Question

Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air? b. What is the E-field amplitude? c. What is the intensity of this wave? What pressure does such wave exert on a totally-absorbing surface?

Answer 1

Answer:

Part a)

$\lambda = 300 m$

Part b)

$E = 2.7 N/C$

Part c)

$I = 9.68 \times 10^{-3} W/m^2$

$P = 3.22 \times 10^{-11} N/m^2$

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

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

$\lambda = \frac{3\times 10^8}{1\times 10^6}$

$\lambda = 300 m$

Part b)

As we know the relation between electric field and magnetic field

$E = Bc$

$E = (9 \times 10^{-9})(3\times 10^8)$

$E = 2.7 N/C$

Part c)

Intensity of wave is given as

$I = \frac{1}{2}\epsilon_0E^2c$

$I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)$

$I = 9.68 \times 10^{-3} W/m^2$

Pressure is defined as ratio of intensity and speed

$P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}$

$P = 3.22 \times 10^{-11} N/m^2$