
At
I = 1 c 0 E 2  1
B2
2
2 0
and
2
2
I AV = 1 c 0 Erms
 1
cBrms
2
2 0
Intensity of an Electromagnetic Wave
The average intensity of an electromagnetic wave is:
2
2
I AV = 1 c 0 Erms
 1
cBrms
2
20
Using E = c B we have:
I AV = c 0 E
2
rms
=
c
0
2
Brms
Using Erms = (1/2) EMAX and Brms = (1/2) BMAX
2
2
I AV = 1 c 0 EMAX
= 1
cBMAX
2
20
At a given instant in time, the electric field in a beam of light
has 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 time
Intensity is power per unit area
U P
I=
=
At A
and
I AV
PAV
=
A
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 bulb
b) The rms values of E and B, 1 m from the bulb
(assume 5% of the dissipated power is converted into light)
A small laser emits a cylindrical beam of light 1 mm in diameter
with and average power of 5 mW.
a) Calculate the average intensity of the beam
b) 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 U
but 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:
Radiation Pressure AV
I AV
=
c
One a sunny day the average intensity of sunlight on earth is
about 1.0x103 W/m2.
Find:
a) The average radiation pressure due to sunlight
b) The average force exerted on a beach towel that is
1 m by 2.5 m in size (assume the towel absorbs the light)
Propagation of Electromagnetic Waves in Matter
Electromagnetic waves are absorbed
as they propagate through matter.
The absorption of radiation is described by:
I(t) = I0 exp [-t]
Where:
I (t)  Intensity at depth t
I0  Initial intensity
t  depth
  absorption coefficient