Solutions
(2):

1) We have:

= (220 V/ m) / (3 x 10

1) We have:

**B**=**E**/c= (220 V/ m) / (3 x 10

**m/s) = 7.3 x 10**^{8}^{-7}T
2)
The intensity of an EM wave is :

I = S(av) = E(max) B(max)/ 2 m

where the area A = 2.4 m x 0.7 m = 1.68 m

Now, S = Power/ area )= ( 25 W/m

And the radiation pressure:

P

= 3.3 x 10

(3) We have: P(av) = 4 kw = 4 x 10

Assume spherical symmetry for area affected so area of the applicable spherical surface is: 4 π r

I = S(av) = E(max) B(max)/ 2 m

_{o}where the area A = 2.4 m x 0.7 m = 1.68 m

**and P**^{2 }_{ }**denotes the radiation pressure, or P**_{R }_{ }**= S/c (for complete absorption)**_{R}Now, S = Power/ area )= ( 25 W/m

**) / (1.68 m**^{ 2}**) = 14.88 W**^{ 2}And the radiation pressure:

P

_{ }**= S(av) / c = 14.88 W/ (3 x 10**_{R}**m/s)**^{8}= 3.3 x 10

**N/m**^{ - 9}^{2}(3) We have: P(av) = 4 kw = 4 x 10

**W**^{3}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

r = 4 mi. =

4 mi (5280 ft./mi)(0.303 m./ft) = 6.4 x 10

Area = 4 π ( 6.4 x 10

Therefore the average Poynting vector associated with the transmission is:

S(av) = P(av)/ A =

(4 x 10

But recall:

S(av) = S(av) = E(max)

*convert to m*:r = 4 mi. =

4 mi (5280 ft./mi)(0.303 m./ft) = 6.4 x 10

**m**^{3}Area = 4 π ( 6.4 x 10

**m)**^{3}**= 5.1 x 10**^{2}**m**^{8}^{2 }Therefore the average Poynting vector associated with the transmission is:

S(av) = P(av)/ A =

(4 x 10

**w)/ (5.1 x 10**^{3}**m**^{8}**) = 7.7 x 10**^{2 }**W/m**^{-6}^{2 }But recall:

S(av) = S(av) = E(max)

**/ 2 m**^{2}_{o }c
Therefore, solving for E(max):

E(max) = [2S(av) m

Then: E(max)=

[2(7.7 x 10

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

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

To get this, the power we need to put in, is:

P(in) = P(out)/ eff = (10

Then:

Area A = P(in)/ S =

E(max) = [2S(av) m

_{o }c]^{½ }Then: E(max)=

[2(7.7 x 10

**W/m**^{-6}**) (4 π x 10**^{2 }**H/m)(3 x 10**^{-7}**m/s)]**^{8}^{ 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

**W/m**^{ 3}^{2}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

**watts is the desired energy to come out, or be produced.**^{6}To get this, the power we need to put in, is:

P(in) = P(out)/ eff = (10

**w)/ 0.3 = 3.33 x 10**^{6}**W**^{6}Then:

Area A = P(in)/ S =

(3.33 x 10

**W)/ (10**^{6}**W/m**^{3}**) = 3.33 x 10**^{2}**m**^{3}^{2}

^{}
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

Therefore:

v = (200A)/ [(7.4 x 10

v = 8.4 x 10

b) E = V

^{ -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/sb) E = V

**/ z1**_{H }
where z1= 0.02m (= 2 cm)

But: V

But: V

**= BI/ new**_{H}
= (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 = V

**/ z1 =**_{H }
2.5 x 10

c) N.B. The answer is already found above as V

^{ -5}V (-**z**^)/ 0.02 m = 1.2 x 10^{ -3}V/m(-**z**^)c) N.B. The answer is already found above as V

**= BI/ new**_{H}
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