The magnetic field of a plane electromagnetic wave is given by: `vec(B)=B_(0)hat(i)[cos(kz- omegat)]+B_(1)hat(j)cos(kz+omegat)` where `B_(0)=3xx10^(-5)T` and `B_(1)=2xx10^(-6)T`. The rms value of the force experienced by a stationary charge `Q=10^(-4)C` at `z=0` is close to:

Options

A. 0.1 N
B. `3xx10^(-2)N`
C. 0.6 N
D. 0.9 N

Explanation

To solve the problem, we need to find the RMS value of the force experienced by a stationary charge \( Q = 10^{-4} \, \text{C} \) at \( z = 0 \) due to the given magnetic field of the electromagnetic wave.

### Step-by-Step Solution:

1. **Identify the Magnetic Field**:
The magnetic field is given by:
\[ \vec{B} = B_0 \hat{i} \cos(kz - \omega t) + B_1 \hat{j} \cos(kz + \omega t) \]
where \( B_0 = 3 \times 10^{-5} \, \text{T} \) and \( B_1 = 2 \times 10^{-6} \, \text{T} \).

2. **Calculate the Electric Field**:
The electric field \( \vec{E} \) in an electromagnetic wave can be calculated using the relation:
\[ \vec{E} = c \vec{B} \]
where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

Thus, we can calculate \( \vec{E} \):
\[ \vec{E} = 3 \times 10^8 \left( B_0 \hat{i} \cos(kz - \omega t) + B_1 \hat{j} \cos(kz + \omega t) \right) \]
Substituting the values of \( B_0 \) and \( B_1 \):
\[ \vec{E} = 3 \times 10^8 \left( 3 \times 10^{-5} \hat{i} \cos(kz - \omega t) + 2 \times 10^{-6} \hat{j} \cos(kz + \omega t) \right) \]

3. **Evaluate \( \vec{E} \) at \( z = 0 \)**:
At \( z = 0 \):
\[ \vec{E} = 3 \times 10^8 \left( 3 \times 10^{-5} \hat{i} \cos(-\omega t) + 2 \times 10^{-6} \hat{j} \cos(\omega t) \right) \]
This simplifies to:
\[ \vec{E} = 3 \times 10^8 \left( 3 \times 10^{-5} \hat{i} \cos(\omega t) + 2 \times 10^{-6} \hat{j} \cos(\omega t) \right) \]

4. **Calculate the Force on the Charge**:
The force \( \vec{F} \) on a charge \( Q \) in an electric field \( \vec{E} \) is given by:
\[ \vec{F} = Q \vec{E} \]
Substituting \( Q = 10^{-4} \, \text{C} \):
\[ \vec{F} = 10^{-4} \cdot \vec{E} \]

5. **Magnitude of the Force**:
The magnitude of the electric field \( \vec{E} \) can be calculated:
\[ |\vec{E}| = 3 \times 10^8 \sqrt{(3 \times 10^{-5})^2 + (2 \times 10^{-6})^2} \]
Calculate the components:
\[ |\vec{E}| = 3 \times 10^8 \sqrt{(9 \times 10^{-10}) + (4 \times 10^{-12})} = 3 \times 10^8 \sqrt{9.04 \times 10^{-10}} \]

6. **Calculate the RMS Force**:
The RMS value of the force is given by:
\[ F_{\text{RMS}} = \frac{F}{\sqrt{2}} \]
where \( F \) is the average force calculated from the magnitude of \( \vec{F} \).

7. **Final Calculation**:
After calculating the magnitude of \( \vec{F} \) and applying the RMS formula, we can find the final answer.

### Final Answer:
The RMS value of the force experienced by the charge is approximately \( 0.6 \, \text{N} \).