Solutions
(2):
1) We have: B = E /c
= (220 V/ m) / (3 x 10 8 m/s) = 7.3 x 10-7 T
1) We have: B = E /c
= (220 V/ m) / (3 x 10 8 m/s) = 7.3 x 10-7 T
2)
The intensity of an EM wave is :
I = S(av) = E(max) B(max)/ 2 mo
where the area A = 2.4 m x 0.7 m = 1.68 m 2 and P R denotes the radiation pressure, or P R = S/c (for complete absorption)
Now, S = Power/ area )= ( 25 W/m 2 ) / (1.68 m 2) = 14.88 W
And the radiation pressure:
P R = S(av) / c = 14.88 W/ (3 x 10 8 m/s)
= 3.3 x 10 - 9 N/m2
(3) We have: P(av) = 4 kw = 4 x 10 3 W
Assume spherical symmetry for area affected so area of the applicable spherical surface is: 4 π r 2
I = S(av) = E(max) B(max)/ 2 mo
where the area A = 2.4 m x 0.7 m = 1.68 m 2 and P R denotes the radiation pressure, or P R = S/c (for complete absorption)
Now, S = Power/ area )= ( 25 W/m 2 ) / (1.68 m 2) = 14.88 W
And the radiation pressure:
P R = S(av) / c = 14.88 W/ (3 x 10 8 m/s)
= 3.3 x 10 - 9 N/m2
(3) We have: P(av) = 4 kw = 4 x 10 3 W
Assume spherical symmetry for area affected so area of the applicable spherical surface is: 4 π r 2
With r the distance (radius) to the
radio station transmitter but convert to m :
r = 4 mi. =
4 mi (5280 ft./mi)(0.303 m./ft) = 6.4 x 103 m
Area = 4 π ( 6.4 x 103 m) 2 = 5.1 x 10 8 m2
Therefore the average Poynting vector associated with the transmission is:
S(av) = P(av)/ A =
(4 x 10 3 w)/ (5.1 x 10 8 m2 ) = 7.7 x 10 -6 W/m2
But recall:
S(av) = S(av) = E(max) 2/ 2 mo c
r = 4 mi. =
4 mi (5280 ft./mi)(0.303 m./ft) = 6.4 x 103 m
Area = 4 π ( 6.4 x 103 m) 2 = 5.1 x 10 8 m2
Therefore the average Poynting vector associated with the transmission is:
S(av) = P(av)/ A =
(4 x 10 3 w)/ (5.1 x 10 8 m2 ) = 7.7 x 10 -6 W/m2
But recall:
S(av) = S(av) = E(max) 2/ 2 mo c
Therefore, solving for E(max):
E(max) = [2S(av) mo c] ½
Then: E(max)=
[2(7.7 x 10 -6 W/m2 ) (4 π x 10 -7 H/m)(3 x 10 8 m/s)] ½
E(max) = 0.076 V/m
Then the emf induced in a 65 cm long (L =0.65m) antenna is:
Emf = E(max) L = (0.076 V/m) x (0.65m) = 0.049 V
(4) We assume: P(solar) = 10 3 W/m2
But because efficiency is relevant we need P(in). Thus,
eff = P (out)/ P(in) = 0.3 = 1 MW/ P(in)
Where 1 MW = 10 6 watts is the desired energy to come out, or be produced.
To get this, the power we need to put in, is:
P(in) = P(out)/ eff = (10 6 w)/ 0.3 = 3.33 x 10 6 W
Then:
Area A = P(in)/ S =
E(max) = [2S(av) mo c] ½
Then: E(max)=
[2(7.7 x 10 -6 W/m2 ) (4 π x 10 -7 H/m)(3 x 10 8 m/s)] ½
E(max) = 0.076 V/m
Then the emf induced in a 65 cm long (L =0.65m) antenna is:
Emf = E(max) L = (0.076 V/m) x (0.65m) = 0.049 V
(4) We assume: P(solar) = 10 3 W/m2
But because efficiency is relevant we need P(in). Thus,
eff = P (out)/ P(in) = 0.3 = 1 MW/ P(in)
Where 1 MW = 10 6 watts is the desired energy to come out, or be produced.
To get this, the power we need to put in, is:
P(in) = P(out)/ eff = (10 6 w)/ 0.3 = 3.33 x 10 6 W
Then:
Area A = P(in)/ S =
(3.33 x 10 6
W)/ (10 3 W/m2) = 3.33 x
10 3 m2
Or
about 3, 330 square meters.
5 ) By definition of current: I = nevA
where v is the drift velocity: v = I/neA
where v is the drift velocity: v = I/neA
Where A = z1 y1 = (2 cm) (1 mm) = 0.02m (0.001 m) =
(2 x 10 -2 m) (1 x 10 -3
m) = 2 x 10 -5 m 2
Therefore:
v = (200A)/ [(7.4 x 10 28 /m )(1.6 x 10 -19 C)(2 x 10 -5 m 2)]
v = 8.4 x 10 -4 (-x^) m/s
b) E = VH / z1
Therefore:
v = (200A)/ [(7.4 x 10 28 /m )(1.6 x 10 -19 C)(2 x 10 -5 m 2)]
v = 8.4 x 10 -4 (-x^) m/s
b) E = VH / z1
where z1= 0.02m (= 2 cm)
But: VH = BI/ new
But: VH = BI/ new
= (1.5T) (200 A) / [(7.4 x 10 28 /m )(1.6 x 10 -19 C)( 1 x 10 -3
m)]
= 2.5 x 10 -5 V (-z^)
Then: E = VH / z1 =
2.5 x 10 -5 V (-z^)/
0.02 m = 1.2 x 10 -3 V/m(-z^)
c) N.B. The answer is already found above as VH = BI/ new
c) N.B. The answer is already found above as VH = BI/ new
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