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 \(\overset{E}\) of the given electromagnetic wave, we will follow these steps:

### Step 1: Identify the given magnetic field
The magnetic field \(\overset{B}\) 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: Extract the amplitude of the magnetic field
The amplitude \(B_0\) of the magnetic field is:
\[ B_0 = 1.6 \times 10^{-6} \, \text{Wb/m}^2 \]

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

### Step 4: Calculate the amplitude of the electric field
Using the above relationship, we can find \(E_0\):
\[ E_0 = c \cdot B_0 = (3 \times 10^8) \cdot (1.6 \times 10^{-6}) = 4.8 \times 10^{2} \, \text{V/m} \]

### 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 direction of \(\overset{B}\) is given by the vector \(2 \hat{i} + \hat{j}\).

### Step 6: Find a unit vector in the direction of the magnetic field
First, we need to find the unit vector in the direction of \(\overset{B}\):
\[ \text{Magnitude of } \overset{B} = \sqrt{(2^2 + 1^2)} = \sqrt{5} \]
The unit vector in the direction of \(\overset{B}\) is:
\[ \hat{b} = \frac{2 \hat{i} + \hat{j}}{\sqrt{5}} \]

### Step 7: Find a direction for the electric field
Since \(\overset{E}\) is perpendicular to \(\overset{B}\), we can find a vector that is perpendicular to \(\hat{b}\). One such vector can be obtained by swapping the components and changing the sign of one:
\[ \overset{E} \propto -\hat{j} + 2\hat{i} \]

### Step 8: Normalize the direction of \(\overset{E}\)
The unit vector in the direction of \(\overset{E}\) can be calculated as:
\[ \hat{e} = \frac{-\hat{j} + 2\hat{i}}{\sqrt{(-1)^2 + 2^2}} = \frac{-\hat{j} + 2\hat{i}}{\sqrt{5}} \]

### Step 9: Write the electric field expression
Thus, the electric field \(\overset{E}\) can be expressed as:
\[ \overset{E} = E_0 \cos(2 \times 10^{7} z + 6 \times 10^{15} t) \hat{e} \]
Substituting \(E_0\) and \(\hat{e}\):
\[ \overset{E} = 4.8 \times 10^{2} \cos(2 \times 10^{7} z + 6 \times 10^{15} t) \left(\frac{-\hat{j} + 2\hat{i}}{\sqrt{5}}\right) \]

### Final Expression
Thus, the associated electric field is:
\[ \overset{E} = \frac{4.8 \times 10^{2}}{\sqrt{5}} \cos(2 \times 10^{7} z + 6 \times 10^{15} t) (-\hat{j} + 2\hat{i}) \, \text{V/m} \]