Problem 2193 - JEE Physics

Question

If the electric field associated with a radiation of frequency \(10 \mathrm{MH} z\) is \(\mathrm{E}=10 \sin (\mathrm{kx}-\omega \mathrm{t}) \mathrm{mV} / \mathrm{m}\) then its energy density is \(\mathrm{Jm}^{-3}\left(\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)\) (A) \(4.425 \times 10^{-10}\) (B) \(6.26 \times 10^{-14}\) (C) \(8.85 \times 10^{-16}\) (D) \(8.85 \times 10^{-14}\)

Step-by-Step Solution

Answer

The energy density of the electromagnetic wave is \(u = 4.425 \times 10^{-16} \, J/m^{3}\). However, none of the given options match this result, indicating a possible error in the options or the question statement.

Step 1: Calculate the angular frequency \(\omega\)

To find the angular frequency \(\omega\), we will use the formula \(\omega = 2\pi f\), where \(f = 10 MHz = 10 \times 10^{6} Hz\). \[\omega = 2\pi (10 \times 10^{6}) = 20\pi \times 10^{6} \, rad/s\] Now, we have found the value of the angular frequency, \(\omega\).

Step 2: Calculate the energy density \(u\)

Next, we will use the given electric field amplitude, \(E = 10 mV/m\), and the permittivity of free space, \(\varepsilon_0 = 8.85\times 10^{-12} C^2/Nm^2\), to find the energy density, \(u\), using the formula \(u = \frac{1}{2} \varepsilon_{0} E^{2}\). First, convert the electric field amplitude from \(mV/m\) to \(V/m\): \[E = 10 mV/m = 10 \times 10^{-3} V/m = 10^{-2} V/m\] Now, use the formula to calculate the energy density, \(u\): \[u = \frac{1}{2} \varepsilon_{0} E^{2} = \frac{1}{2}(8.85 \times 10^{-12} \, C^2/Nm^2)(10^{-2} \, V/m)^{2}\] \[u = \frac{1}{2}(8.85 \times 10^{-12})(10^{-4}) \, J/m^{3}\] \[u = 4.425 \times 10^{-16} \, J/m^{3}\] Finally, the energy density of the electromagnetic wave is \(4.425 \times 10^{-16} \, J/m^{3}\). Comparing our calculated value with the given options, we see that none of the choices match our result. It seems that there might be an error in the given options or the question statement.

Key Concepts

Frequency Calculation

In the world of electromagnetic waves, understanding frequency calculation is crucial. Frequency, often denoted as \( f \), describes how often the wave oscillates in one second. The unit of frequency is Hertz, abbreviated as \( Hz \). In this exercise, the frequency given is \( 10 \, \text{MHz} \) which converts to \( 10 \times 10^6 \text{Hz} \).
To find the angular frequency \( \omega \), which is directly tied to how fast the wave cycles occur, we use the relationship:

• \( \omega = 2\pi f \)
• where \( 2 \pi \) converts the linear frequency to an angular quantity, accounting for the circular nature of wave oscillations.

Therefore, plugging in the values gives us:

• \( \omega = 2\pi (10 \times 10^6) = 20\pi \times 10^6 \) rad/s.

This angular frequency reflects the rapidity of the wave's oscillations in radians per second.

Energy Density

Energy density is an integral concept in understanding electromagnetic waves, signifying how much energy is stored per unit volume in space. For electromagnetic waves, the energy density associated with the electric field, \( E \), can be calculated using the formula:

• \( u = \frac{1}{2} \varepsilon_0 E^2 \)

Here, \( \varepsilon_0 \) is the permittivity of free space and \( E \) is the electric field amplitude.
In our exercise scenario:

• The given electric field amplitude is \( 10 \, \text{mV/m} \).
• Convert this to \( \text{V/m} \) by dividing by 1000, so \( 10^{-2} \, \text{V/m} \).

Plugging these into the energy density formula gives:

• \( u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (10^{-2})^2 \)
• This simplifies to \( 4.425 \times 10^{-16} \, \text{J/m}^3 \).

Energy density hence quantifies the electromagnetic wave's energy in terms of both the field strength and the permittivity constant.

Permittivity of Free Space

The permittivity of free space, \( \varepsilon_0 \), is a fundamental constant in electromagnetism, representing the ability of a vacuum to permit electric field lines. Essentially, it quantifies how much electric charge can be stored in free space.

• The numeric value is \( 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \).
• This constant is intrinsic to the formulation of electric field equations and plays a pivotal role in the calculation of energy density.

In context, \( \varepsilon_0 \) enables the determination of energy density when combined with the electric field through the relation:

• \( u = \frac{1}{2} \varepsilon_0 E^2 \).

This illustrates how \( \varepsilon_0 \) affects the behavior of electric fields in a vacuum, emphasizing its indispensable nature in understanding electromagnetic phenomena.