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Homework 12
chapter 34: 9, 51
chapter 35: 31, 55
Problem 34.9
Figure 34.4b shows a plane electromagnetic sinusoidal wave propagating in the
x-direction. Suppose that the wavelength is 50 m, and the electric field vibrates in
the xy plane with an amplitude of 22 V/m. Calculate (a) the frequency of the wave
and (b) the magnitude and direction of the magnetic field when the electric field
has its maximum value in the negative y-direction. (c) Write an expression for the
magnetic field with the correct unit vector, with numerical values for Bmax, k, and
ω, and its magnitude in the form
B = Bmax cos (kx-ωt).
y
λ
E
Em
c
ΦE
dy
ΦB
dz
B
x
dx
z
a) Solution 1 (formal)
Wave phase speed is determined by the relation between these
two quantities. It moves with such a way that the phase remains
constant value
kx (t ) − ωt = const.
From which
c=
dx ω
=
dt k
1
Wavelength of a wave is directly related to the magnitude of the
propagation vector
λ=
2π
k
while frequency is related to the angular frequency of the wave
ω = 2πf
Therefore
f=
ω ck ck 2πc c
=
=
=
= =
2π 2π 2π 2πλ λ
m
s = 6MHz
50m
3 ⋅ 108
Solution 2
During one oscillation (one period) the wave moves by a
distance equal to the wavelength of the wave. Therefore
T=
λ
c
If one oscillation take time T, in one second the number of
oscillations will be
f=
1 c
= =
T λ
m
s = 6MHz
50m
3 ⋅108
b) Solution 1 (from Faraday’s law)
The magnetic flux through the vertical, differential surface indicated on
the figure is proportional to the magnitude of the magnetic field.
Therefore the rate at which this magnetic flux changes is related only to
the rate at which the magnetic field varies.
2
∂B
dΦ B d
= (B ⋅ dxdy ) =
⋅ dxdy
∂t
dt
dt
The linear integral of the electric field vector can be related directly to
the electric field strength at the location of the differential surface
⎡
⎣
∫ E ⋅ ds = 0 + E( x + dx ) ⋅ dy + 0 − E( x ) ⋅ dy = ⎢E(x ) +
∂E
∂E
⎤
dx − E( x )⎥ ⋅ dy =
dxdy
∂x
∂x
⎦
Therefore, from Faraday’s law (using the general form for a sinusoidal
wave), the above equation requires that
∂E
∂B
− E m k sin (kx − ωt ) =
=−
= − B m ω sin (kx − ωt )
∂t
∂x
From which the magnitude of the magnetic field is
V
22
k
2π
E
m = 73.3nT
Bm = E m =
Em = m =
m
ω
2πfλ
c
3 ⋅ 108
s
Using the right-hand rule, when the y-component of the electric field is
negative, the z-component of the magnetic field is negative too.
Therefore
B = - 73.3nT ⋅ kˆ
Solution 2 (From Ampere-Maxwell’s law)
From the definition, the displacement current through the horizontal,
differential surface is related to the rate of change in the electric flux
dΦ E
d
∂E
Id = ε0
= ε 0 (E ⋅ dxdz ) = ε 0
⋅ dxdy
∂t
dt
dt
The linear integral of the magnetic field vector can be related directly to
the magnetic field strength at the location of the differential surface
3
⎡
⎣
∫ B ⋅ ds = 0 − B( x + dx) ⋅ dz + 0 + B( x ) ⋅ dz = ⎢− B(x ) −
∂B
∂B
⎤
dx + B( x )⎥ ⋅ dz = − dxdz
∂x
∂x
⎦
Ampere-Maxwell’s law yields
∂B
∂E
= −μ 0 ε 0
= −μ 0 ε 0 E m ω sin (kx − ωt )
− B m k sin (kx − ωt ) =
∂x
∂t
From which
1
E
ω
B m = μ 0 ε 0 E m = 2 cE m = m
k
c
c
which leads to the same answer.
c) From the frequency of the wave, its angular frequency is
ω = 2πf = 2π ⋅ 6MHz = 37.7MHz
From the given wavelength, the value of the x-component of the
propagation vector is
2π 2π
k=
=
= 0.126m −1
λ 50m
(The peak value of the magnetic field was found in part (c)).
Hence
(
)
B(x, t ) = Bm cos(kx − ωt ) ⋅ kˆ = 73.3nT ⋅ cos 0.126m −1 ⋅ x − 37.7MHz ⋅ t ⋅ kˆ
4
Problem 34.51
A dish antenna having a diameter of 20 m receives (at normal incidence) a radio
signal fro a distant source as shown in the
figure. The radio signal is a continuous
Em = 0.2 μV/m
20 m
sinusoidal wave with amplitude Emax = 0.2
μV/m. Assume the antenna absorbs all the
radiation the falls on the dish. (a) What is the
amplitude of the magnetic field in this wave?
(b) What is the intensity of the radiation
received by the antenna? (d) What force is
exerted by the radio waves on the antenna?
a) The magnitude of the electric field vector and the magnitude
of the magnetic field vector are proportional to each other
E m 0.2 ⋅ 10 −6 V / m
Bm =
=
= 6.7 ⋅ 10 −16 T
8
c
3 ⋅ 10 m / s
b)By definition, the intensity of electromagnetic wave is equal to
the average value of the magnitude of the Poynting vector. It can
also be expressed in terms of the magnitude of the electric field
vector or the magnitude of the magnetic field vector.
(
(
)
2
E 2m
B2m
6.7 ⋅ 10 −16 T
W
I = Sav =
=
=
=
5
.
31
2μ 0 c 2μ 0 2 ⋅ 4π ⋅ 10 − 7 Tm / A
m2
)
c) The power received by the antenna is related to the size of the
antenna and the intensity of the approaching wave
πD 2
W π ⋅ (20m )
Pav = I ⋅
= 5.31 2 ⋅
= 1.67 ⋅10 −14 W
4
4
m
2
d) The force exerted on the antenna is equal to product of the
antena’s area and the wave pressure, related to the magnitude
of the Poyting vector
Sav πD 2 5.31W / m 2 π ⋅ (20m )
F = P⋅A =
⋅
=
⋅
= 5.56 ⋅ 10 −23 N
8
c
4
4
3 ⋅ 10 m / s
2
5
Problem 35.31
The index of refraction for violet in silica flint glass is 1.66, and for red light is
1.62. What is the angular dispersion of visible light passing through a prism of
apex 60° if the angle of incidence 60°?
deviation
angle
60°
β
dispersion
angle
50°
α
γ
θ
ϕ
The right-hand side drawing above traces a monochromatic ray
through the prism. From the geometrical considerations
and
β = 90° − α
γ = 180° − 60° − β
Combining these two equations gives
γ = 180° − 60° − 90° + α = 30° + α
Also, observe that
θ = 90° − γ = 90° − 30° − α = 60° − α
Note that in the previous equation, α is the angle of refraction at
the first surface and θ is the angle of incidence at the second
surface.
From this point it is easier to use numerical calculations
rather than algebraic transformations. For red light, the index
of refraction of glass is nr = 1.62. Applying Snell’s law to the
refraction of light at the first surface,
6
⎞
⎛ 1
α r = arcsin⎜
sin 50° ⎟ = 28.22°
⎝ 1.62
⎠
Thus, the angle of incidence at the second surface (of the red
light) is
θ r = 60° − α r = 60° − 28.22° = 31.78° .
Applying Snell’s law to the refraction at the second surface
yields
⎞
⎛ 1.62
ϕr = arcsin⎜
sin 31.8 ⎟ = 58.56°
⎝ 1.00
⎠
Repeating the calculation for the violet light, at which, the
index of refraction of glass is nr = 1.66 we obtain the refraction
agle at the first surface to be
⎛ 1
⎞
α v = arcsin⎜
sin 50° ⎟ = 27.48°
⎝ 1.66
⎠
which results in the angle if incidence at the second surface (of
the violet light) equal to
θ v = 60° − α v = 60° − 27.48° = 32.52° .
Applying Snell’s law to the refraction at the second surface
again gives
⎛ 1.66
⎞
ϕ v = arcsin⎜
sin 31.8 ⎟ = 63.17°
⎝ 1.00
⎠
The dispersion of the visible light is therefore
Δϕ = 63.17° − 58.56° = 4.61°
7
Problem 35.55
A shallow glass dish is 4.00 cm wide at the bottom, as shown in Figure P35.63.
When an observer’s eye is positioned in as shown, the observer sees the edge of
the bottom of the empty dish. When the dish is filled with water, the observer sees
the center of the bottom of the dish. Find the height of the dish.
θ
2
θ
2
θ
1
h
2r
r
empty
filled
The observer can only see those points from which the light
enters the observer's eye. When the glass is empty, the angle at
which the observer sees the point, the height and the radius of
the glass are related. A simple geometrical analysis gives:
2r
(1)
tan θ2 =
h
When the glass is filled with water n1 = 1.33, Snell's law
relates the angle of incidence with the angle of refraction
n1 sin θ1 = n 2 sin θ2
(2)
where n 2 is the index of refraction of air.
The angle of incidence, the radius of the glass and its
height are again geometrically related:
r
(3)
tan θ1 =
h
8
We obtained three equations with three unknowns (h,θ1,θ2).
The rest is math. Substituting for the trigonometric functions in
equation (2) using equations (1) and (3) we obtain
n1
(hr )2
2
1 + (hr )
= n2
(2hr )2
2
1 + (2hr )
Squaring both sides and rearranging terms leads to the final
equation
⎛
n12 ⎜
r⎞
⎟
⎝h⎠
2
2
⎛ ⎛ 2r ⎞ 2 ⎞
⎛ ⎛ r ⎞2 ⎞
2
r
⎛
⎞
2
⋅ ⎜1 + ⎜ ⎟ ⎟ = n 2 ⎜ ⎟ ⋅ ⎜1 + ⎜ ⎟ ⎟
⎜ ⎝h⎠ ⎟
⎝ h ⎠ ⎜⎝ ⎝ h ⎠ ⎟⎠
⎝
⎠
4
(
n12
− n 22
)
2
⎛r⎞
2
2
⎜ ⎟ = 4n 2 − n 1
⎝h⎠
Finally
n12 − n 22
1.332 − 12
= 2. 4cm
= 2 ⋅ 2cm ⋅
h = 2r
2
2
2
2
4 ⋅1 − 1.33
4 n 2 − n1
9