Question

The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 440 mW/m2. The maximum value of the magnetic field in the wave is closest:

Answer 1

The maximum value of the magnetic field in the wave can be calculated by multiplying the average intensity of the wave by the speed of light. In this case, the maximum value is approximately 132 T.

Step-by-step explanation:

The average intensity of an electromagnetic wave (Iave) can be expressed in terms of the magnetic field strength (B) by using the relationship B = E/c, where c is the speed of light. The energy flow per unit time per unit area (S) of the wave is given by S = Iave. The maximum value of the magnetic field (B) in the wave can be calculated by rearranging the equation as B = S * c.

In this case, the average intensity is given as 440 mW/m2 or 440 * 10-3 W/m2. The speed of light (c) is approximately 3 * 108 m/s. Plugging in these values, we get B = 440 * 10-3 * 3 * 108. The maximum value of the magnetic field is approximately 132 T.

Answer 2

To solve this problem we will proceed to convert the Intensity Units given to a normal system, that is, to standardize their value. Later we will use the value of the electric field according to the RMS electric field, to find it.

Finally the magnetic field would be given based on the previously found electric field, depending on the speed of light:

The value of the intensity is:

`I = 440mW/m^2 = 0.440W/m^2`

The maximum value of the electric field is,

`E_(max) = √(2)E_(rms)`

`E_(max) = √(2)√(Ic\mu_0)`

Where,

I = Intensity,

`\mu`= Permeability constant

c = Speed of light

Replacing,

`E_(max) = √(2)\sqrt{(0.440)(3*10^8)(4\pi*10^(-7))}`

`E_(max) = 18.2V/m`

The maximum value of the magnetic field is,

`B_(max) = (E_(max))/(c)`

`B_(max) = (18.2)/(3*10^8)`

`B_(max) = 6.06*10^(-8)T`

Therefore the maximum magnetic field is
`7.10*10^(-8)T`