Answer:
[tex]E_{max}[/tex] = 2309 V/M
Explanation:
The average power emitted by the electromagnetic wave is:
[tex]P_{avg }=SA[/tex]
where:
S = intensity of the electromagnetic wave
A = cross sectional area of the beam
[tex]A = \pi r^2[/tex]
Replacing A with [tex]\pi r^2[/tex] and making S the subject of the formula; we have:
[tex]S = \frac{P_{avg}}{\pi r^2}[/tex]
[tex]S = \frac{4.5*10^{-3} \ W}{ \pi (\frac{1}{2}(0.9*10^{-3}m)^2}}[/tex]
[tex]S = 7073.55 \ W/m^2[/tex]
Intensity of electromagnetic wave in terms of electric field is;
[tex]S = \frac{1}{2 \mu_o c}E_{max}^2[/tex]
where [tex]E_{max}[/tex] is the maximum value of the electric field
Making [tex]E_{max}[/tex] the subject of the formula; we have:
[tex]E_{max}[/tex] [tex]= \sqrt{2 \mu_oCS}[/tex]
[tex]E_{max}[/tex] = [tex]\sqrt{2 (4 \pi *10^{-7} \ H/m) (3*10^8 \ m/s) (7073.55 \ W/m^2)[/tex]
[tex]E_{max}[/tex] = 2309 V/M
The amplitude of the electric field will be:
"2309 V/M".
According to the question,
Diameter, d = 0.900 mm
Total power output, [tex]P_{avg}[/tex] = 4.50 mW or,
= 4.5 × 10⁻³ W
We know that,
Cross sectional area (A):
= πr²
The average power emitted be:
→ [tex]P_{avg}[/tex] = SA
or,
S = [tex]\frac{P_{avg}}{\pi r^2}[/tex]
By substituting the values,
= [tex]\frac{4.5\times 10^{-3}}{\pi\times \frac{1}{2}(0.9\times 10^{-3})^2 }[/tex]
= 7073.55 W/m²
hence,
The intensity of electromagnetic wave:
→ S = [tex]\frac{1}{2 \mu_o C}[/tex] [tex]E_{max}^2[/tex]
or,
[tex]E_{max}[/tex] = [tex]\sqrt{2 \mu_o CS}[/tex]
By putting the values,
= [tex]\sqrt{2\times 4 \pi\times 10^{-7}\times 3\times 10^{8}\times 7073.55}[/tex]
= 2309 V/M
Thus the above answer is correct.
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