An EM wave travels in vacuum along z direction:
`vecE=(E_1hati+E_2hatj)cos (kz-omega t)`. Choose the correct option from the following :

Options

A. The associated magnetic field is given as
`B=1/C (E_(1)hati -E_(2)hatj) " cos " (kz - omegat)`
B. The associated magnetic field is given as
`B=1/c (E_(1) hati -E_(2) hatj) " cos "(kz- omegat)`
C. The given electromagnetic field is circulary polarised
D. The given electromagnetic wave is plane polarised

Explanation

To solve the problem regarding the electromagnetic wave traveling in vacuum along the z-direction with the given electric field equation, we can follow these steps:

### Step 1: Understand the Given Electric Field
The electric field is given by:
\[ \vec{E} = (E_1 \hat{i} + E_2 \hat{j}) \cos(kz - \omega t) \]
This indicates that the electric field has components in both the x-direction (i.e., along \(\hat{i}\)) and the y-direction (i.e., along \(\hat{j}\)).

### Step 2: Identify the Direction of Wave Propagation
Since the wave is traveling along the z-direction, we can denote the wave vector \(\vec{k}\) as:
\[ \vec{k} = k \hat{z} \]
The wave is propagating in the positive z-direction.

### Step 3: Determine the Magnetic Field Direction
In an electromagnetic wave, the electric field \(\vec{E}\), magnetic field \(\vec{B}\), and the direction of wave propagation are mutually perpendicular. Therefore, we need to find the direction of the magnetic field \(\vec{B}\).

### Step 4: Use the Right-Hand Rule
To find the direction of the magnetic field, we can use the right-hand rule:
- Point your fingers in the direction of \(\vec{E}\) (which lies in the x-y plane).
- Curl your fingers in the direction of wave propagation (along the z-axis).
- Your thumb will point in the direction of \(\vec{B}\).

Since \(\vec{E}\) lies in the x-y plane, \(\vec{B}\) will also lie in the x-y plane but will be perpendicular to \(\vec{E}\).

### Step 5: Calculate the Magnitude of the Magnetic Field
The relationship between the magnitudes of the electric field and magnetic field in vacuum is given by:
\[ B = \frac{E}{c} \]
where \(c\) is the speed of light. The magnitude of the electric field can be calculated as:
\[ |\vec{E}| = \sqrt{E_1^2 + E_2^2} \]
Thus, the magnitude of the magnetic field will be:
\[ |\vec{B}| = \frac{\sqrt{E_1^2 + E_2^2}}{c} \]

### Step 6: Write the Expression for the Magnetic Field
The magnetic field can be expressed as:
\[ \vec{B} = \frac{1}{c} (E_1 \hat{i} - E_2 \hat{j}) \cos(kz - \omega t) \]
This indicates that the magnetic field has components in the x and y directions, but with opposite signs for the y-component compared to the electric field.

### Step 7: Determine Polarization
Since the electric field oscillates in the x-y plane, it is classified as plane polarized. The oscillations are perpendicular to the direction of wave propagation (z-direction).

### Final Answer
The correct options based on the analysis are:
1. The associated magnetic field is given by \(\vec{B} = \frac{1}{c} (E_1 \hat{i} - E_2 \hat{j}) \cos(kz - \omega t)\).
2. The electromagnetic wave is plane polarized.