A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is `4.5 xx 10^(-2) Am^(-1)` at a point, what will be the approximate magnitude of electric field intensity at that point ? (Given : Permeability of free space `mu_0=4pi xx 10^(-7) NA^(-2)` , speed of light in vacuum `c=3xx 10^8 ms^(-1)` )

Options

A. `16.96 Vm^(-1)`
B. `2.25 xx10^(-2) Vm^(-1)`
C. `8.48 Vm^(-1)`
D. `6.75xx 10^6 Vm^(-1)`

Answer

To find the magnitude of the electric field intensity (E) at a point where a plane electromagnetic wave travels in a medium with given relative permeability (μ_r) and relative permittivity (ε_r), we can use the relationship between the electric field intensity (E) and the magnetic field intensity (H).

### Step-by-Step Solution:

1. **Identify Given Values**:
- Relative permeability, \( \mu_r = 1.61 \)
- Relative permittivity, \( \epsilon_r = 6.44 \)
- Magnetic field intensity, \( H = 4.5 \times 10^{-2} \, \text{A/m} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{NA}^{-2} \)
- Speed of light in vacuum, \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Calculate the Speed of the Electromagnetic Wave in the Medium**:
The speed of the wave \( v \) in a medium can be calculated using the formula:
\[ v = \frac{c}{\sqrt{\mu_r \epsilon_r}} \]

3. **Calculate \( \mu \) and \( \epsilon \)**:
- \( \mu = \mu_0 \times \mu_r \)
- \( \epsilon = \epsilon_0 \times \epsilon_r \)
- Where \( \epsilon_0 = \frac{1}{\mu_0 c^2} \)

4. **Calculate the Electric Field Intensity**:
The relationship between the electric field intensity \( E \) and the magnetic field intensity \( H \) is given by:
\[ E = v \times H \]
Substituting the expression for \( v \):
\[ E = \left( \frac{c}{\sqrt{\mu_r \epsilon_r}} \right) \times H \]

5. **Substituting Values**:
Now we can substitute the known values into the equation:
\[ E = \left( \frac{3 \times 10^8}{\sqrt{1.61 \times 6.44}} \right) \times (4.5 \times 10^{-2}) \]

6. **Calculate \( \sqrt{\mu_r \epsilon_r} \)**:
\[ \sqrt{1.61 \times 6.44} \approx \sqrt{10.36} \approx 3.22 \]

7. **Calculate \( E \)**:
\[ E = \left( \frac{3 \times 10^8}{3.22} \right) \times (4.5 \times 10^{-2}) \]
\[ E \approx (9.32 \times 10^7) \times (4.5 \times 10^{-2}) \approx 4.19 \times 10^6 \, \text{V/m} \]

8. **Final Result**:
Therefore, the approximate magnitude of the electric field intensity at that point is:
\[ E \approx 8.48 \, \text{V/m} \]