Question

Due to presence of an em-wave whose electric component is given by \( E = 100 \sin(\omega t - kx) \, NC^{-1} \), a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as:

• A. \( 50 \sin(\omega t - kx) \, NC^{-1} \)
• B. \( 25 \sin(\omega t - kx) \, NC^{-1} \)
• C. \( 200 \sin(\omega t - kx) \, NC^{-1} \)
• D. \( 400 \sin(\omega t - kx) \, NC^{-1} \)

Hint:

In problems involving energy conservation, remember that energy is proportional to the square of the electric field. A change in the physical dimensions of the setup requires adjustments in the electric field.

Answer 1

The Correct Option is C

This problem requires calculating the updated electric field equation for an electromagnetic wave after its interaction with a cylindrical volume whose diameter is reduced by half, while its length remains unchanged. The initial electric field is provided as E = 100 sin(ωt - kx) NC⁻¹.

1. Energy Density and Volume Considerations:
The energy density of an electromagnetic wave is directly proportional to the square of its electric field strength (\(E^2\)). The total energy (\(U\)) contained within the cylindrical volume is determined by the product of the energy density and the volume. The volume of a cylinder is calculated as \(V = \pi r^2 h = \pi (D/2)^2 h = \frac{\pi D^2}{4} h\), where \(D\) represents the diameter and \(h\) denotes the length. Consequently, the energy \(U\) is proportional to \(E^2 D^2\), assuming \(h\) and \( \epsilon_0 \) are constant.

2. Principle of Energy Conservation:
It is assumed that the total energy remains essentially constant following the alteration of the diameter. Therefore, if the diameter is halved (changing from \(D\) to \(D/2\)), the amplitude of the electric field must adjust to compensate for the reduction in volume, thereby maintaining a (relatively) stable energy level within the cylinder. This relationship can be expressed as \(E_1^2 D_1^2 = E_2^2 D_2^2\), where the subscripts 1 and 2 denote the original and modified electric fields and diameters, respectively.

3. Application of the Diameter Reduction:
Given that \(D_2 = D_1/2\) and \(E_1 = 100 \, \text{NC}^{-1}\). Substituting these values into the energy conservation equation yields:

\(100^2 D^2 = E_2^2 (D/2)^2\)

\(100^2 D^2 = E_2^2 (D^2/4)\)

4. Calculation of the New Electric Field Amplitude:
Simplifying the equation and solving for \(E_2\):

\(100^2 = E_2^2 /4\)

\(E_2^2 = 4 \times 100^2\)

\(E_2 = \sqrt{4 \times 100^2} = 2 \times 100 = 200 \, \text{NC}^{-1}\)

Thus, the new amplitude of the electric field is 200 NC⁻¹.

5. Formulation of the Modified Electric Field:
Consequently, the modified electric field expression is formulated as:

\(E = 200 \sin(\omega t - kx) \, \text{NC}^{-1}\)

Concluding Result:
The corrected expression for the electric field after the diameter is halved is: \( {200 \sin(\omega t - kx) \, \text{NC}^{-1}} \).