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

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Determine the Magnetic Field at z = 0

The magnetic field is given by:
\[ \vec{B} = B_0 \hat{i} \cos(kz - \omega t) + B_1 \hat{j} \cos(kz + \omega t) \]
At \( z = 0 \):
\[ \vec{B} = B_0 \hat{i} \cos(-\omega t) + B_1 \hat{j} \cos(\omega t) \]
Using the property \( \cos(-\theta) = \cos(\theta) \):
\[ \vec{B} = B_0 \hat{i} \cos(\omega t) + B_1 \hat{j} \cos(\omega t) \]
Substituting the values \( B_0 = 3 \times 10^{-5} \, \text{T} \) and \( B_1 = 2 \times 10^{-6} \, \text{T} \):
\[ \vec{B} = (3 \times 10^{-5} \hat{i} + 2 \times 10^{-6} \hat{j}) \cos(\omega t) \]

### Step 2: Calculate the Peak Value of the Magnetic Field

The peak value of the magnetic field \( B_{\text{peak}} \) can be calculated as:
\[ B_{\text{peak}} = \sqrt{B_0^2 + B_1^2} \]
Calculating:
\[ B_{\text{peak}} = \sqrt{(3 \times 10^{-5})^2 + (2 \times 10^{-6})^2} \]
\[ = \sqrt{9 \times 10^{-10} + 4 \times 10^{-12}} = \sqrt{9.04 \times 10^{-10}} = 3.01 \times 10^{-5} \, \text{T} \]

### Step 3: Calculate the Peak Value of the Electric Field

The relationship between the electric field \( E \) and the magnetic field \( B \) in an electromagnetic wave is given by:
\[ E = cB \]
where \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \). Therefore:
\[ E_{\text{peak}} = c \cdot B_{\text{peak}} = 3 \times 10^8 \cdot 3.01 \times 10^{-5} \]
Calculating:
\[ E_{\text{peak}} \approx 9.03 \times 10^3 \, \text{V/m} \]

### Step 4: Calculate the RMS Value of the Electric Field

The RMS value of the electric field is given by:
\[ E_{\text{rms}} = \frac{E_{\text{peak}}}{\sqrt{2}} = \frac{9.03 \times 10^3}{\sqrt{2}} \approx 6.38 \times 10^3 \, \text{V/m} \]

### Step 5: Calculate the RMS Force on the Charge

The force on a stationary charge \( Q \) in an electric field \( E \) is given by:
\[ F = Q \cdot E \]
Substituting \( Q = 10^{-4} \, \text{C} \):
\[ F = 10^{-4} \cdot 6.38 \times 10^3 \]
Calculating:
\[ F \approx 0.638 \, \text{N} \]

### Conclusion

The RMS value of the force experienced by the stationary charge is approximately:
\[ \boxed{0.6 \, \text{N}} \]