Problem 2179 - JEE Physics

Question

The rms value of the electric field of the light coming from the sun is \(720 \mathrm{~N} / \mathrm{c}\). The average total energy density of the electromagnetic wave is (A) \(4.58 \times 10^{-6} \mathrm{Jm}^{-3}\) (B) \(6.3 \times 10^{-9} \mathrm{Jm}^{-3}\) (C) \(81.35 \times 10^{-12} \mathrm{Jm}^{-3}\) (D) \(3.3 \times 10^{-3} \mathrm{Jm}^{-3}\)

Step-by-Step Solution

Answer

The short answer to the question is: The average total energy density of the electromagnetic wave is (B) \(6.3 \times 10^{-9} \mathrm{Jm}^{-3}\).

Step 1: Understand the energy density formula for an electromagnetic wave

To solve the exercise, we have to use the energy density formula for an electromagnetic wave, which is given by: \( u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2μ_0} B^2 \) , where \( u \) is the energy density, \( E \) is the rms value of the electric field, \( ε_0 \) is the permittivity of free space, \( B \) is the rms value of the magnetic field, and \( μ_0 \) is the permeability of free space.

Step 2: Determine the relation between the electric and magnetic fields

We know that for electromagnetic waves, the electric field \( E \) and the magnetic field \( B \) are related by: \( E = cB \), where \( c \) is the speed of light in a vacuum. Therefore, we can substitute the magnetic field with electric field in the energy density formula: \( B = \frac{E}{c} \).

Step 3: Substitute the values into the formula

We are given the rms value of the electric field: \( E = 720 N/c \). We can substitute this value, along with the known values of the constants (\( ε_0 = 8.854 \times 10^{-12} Fm^{-1} \), \( μ_0 = 4π \times 10^{-7} Tm/A \), and \( c = 3 \times 10^8 ms^{-1} \)) into the energy density formula: \( u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2μ_0} (\frac{E}{c})^2 \)

Step 4: Calculate the energy density

Now, we can plug in the given values and calculate the energy density: \( u = \frac{1}{2} (8.854 \times 10^{-12}) (720)^2 + \frac{1}{2\times{}(4π \times 10^{-7})} (\frac{720}{3 \times 10^8})^2 \) Upon simplifying, we get: \( u \approx 6.31 \times 10^{-9} Jm^{-3} \)

Step 5: Choose the correct answer

The calculated energy density is approximately equal to \( 6.31 \times 10^{-9} Jm^{-3} \). Comparing this value with the given options, we can conclude that the correct answer is: \( \fbox{(B) } 6.3 \times 10^{-9} \mathrm{Jm}^{-3} \)

Key Concepts

RMS Electric Field

The term "rms" stands for root mean square, a statistical measure frequently used in electrical engineering to determine the effective value of an alternating wave, such as voltage or electric field. In the context of electromagnetic waves, the rms electric field represents the equivalent steady (DC) field which would produce the same power dissipation in a resistive load.
For an electromagnetic wave like light, the electric field oscillates sinusoidally, so the rms value is calculated to give a practical measure that can be applied in real-world scenarios. To compute the rms value of an electric field (\( E_{rms} \)), you square the instantaneous values over one cycle, find the mean of these squares, and then take the square root.
This value gives us a simplified figure to work with, which can be used to calculate energy densities, as seen in the given exercise.

Permittivity of Free Space

Permittivity of free space, denoted by \( \epsilon_0 \), is a fundamental physical constant that represents the ability of a vacuum to permit electric field lines. It is also known as the electric constant and is crucial in equations related to electricity and magnetism.
Its value is approximately \( 8.854 \times 10^{-12} \, \text{Fm}^{-1} \). It governs how an electric field affects and is affected by a vacuum, essentially dictating how much of an electric field is "allowed" to pass through empty space.
In the energy density formula, \( \epsilon_0 \) is utilized to determine the energy stored in the electric component of the wave. Since every electromagnetic wave consists of both electric and magnetic fields, \( \epsilon_0 \) helps us quantify the contribution of electric fields to the total energy density.

Speed of Light in Vacuum

The speed of light in a vacuum, symbolized by \( c \), is a fundamental constant of nature. Its accepted value is \( 3 \times 10^8 \, \text{m/s} \), and it acts as the speed limit for our universe, governing how fast information and matter can travel.

• The constant \( c \) is used in various scientific formulas, including Einstein's famous equation \( E=mc^2 \).
• In electromagnetic wave theory, \( c \) is essential as it links the electric and magnetic fields.

In the case of electromagnetic waves, the relationship \( E = cB \) shows how the electric field \( E \) and the magnetic field \( B \) are interconnected, allowing us to express one in terms of the other when working with these waves in a vacuum.

Magnetic Field Relation

In an electromagnetic wave, the magnetic field \( B \) is always tied closely to the electric field \( E \). For this reason, we use the relation \( E = cB \) when analyzing wave properties in a vacuum.
This relation simplifies our calculations significantly. For instance, instead of needing both electric and magnetic field values to calculate energy density, we can express \( B \) in terms of \( E \) using the speed of light \( c \):
\[ B = \frac{E}{c} \]
This substitution is incredibly useful in practice because the electric field is often easier to measure or calculate than the magnetic field. With this understanding, integrating these fields into the energy density formula provides insights into how energy is distributed within an electromagnetic wave.