The electric field component of a monochromatic radiation is given by
`vecE=2E_(0)hati cos kz cos omega t`
Its magnetic field `vecB` is then given by :

Options

A. `(2E_(o))/(c)hatj sin kz cos omegat`
B. `-(2E_(o))/(c)hatj sin kz sin omegat`
C. `(2E_(o))/(c)hatj sin kz sin omegat`
D. `(2E_(o))/(c)hatjcos kz cos omegat`

Answer

To find the magnetic field vector \(\vec{B}\) corresponding to the given electric field vector \(\vec{E} = 2E_0 \hat{i} \cos(kz) \cos(\omega t)\), we can follow these steps:

### Step 1: Understand the Relationship Between Electric and Magnetic Fields
The electric field \(\vec{E}\) and magnetic field \(\vec{B}\) in electromagnetic waves are always perpendicular to each other. Additionally, they are related by the equation:
\[ \vec{B} = \frac{\vec{E}}{c} \]
where \(c\) is the speed of light in vacuum.

### Step 2: Identify the Direction of Propagation
From the given electric field, we can see that it oscillates in the \(\hat{i}\) direction (x-direction) and propagates in the z-direction (since it depends on \(z\)). Therefore, the magnetic field \(\vec{B}\) will oscillate in the y-direction (perpendicular to both \(\hat{i}\) and the direction of propagation \(\hat{k}\)).

### Step 3: Calculate the Magnitude of the Magnetic Field
Using the relationship between the electric field and magnetic field, we can express \(\vec{B}\) as:
\[ \vec{B} = \frac{\vec{E}}{c} \]
Substituting the expression for \(\vec{E}\):
\[ \vec{B} = \frac{2E_0 \hat{i} \cos(kz) \cos(\omega t)}{c} \]

### Step 4: Determine the Direction of the Magnetic Field
Since \(\vec{E}\) is in the \(\hat{i}\) direction, and \(\vec{B}\) must be perpendicular to \(\vec{E}\) and in the direction of propagation (which is along \(\hat{k}\)), we can express \(\vec{B}\) in the \(\hat{j}\) direction (y-direction):
\[ \vec{B} = \frac{2E_0}{c} \hat{j} \cos(kz) \cos(\omega t) \]

### Step 5: Final Expression for the Magnetic Field
Thus, the magnetic field vector \(\vec{B}\) can be written as:
\[ \vec{B} = \frac{2E_0}{c} \hat{j} \cos(kz) \cos(\omega t) \]

### Summary
The magnetic field corresponding to the given electric field is:
\[ \vec{B} = \frac{2E_0}{c} \hat{j} \cos(kz) \cos(\omega t) \]