The magnetic field of an electromagnetic wave is given by:
`oversetB=1.6xx10^(-6)cos(2xx10^(7)z+6xx10^(15)t)(2hati+hatj)(Wb)/m^(2)`
The associated electric field will be:

Options

A. `oversettoB=1.6xx10^(-6)cos(2xx10^(7)z-6xx10^(15)t)(2hati+hatj)v/m`
B. `oversettoE=4.8xx10^(2)cos(2xx10^(7)z-6xx10^(15)t)(-2hatj+hati)v/m`
C. `oversettoE=4.8xx10^(2)cos(2xx10^(7)z-6xx10^(15)t)(-2hatj+hati)v/m`
D. `oversettoE=4.8xx10^(2)cos(2xx10^(7)z-6xx10^(15)t)(2hatj+hati)v/m`

Answer

To find the associated electric field of the given electromagnetic wave, we can follow these steps:

### Step 1: Identify the given magnetic field
The magnetic field of the electromagnetic wave is given as:
\[ \overset{B} = 1.6 \times 10^{-6} \cos(2 \times 10^{7} z + 6 \times 10^{15} t) (2 \hat{i} + \hat{j}) \, \text{Wb/m}^2 \]

### Step 2: Determine the magnitude of the magnetic field
The magnitude of the magnetic field \( B_0 \) can be extracted from the expression:
\[ B_0 = 1.6 \times 10^{-6} \, \text{Wb/m}^2 \]

### Step 3: Use the relationship between electric field and magnetic field
In an electromagnetic wave, the magnitudes of the electric field \( E_0 \) and magnetic field \( B_0 \) are related by the equation:
\[ E_0 = c B_0 \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step 4: Calculate the electric field magnitude
Substituting the values into the equation:
\[ E_0 = (3 \times 10^8) \times (1.6 \times 10^{-6}) = 4.8 \times 10^2 \, \text{V/m} \]

### Step 5: Determine the direction of the electric field
The direction of the electric field \( \overset{E} \) is perpendicular to the direction of the magnetic field \( \overset{B} \). The magnetic field vector \( \overset{B} \) has components along \( \hat{i} \) and \( \hat{j} \):
\[ \overset{B} = 2 \hat{i} + \hat{j} \]
To find a vector \( \overset{E} \) that is perpendicular to \( \overset{B} \), we can use the cross product or find a vector that satisfies the dot product condition:
\[ \overset{E} \cdot \overset{B} = 0 \]

### Step 6: Find a suitable electric field vector
Assuming the electric field vector has the form:
\[ \overset{E} = E_0 (\hat{a} \hat{i} + \hat{b} \hat{j}) \]
We need to find \( \hat{a} \) and \( \hat{b} \) such that:
\[ (2 \hat{i} + \hat{j}) \cdot (a \hat{i} + b \hat{j}) = 0 \]
This gives us the equation:
\[ 2a + b = 0 \quad \Rightarrow \quad b = -2a \]

### Step 7: Choose a suitable value for \( a \)
Let’s choose \( a = 1 \):
\[ b = -2 \quad \Rightarrow \quad \overset{E} = 4.8 \times 10^2 (1 \hat{i} - 2 \hat{j}) \, \text{V/m} \]

### Final Answer
Thus, the associated electric field is:
\[ \overset{E} = 4.8 \times 10^2 (1 \hat{i} - 2 \hat{j}) \, \text{V/m} \]