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

Part a)

`\lambda = 300 m`

Part b)

`E = 2.7 N/C`

Part c)

`I = 9.68 * 10^(-3) W/m^2`

`P = 3.22 * 10^(-11) N/m^2`

Step-by-step 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 = (c)/(f)`

`\lambda = (3* 10^8)/(1* 10^6)`

`\lambda = 300 m`

Part b)

As we know the relation between electric field and magnetic field

`E = Bc`

`E = (9 * 10^(-9))(3* 10^8)`

`E = 2.7 N/C`

Part c)

Intensity of wave is given as

`I = (1)/(2)\epsilon_0E^2c`

`I = (1)/(2)(8.85 * 10^(-12))(2.7)^2(3* 10^8)`

`I = 9.68 * 10^(-3) W/m^2`

Pressure is defined as ratio of intensity and speed

`P = (I)/(c) = (9.68* 10^(-3))/(3* 10^8)`

`P = 3.22 * 10^(-11) N/m^2`