Question

The electric field of a plane electromagnetic wave in vacuum is \(E_{y}=0.5 \cos \left[2 \pi \times 10^{8}\left\\{t-\frac{x}{c}\right\\}\right] V / m ; E_{z}=E_{x}=0\) Then select correct alternative. (1) Direction of propagation of wave is along +ve \(x\)-axis (2) Plane of polarization is \(\mathrm{x}\)-y plane (3) Amplitude of magnetic field is \(1.66 \times 10^{-9} \mathrm{~T}\) (4) All of above

Short answer

(4) All of above

Step-by-step solution

Understand the Given Information

The electric field of the wave is given by \(E_{y}=0.5 \cos \[2 \pi \times 10^{8}\left\{t-\frac{x}{c}\right\}\] \ V / m \) and \( E_{z}=E_{x}=0 \).

Determine Propagation Direction

The wave equation \( E_{y}=0.5 \cos \[2 \pi \times 10^{8}\left\{t-\frac{x}{c}\right\}\] \ V / m \) indicates that the wave propagates in the positive x-direction because of the term \( t-\frac{x}{c} \). This means option (1) is correct.

Identify the Plane of Polarization

The electric field orientation and the fact that \(E_{y} \eq 0 \) while \( E_{z}=E_{x}=0 \) indicates that the electric field oscillates in the y-direction and not in the z-direction. Thus, the plane of polarization is the x-y plane. Hence, option (2) is correct.

Calculate the Magnetic Field Amplitude

Use the relationship between the electric field amplitude \( E_{0} \) and the magnetic field amplitude \( B_{0} \) in a plane wave: \( E_{0} = cB_{0} \). Given \( E_{0} = 0.5 \ V/m \) and the speed of light \( c = 3 \times 10^8 \ m/s \), solve for \( B_{0} \): \( B_{0} = \frac{E_{0}}{c} = \frac{0.5 \ V/m}{3 \times 10^8 \ m/s} = 1.67 \times 10^{-9} \ T \). This indicates that option (3) is correct.

Verify All Options

Since option (1), (2), and (3) are all verified as correct, the correct answer must be (4) All of above.

Key concepts

Electric Field Propagation

In this exercise, we are dealing with a plane electromagnetic wave characterized by an electric field equation:

\(E_{y}=0.5 \, \cos \left[2 \pi \times 10^{8}\left\{t-\frac{x}{c}\right\}\right]\, V / m\).

Electric field propagation in this context refers to the direction in which the wave travels.

The term \(t - \frac{x}{c}\) in the wave equation provides important information about the wave's movement.

It suggests that the wave propagates in the positive x-direction because the position variable x decreases the value of the phase when moving positively. Hence,

• The wave travels along the positive x-axis.

Understanding the direction of propagation is crucial for analyzing how electric and magnetic fields interrelate in space and time.

Plane of Polarization

Polarization describes the orientation of the electric field in an electromagnetic wave. In the equation specified in this exercise, we have:

\(E_{y}=0.5 \, \cos \left[2 \pi \times 10^{8}\left\{t-\frac{x}{c}\right\}\right]\, V / m\)

and \(E_{z}=E_{x}=0\).

This tells us that the electric field oscillates only in the y-direction while there are no oscillations in the z or x directions.

The absence of components in the z and x directions informs us that the oscillations occur in the plane formed by the x and y axes.

• This plane is referred to as the x-y plane.

Thus, the plane of polarization is the x-y plane, where the electric field vectors lie and oscillate with the time variable t.

Magnetic Field Amplitude Calculation

An important relationship in electromagnetic waves is the connection between the electric field (E) and the magnetic field (B). This relationship is given by:

\(E_{0} = c B_{0}\)

Where \(c\) is the speed of light in a vacuum (\(3 \times 10^8 \ m/s\)).

Given the amplitude of the electric field \(E_{0} = 0.5 \ V/m\), we calculate the magnetic field amplitude \(B_{0}\) using:

\[\begin{equation} B_{0} = \frac{E_{0}}{c} = \frac{0.5 \ V/m}{3 \times 10^8 \ m/s}\end{equation}\]This simplifies to:

\[\begin{equation} B_{0} = 1.67 \times 10^{-9} \ T.\end{equation}\]

This calculation shows that the magnetic field amplitude is tiny, reaffirming the wave's extremely high frequency and energy.

Therefore, it is essential to remember that:

• Magnetic field amplitude can be directly calculated from the electric field amplitude and the speed of light.