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 magentic 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 (U_B) for a plane electromagnetic wave with a given electric field amplitude, we can follow these steps:

### Step 1: Understand the relationship between electric and magnetic fields in an electromagnetic wave.
In an electromagnetic wave, the electric field (E) and the magnetic field (B) are related by the speed of light (c):
\[ B = \frac{E}{c} \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step 2: Calculate the RMS value of the magnetic field (B_rms).
The RMS (Root Mean Square) value of the magnetic field can be calculated using the relationship:
\[ B_{\text{rms}} = \frac{B_0}{\sqrt{2}} \]
where \( B_0 \) is the amplitude of the magnetic field. First, we need to find \( B_0 \).

Using the amplitude of the electric field (E_0 = 2 V/m):
\[ B_0 = \frac{E_0}{c} = \frac{2 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]
Calculating this gives:
\[ B_0 = \frac{2}{3 \times 10^8} \approx 6.67 \times 10^{-9} \, \text{T} \]

Now, calculate \( B_{\text{rms}} \):
\[ B_{\text{rms}} = \frac{6.67 \times 10^{-9}}{\sqrt{2}} \approx 4.71 \times 10^{-9} \, \text{T} \]

### Step 3: Calculate the average energy density of the magnetic field (U_B).
The average energy density of the magnetic field is given by the formula:
\[ U_B = \frac{1}{2\mu_0} B_{\text{rms}}^2 \]
where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

Substituting the values:
\[ U_B = \frac{1}{2 \times (4\pi \times 10^{-7})} (4.71 \times 10^{-9})^2 \]

Calculating this:
1. Calculate \( (4.71 \times 10^{-9})^2 \):
\[ (4.71 \times 10^{-9})^2 \approx 2.22 \times 10^{-17} \, \text{T}^2 \]
2. Now substitute into the formula:
\[ U_B = \frac{1}{2 \times (4\pi \times 10^{-7})} (2.22 \times 10^{-17}) \]
\[ U_B = \frac{2.22 \times 10^{-17}}{8\pi \times 10^{-7}} \]

Calculating the denominator:
\[ 8\pi \approx 25.13 \]
Thus:
\[ U_B \approx \frac{2.22 \times 10^{-17}}{25.13 \times 10^{-7}} \approx 8.83 \times 10^{-12} \, \text{J/m}^3 \]

### Final Answer:
The average energy density of the magnetic field is approximately \( 8.83 \times 10^{-12} \, \text{J/m}^3 \).

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