if `vecE=E_(0)cos(kz)cos(omegat)hati` then `vec(B)` for electromagentic wave is

Options

A. `B=(E_(0))/(C)hat(k)`
B. `B=(E_(0))/(C)sin(k_(2)) sin(omegat)hat(j)`
C. `B=(E_(0))/(C)sin (kz)cos(omegat)hat(j)`
D. `B=(E_(0))/(C)cos(kz)sin(omegat)hat(j)`

Answer

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

### Step 1: Understand the relationship between \(\vec{E}\) and \(\vec{B}\)
In electromagnetic waves, the electric field \(\vec{E}\) and the magnetic field \(\vec{B}\) are related through the speed of light \(c\). The relationship can be expressed as:

\[ |\vec{B}| = \frac{|\vec{E}|}{c} \]

### Step 2: Differentiate \(\vec{E}\) with respect to \(z\)
We will differentiate the electric field \(\vec{E}\) with respect to \(z\) to find \(\frac{dE}{dz}\):

\[ \vec{E} = E_0 \cos(kz) \cos(\omega t) \hat{i} \]

Differentiating with respect to \(z\):

\[ \frac{dE}{dz} = -E_0 k \sin(kz) \cos(\omega t) \hat{i} \]

### Step 3: Differentiate \(\vec{B}\) with respect to \(t\)
Using the relationship \( \frac{dE}{dz} = -\frac{dB}{dt} \), we can express \(\frac{dB}{dt}\):

\[ -\frac{dB}{dt} = -E_0 k \sin(kz) \cos(\omega t) \]

Thus,

\[ \frac{dB}{dt} = E_0 k \sin(kz) \cos(\omega t) \]

### Step 4: Integrate \(\frac{dB}{dt}\) to find \(\vec{B}\)
To find \(\vec{B}\), we integrate \(\frac{dB}{dt}\) with respect to \(t\):

\[ B = \int E_0 k \sin(kz) \cos(\omega t) dt \]

Using the integral of \(\cos(\omega t)\):

\[ B = \frac{E_0 k}{\omega} \sin(kz) \sin(\omega t) + C \]

Where \(C\) is a constant of integration. However, we can ignore the constant for the purpose of electromagnetic waves.

### Step 5: Substitute \(k\) in terms of \(\omega\) and \(c\)
We know that \(k = \frac{\omega}{c}\). Thus, we can substitute \(k\) in the expression for \(\vec{B}\):

\[ B = \frac{E_0}{c} \sin(kz) \sin(\omega t) \hat{j} \]

### Final Expression for \(\vec{B}\)
The magnetic field vector \(\vec{B}\) can be expressed as:

\[ \vec{B} = \frac{E_0}{c} \sin(kz) \sin(\omega t) \hat{j} \]

### Conclusion
Thus, the magnetic field corresponding to the given electric field is:

\[ \vec{B} = \frac{E_0}{c} \sin(kz) \sin(\omega t) \hat{j} \]