In an apparatus the electric field was found to oscillate with an amplitude of `18` `V//M`. The magnitude of the oscillating magnetic field will be

Options

A. `4xx10^(-6)T`
B. `6xx10^(-8)T`
C. `9xx10^(-9)T`
D. `11xx10^(-11)T`

Explanation

To find the magnitude of the oscillating magnetic field (B) given the amplitude of the electric field (E), we can use the relationship between the electric field and the magnetic field in electromagnetic waves. The formula that relates them is:

\[ C = \frac{E}{B} \]

Where:
- \( C \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s.
- \( E \) is the electric field amplitude.
- \( B \) is the magnetic field amplitude.

### Step-by-Step Solution:

1. **Identify the given values:**
- The amplitude of the electric field \( E = 18 \, \text{V/m} \).

2. **Use the formula to express B in terms of E and C:**
\[ B = \frac{E}{C} \]

3. **Substitute the known values into the equation:**
- We know \( C = 3 \times 10^8 \, \text{m/s} \).
\[ B = \frac{18 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]

4. **Calculate B:**
\[ B = \frac{18}{3 \times 10^8} = \frac{18}{3} \times 10^{-8} = 6 \times 10^{-8} \, \text{T} \]

5. **Final Result:**
- The magnitude of the oscillating magnetic field \( B \) is \( 6 \times 10^{-8} \, \text{T} \).