In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.5xx10^10Hz` and amplitued `480V//m`. The amplitude of oscillating magnetic field will be

Options

A. `1.52xx10^-8Wb//m^2`
B. `1.52xx10^-7Wb//m^2`
C. `1.6xx10^-6Wb//m^2`
D. `1.6xx10^-7Wb//m^2`

Answer

To find the amplitude of the oscillating magnetic field in a plane electromagnetic wave, we can use the relationship between the electric field and the magnetic field in electromagnetic waves. The relationship is given by:

\[ \frac{E_0}{B_0} = c \]

where:
- \(E_0\) is the amplitude of the electric field,
- \(B_0\) is the amplitude of the magnetic field,
- \(c\) is the speed of light in vacuum, approximately \(3 \times 10^8 \, \text{m/s}\).

### Step-by-Step Solution

1. **Identify the given values:**
- Amplitude of the electric field, \(E_0 = 480 \, \text{V/m}\)
- Speed of light, \(c = 3 \times 10^8 \, \text{m/s}\)

2. **Use the relationship to find \(B_0\):**
\[ B_0 = \frac{E_0}{c} \]

3. **Substitute the known values into the equation:**
\[ B_0 = \frac{480 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]

4. **Calculate \(B_0\):**
\[ B_0 = \frac{480}{3 \times 10^8} \]
\[ B_0 = \frac{480}{300000000} \]
\[ B_0 = 1.6 \times 10^{-6} \, \text{T} \]

5. **Final Result:**
The amplitude of the oscillating magnetic field is:
\[ B_0 = 1.6 \, \mu\text{T} \, \text{(microtesla)} \]