Version PREVIEW – HW12 – feng – (58230)
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Poynting Vector
001
10.0 points
A
radio
transmitter
in
space
transmits
isotropically
with
an
average
power
of
630 kW.
What
is
the
average
magnitude
of
the
Poynting vector 45 miles from the radio trans
mitter?
Correct answer: 9
.
56299
×
10
−
6
W
/
m
2
.
Explanation:
Let :
P
= 630 kW = 6
.
3
×
10
5
W
,
and
d
= 45 miles = 72405 m
.
The average value of the Poynting vector is
the intensity of the wave; intensity is power
per unit area.
Since the wave is isotropical,
the intensity is the same in all directions,
and therefore the power distributes uniformly
over a surface 4
π d
2
at a distance
d
around
the transmitter. This means that
(
S
)
=
I
=
P
4
π d
2
=
6
.
3
×
10
5
W
4
π
(72405 m)
2
=
9
.
56299
×
10
−
6
W
/
m
2
.
The conversion from miles to meters is
given by
d
m
= 1609 m
/
mi
×
d
mi
.
Poynting Vector for a Wire
002
10.0 points
A long, straight wire of resistance 2 Ω, radius
0
.
5 mm, and length 2 m carries a constant
current 2
.
6 A.
Calculate the Poynting vector at the surface
of the wire.
Correct answer: 2151
.
77 W
/
m
2
.
Explanation:
Let :
R
= 2 Ω
,
a
= 0
.
5 mm = 0
.
0005 m
,
ℓ
= 2 m
,
and
I
= 2
.
6 A
.
The surface area of the wire is
A
= 2
π
(0
.
0005 m) (2 m) = 0
.
00628319 m
2
.
Let us find the magnitude of the electric
field
vector
E
along the wire.
If
V
is the potential
difference across the ends of the wire, then
V
=
IR
and
E
=
V
ℓ
=
I R
ℓ
.
Applying Amp`
ere’s law, we find that that the
magnetic field at the surface of the wire is
B
=
μ
0
I
2
π a
.
The vectors
vector
E
and
vector
B
are mutually perpendic
ular, so

vector
E
×
vector
B

=
E B
. Hence, the Poynting
vector
vector
S
has a magnitude
S
=
E B
μ
0
=
1
μ
0
I R
ℓ
μ
0
I
2
π a
=
I
2
R
2
π a ℓ
=
I
2
R
A
=
(2
.
6 A )
2
(2 Ω)
0
.
00628319 m
2
=
2151
.
77 W
/
m
2
.
Form this result, we see that
S A
=
I
2
R ,
that is,
the rate SA at which electromagnetic
energy flows into the wire equals the power
I
2
R
(
i.e., rate of energy) dissipated as Joule
heat.
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Version PREVIEW – HW12 – feng – (58230)
2
Cable Energy Flux
003
(part 1 of 2) 10.0 points
The cable is carrying the current
I
(
t
).
Evaluate the electromagnetic energy flux
S
at the surface of a long transmission cable of
resistivity
ρ
, length
ℓ
and radius
a
, using the
expression
vector
S
=
1
μ
0
vector
E
×
vector
B .
1.
S
=
μ
0
ǫ
0
ρ I
2
π a
2
2.
S
=
μ
0
ρ I
2
π a
2
3.
S
=
ρ ℓ I
2
π a
2
4.
S
=
μ
0
c I
2
4
π a
2
ℓ
5.
S
=
μ
0
I
2
2
π a
2
6.
S
=
ρ I
2
2
π
2
a
3
correct
7.
None of these.
8.
S
=
π a
2
I
2
ρ ℓ
9.
S
=
I
2
4
π μ
0
c ℓ
10.
S
=
ǫ
0
μ
0
I
2
2
π a
2
Explanation:
The basic expression for the Poynting Vec
tor is
vector
S
=
1
μ
0
vector
E
×
vector
B .
Based on Ampere’s law,
B
=
μ
0
I
2
π a
,
and Ohm’s law is
E
≡
V
ℓ
=
I R
ℓ
=
ρ I
π a
2
,
since
R
≡
ρ ℓ
π a
2
.
It is easy to see that
vector
E
and
vector
B
are perpen
dicular to each other.
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 Spring '08
 Turner
 Physics, Angle of Incidence, Snell's Law, Total internal reflection, Geometrical optics

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