Respuesta :
Intensity of electromagnetic wave is given as
[tex]I = 2[\frac{B_{rms}^2}{2\mu_0}\times c][/tex]
given that
[tex]I = 0.0275 W/m^2[/tex]
[tex]0.0275 = \frac{B_{rms}^2}{\mu_0} \times c[/tex]
here we know that
[tex]\mu_0 = 4\pi \times 10^{-7}[/tex]
[tex]c = 3 \times 10^8 m/s[/tex]
now we have
[tex]0.0275 = \frac{B_{rms}^2}{4 \pi \times 10^{-7}}(3 \times 10^8)[/tex]
[tex]B_{rms} = 1.07 \times 10^{-8} T[/tex]
now we will have
[tex]B_0 = \sqrt2 B_{rms}[/tex]
[tex]B_0 = 1.52 \times 10^{-8} T[/tex]
frequency of wave is given as
[tex]f = \frac{c}{\lambda}[/tex]
[tex]f = \frac{3 \times 10^8}{6.90} = 4.35 \times 10^7 Hz[/tex]
now the induced EMF is given as
[tex]EMF = B_0A2\pi f[/tex]
[tex]EMF = 1.52 \times 10^{-8} \times \pi(0.075)^2 \times (2\pi \times 4.25 \times 10^7)[/tex]
[tex]EMF = 0.0733 Volts[/tex]
I = intensity of the wave at the location of the loop = 0.0275 W/m²
B₀ = amplitude of magnetic field = ?
Intensity is given as
I = B²₀ c/(2μ₀)
inserting the values
0.0275 = B²₀ (3 x 10⁸)/(2(12.56 x 10⁻⁷))
B₀ = 1.52 x 10⁻⁸ T
λ = wavelength of the wave = 6.90 m
frequency of the wave is given as
f = c/λ = (3 x 10⁸)/(6.90) = 4.35 x 10⁷ Hz
angular frequency is given as
w = 2πf = 2 x 3.14 x 4.35 x 10⁷ = 2.73 x 10⁸ rad/s
r = radius of the loop = 7.50 cm = 0.075 m
A = area of the loop = πr² = (3.14) (0.075)² = 0.0177 m²
maximum induced emf is given as
E = B₀ A w
inserting the values
E = (1.52 x 10⁻⁸) (0.0177) (2.73 x 10⁸)
E = 0.0735 Volts
E = 73.5 mV
