Question

A circular wire loop has a radius of 7.50 cm. a sinusoidal electromagnetic plane wave traveling in air passes throughthe loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. the intensity of the wave atthe location of the loop is 0.0275 w>m2, and the wavelength of thewave is 6.90 m. what is the maximum emf induced in the loop

Answer 1

Intensity of electromagnetic wave is given as

$I = 2[\frac{B_{rms}^2}{2\mu_0}\times c]$

given that

$I = 0.0275 W/m^2$

$0.0275 = \frac{B_{rms}^2}{\mu_0} \times c$

here we know that

$\mu_0 = 4\pi \times 10^{-7}$

$c = 3 \times 10^8 m/s$

now we have

$0.0275 = \frac{B_{rms}^2}{4 \pi \times 10^{-7}}(3 \times 10^8)$

$B_{rms} = 1.07 \times 10^{-8} T$

now we will have

$B_0 = \sqrt2 B_{rms}$

$B_0 = 1.52 \times 10^{-8} T$

frequency of wave is given as

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^8}{6.90} = 4.35 \times 10^7 Hz$

now the induced EMF is given as

$EMF = B_0A2\pi f$

$EMF = 1.52 \times 10^{-8} \times \pi(0.075)^2 \times (2\pi \times 4.25 \times 10^7)$

$EMF = 0.0733 Volts$

Answer 2

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