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

Options

A. `1.52 xx 10^(-8) Wb m^(-2)`
B. `1.52 xx 10^(-7) Wb m^(-2)`
C. `1.6 xx 10^(-6) Wb m^(-2)`
D. `1.6 xx 10^(-76) Wb m^(-2)`

Answer

To find the amplitude of the oscillating magnetic field in an electromagnetic (EM) wave, we can use the relationship between the electric field (E) and the magnetic field (B) in an EM wave. 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 a vacuum, approximately \( 3 \times 10^8 \) m/s.

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency of the EM wave, \( f = 2.5 \times 10^{10} \) Hz (not directly needed for this calculation).
- 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 between electric and magnetic fields:**
\[ B_0 = \frac{E_0}{c} \]

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

4. **Calculate \( B_0 \):**
\[ B_0 = \frac{480}{3 \times 10^8} = \frac{480}{300000000} = 1.6 \times 10^{-6} \, \text{T} \, (\text{or} \, \text{Wb/m}^2) \]

5. **Final answer:**
The amplitude of the oscillating magnetic field is:
\[ B_0 = 1.60 \times 10^{-6} \, \text{T} \]