The electric field of a plane electromagnetic wave varies with time of amplitude `2Vm^-1` propagating along z-axis. The average energy density of the magnetic field is (in `Jm^-3`)

Options

A. `13.29xx10^(-12)`
B. `8.86xx10^(-12)`
C. `17.72xx10^(-12)`
D. `4.43xx10^(-12)`

Explanation

To find the average energy density of the magnetic field of a plane electromagnetic wave with a given electric field amplitude, we can follow these steps:

### Step-by-Step Solution

1. **Identify Given Values**:
- The amplitude of the electric field \( E_0 = 2 \, \text{V/m} \).

2. **Use the Formula for Average Energy Density of the Magnetic Field**:
The average energy density \( u_B \) of the magnetic field is given by the formula:
\[ u_B = \frac{B_0^2}{2\mu_0} \]
where \( B_0 \) is the amplitude of the magnetic field and \( \mu_0 \) is the permeability of free space.

3. **Relate Electric Field and Magnetic Field**:
The relationship between the electric field \( E_0 \) and the magnetic field \( B_0 \) in an electromagnetic wave is given by:
\[ \frac{E_0}{B_0} = c \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \). Thus, we can express \( B_0 \) as:
\[ B_0 = \frac{E_0}{c} \]

4. **Substitute \( B_0 \) into the Energy Density Formula**:
Substituting \( B_0 \) into the energy density formula gives:
\[ u_B = \frac{1}{2\mu_0} \left(\frac{E_0}{c}\right)^2 \]

5. **Use the Value of \( \mu_0 \)**:
The permeability of free space \( \mu_0 \) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

6. **Calculate \( u_B \)**:
Now substituting the values:
\[ u_B = \frac{1}{2 \times (4\pi \times 10^{-7})} \left(\frac{2}{3 \times 10^8}\right)^2 \]
Simplifying this:
\[ u_B = \frac{1}{8\pi \times 10^{-7}} \left(\frac{4}{9 \times 10^{16}}\right) \]
\[ u_B = \frac{4}{72\pi \times 10^9} = \frac{1}{18\pi \times 10^9} \]

7. **Final Calculation**:
Using \( \pi \approx 3.14 \):
\[ u_B \approx \frac{1}{18 \times 3.14 \times 10^9} \approx \frac{1}{56.52 \times 10^9} \approx 1.77 \times 10^{-11} \, \text{J/m}^3 \]

### Conclusion
The average energy density of the magnetic field is approximately \( 1.77 \times 10^{-11} \, \text{J/m}^3 \).