The electric field of a plane electromagnetic wave is given by
`vec(E)=E_0haticos(kz)cos(omegat)`
The corresponding magnetic field `vec(B)` is then given by :

Options

A. `vec(B)=(E_0)/Chatj sin (kz) sin(omegat)`
B. `vec(B)=(E_0)/Chatj sin (kz) cos(omegat)`
C. `vec(B)=(E_0)/Chatj cos (kz) sin(omegat)`
D. `vec(B)=(E_0)/Chatk sin (kz) cos(omegat)`

Answer

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

### Step 1: Identify the Direction of Propagation
The electric field is given in the form \(\vec{E} = E_0 \hat{i} \cos(kz) \cos(\omega t)\). Here, the wave is propagating in the \(z\) direction (as indicated by the \(kz\) term). The electric field oscillates in the \(x\) direction (as indicated by the \(\hat{i}\)).

### Step 2: Determine the Direction of the Magnetic Field
In an electromagnetic wave, the electric field \(\vec{E}\), magnetic field \(\vec{B}\), and the direction of propagation (wave vector \(\vec{k}\)) are mutually perpendicular. Since \(\vec{E}\) is in the \(x\) direction and the wave is propagating in the \(z\) direction, the magnetic field \(\vec{B}\) must be in the \(y\) direction.

### Step 3: Use Maxwell's Equations
According to Maxwell's equations, the relationship between the electric field and magnetic field in a plane wave can be expressed as:
\[ \vec{E} \times \vec{B} = \vec{k} \]
where \(\vec{k}\) is the wave vector in the direction of propagation.

### Step 4: Calculate the Magnetic Field
We know that the magnitudes of the electric field and magnetic field are related by:
\[ \frac{E}{B} = c \]
where \(c\) is the speed of light. Thus, we can express the magnetic field as:
\[ B = \frac{E}{c} \]

### Step 5: Substitute the Electric Field
Substituting the expression for the electric field into the equation for the magnetic field gives:
\[ B = \frac{E_0 \cos(kz) \cos(\omega t)}{c} \]

### Step 6: Determine the Direction of the Magnetic Field
Since the magnetic field oscillates in the \(y\) direction, we can write:
\[ \vec{B} = B_0 \hat{j} \cos(kz) \cos(\omega t) \]
where \(B_0 = \frac{E_0}{c}\).

### Final Expression
Thus, the corresponding magnetic field is:
\[ \vec{B} = \frac{E_0}{c} \hat{j} \cos(kz) \cos(\omega t) \]

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