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`

Explanation

To solve the problem of finding the magnitude of the oscillating magnetic field given the amplitude of the electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Electric and Magnetic Fields**:
In an electromagnetic wave, the ratio of the amplitude of the electric field (E₀) to the amplitude of the magnetic field (B₀) is constant and is equal to the speed of light (c). This can be expressed mathematically as:
\[ \frac{E_0}{B_0} = c \]
where \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

2. **Rearrange the Formula to Find B₀**:
From the above relationship, we can rearrange the formula to find the amplitude of the magnetic field:
\[ B_0 = \frac{E_0}{c} \]

3. **Substitute the Given Values**:
We know that the amplitude of the electric field \( E_0 \) is given as \( 18 \, \text{V/m} \). Now, substituting the values into the equation:
\[ B_0 = \frac{18 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]

4. **Calculate B₀**:
Performing the calculation:
\[ B_0 = \frac{18}{3 \times 10^8} = 6 \times 10^{-8} \, \text{T} \]

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

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

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