Grade 12General Physics

The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is ?

sudhanshu

12 Years agoGrade 12

1 Answer

Askiitians Tutor Team

Approved Tutor Answer1 Year ago

To find the average total energy density of the electromagnetic wave, we can use a fundamental relationship in electromagnetism. The energy density \( u \) of an electromagnetic wave is given by the formula:

Energy Density Formula

The energy density \( u \) can be expressed as:

However, for electromagnetic waves in free space, we can simplify this. The electric field \( E \) and the magnetic field \( B \) are related by the speed of light \( c \), where \( B = \frac{E}{c} \). Thus, we can express the energy density solely in terms of the electric field:

Using the Electric Field

Substituting \( B \) into the energy density formula, we get:

Since \( \mu_0 = \frac{1}{c^2 \epsilon_0} \), we can rewrite the second term:

Calculating Energy Density

Now, we can calculate the average total energy density using the given rms value of the electric field, which is \( E_{rms} = 720 \, \text{N/C} \). The average energy density \( u \) is then:

Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \). Plugging in the values:

Step-by-Step Calculation

Calculate \( E_{rms}^2 \):

720 N/C = 720^2 = 518400 \, \text{(N/C)}^2

Multiply by \( \epsilon_0 \):

u = 8.85 \times 10^{-12} \, \text{F/m} \times 518400 \, \text{(N/C)}^2

Perform the multiplication:

u ≈ 4.59 \times 10^{-6} \, \text{J/m}^3

Final Result

The average total energy density of the electromagnetic wave from the Sun is approximately \( 4.59 \times 10^{-6} \, \text{J/m}^3 \). This value represents the energy stored per unit volume in the electromagnetic field, illustrating how much energy is carried by the light that reaches us from the Sun.