The rms value of electric field of light coming from sun is $720 \mathrm{~N} / \mathrm{c}$. The average total energy density of electromagnetic wave is (A) \(6.37 \times 10^{-9} \mathrm{~J} / \mathrm{m}^{3}\) (B) \(4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^{3}\) (C) \(81.35 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}\) (A) \(3.3 \times 10^{-3} \mathrm{~J} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The average total energy density of the electromagnetic wave is (B) \(4.58 \times 10^{-6} \frac{J}{m^3}\).

Step by step solution

01

First, remember that the rms value of electric field (E) is given to us: \( E_{rms} = 720 \frac{N}{C} \). #Step 2: Calculate the magnetic field (B) rms value#

Now, we need to find the corresponding rms value of the magnetic field (B) in the electromagnetic wave. We can use the equation: \( B_{rms} = \frac{E_{rms}}{c} \), where c is the speed of light \((3 \times 10^8 \frac{m}{s})\). So, \[ B_{rms} = \frac{720 \frac{N}{C}}{3 \times 10^8 \frac{m}{s}} = 2.4 \times 10^{-6} \frac{T (Tesla)}{m} \] #Step 3: Calculate average energy density#
02

Next, let's calculate the average total energy density (u) of the electromagnetic wave. The formula for energy density is: \(u = \frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{1}{2} \frac{B_{rms}^2}{\mu_0}\), where \(\epsilon_0\) is the vacuum permittivity \((8.854 \times 10^{-12} \frac{C^2}{N m^2})\) and \(\mu_0\) is the vacuum permeability \((4 \pi \times 10^{-7} \frac{T m}{A})\). #Step 4: Insert values and calculate the answer#

Let's plug in the given values and calculate u: \[u = \frac{1}{2} \times (8.854 \times 10^{-12} \frac{C^2}{N m^2}) \times (720 \frac{N}{C})^2 + \frac{1}{2} \times \frac{(2.4 \times 10^{-6} \frac{T}{m})^2}{(4 \pi \times 10^{-7} \frac{T m}{A})}\] After calculating, we get: \[u = 4.58 \times 10^{-6} \frac{J}{m^3}\] So the correct answer is (B) \(4.58 \times 10^{-6} \frac{J}{m^3}\).

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