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

Options

A. `4 xx 10^(-6) T`
B. `6 xx 10^(-8) T`
C. `9 xx 10^(-9)` T
D. `11 xx 10^(-11)` T

Answer

To find the magnitude of the oscillating magnetic field when the electric field oscillates with an amplitude of 18 V/m, we can use the relationship between the electric field (E) and the magnetic field (B) in electromagnetic waves. The speed of electromagnetic waves in free space (c) is approximately \(3 \times 10^8\) m/s.

### Step-by-Step Solution:

1. **Identify the relationship between electric field and magnetic field**:
The relationship between the electric field (E) and the magnetic field (B) in electromagnetic waves is given by the equation:
\[ c = \frac{E}{B} \]
where \(c\) is the speed of light in a vacuum, \(E\) is the electric field strength, and \(B\) is the magnetic field strength.

2. **Rearrange the equation to find B**:
We can rearrange the equation to solve for the magnetic field (B):
\[ B = \frac{E}{c} \]

3. **Substitute the known values**:
We know that:
- The amplitude of the electric field, \(E = 18 \, \text{V/m}\)
- The speed of light, \(c = 3 \times 10^8 \, \text{m/s}\)

Substituting these values into the equation:
\[ B = \frac{18 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]

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

5. **Final answer**:
Therefore, the magnitude of the oscillating magnetic field is:
\[ B = 6 \times 10^{-8} \, \text{T} \]