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

The problem asks us to determine the modified electric field expression of an electromagnetic wave after the diameter of the cylindrical volume it interacts with is halved, while the length remains constant. The original electric field is given as E = 100 sin(ωt - kx) NC⁻¹.

1. Energy Density and Volume:
The energy density of an electromagnetic wave is proportional to the square of the electric field strength (\(E^2\)). The total energy (\(U\)) within the cylindrical volume is given by the product of the energy density and the volume. The volume of a cylinder is \(V = \pi r^2 h = \pi (D/2)^2 h = \frac{\pi D^2}{4} h\), where \(D\) is the diameter and \(h\) is the length. Therefore, the energy \(U \propto E^2 D^2\), since \(h\) and \( \epsilon_0 \) are constant.

2. Energy Conservation:
It is assumed that the total energy remains approximately the same after the diameter change. Therefore, if the diameter is halved (from \(D\) to \(D/2\)), the electric field amplitude must change to compensate for this reduction in volume to maintain a (relatively) constant energy level inside the cylinder. Thus \(E_1^2 D_1^2 = E_2^2 D_2^2\), where the subscripts 1 and 2 refer to the original and modified electric fields and diameters, respectively.

3. Applying the Diameter Change:
We have \(D_2 = D_1/2\) and \(E_1 = 100 \, \text{NC}^{-1}\). Substituting these values into the energy conservation equation, we have:

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

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

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

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

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

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

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

5. Modified Electric Field Expression:
The modified electric field expression is therefore:

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

Final Answer:
The correct electric field expression after diameter halving is: \( {200 \sin(\omega t - kx) \, \text{NC}^{-1}} \).