An electromagnetic wave is propagating along x-axis. At `x=1m` and `t=10s`, its electric vector `vecE = 3.6 V//m`. then, the magnitude of its magnetic vector is:

Options

A. `3 xx 10^(-8) T`
B. `1.2 xx 10^(-8) T`
C. `2 xx 10^(-8) T`
D. `2.8 xx 10^(-8) T`

Explanation

To find the magnitude of the magnetic vector \( \vec{B} \) of an electromagnetic wave given the electric vector \( \vec{E} \), we can use the relationship between the electric field and the magnetic field in an electromagnetic wave. The relationship is given by:

\[ |\vec{E}| = c |\vec{B}| \]

where:
- \( |\vec{E}| \) is the magnitude of the electric field,
- \( |\vec{B}| \) is the magnitude 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**:
- The electric field \( |\vec{E}| = 3.6 \, \text{V/m} \).
- The speed of light \( c = 3 \times 10^8 \, \text{m/s} \).

2. **Rearrange the formula to solve for \( |\vec{B}| \)**:
\[ |\vec{B}| = \frac{|\vec{E}|}{c} \]

3. **Substitute the known values into the equation**:
\[ |\vec{B}| = \frac{3.6 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]

4. **Perform the calculation**:
\[ |\vec{B}| = \frac{3.6}{3 \times 10^8} = 1.2 \times 10^{-8} \, \text{T} \]

5. **Conclusion**:
The magnitude of the magnetic vector \( |\vec{B}| \) is \( 1.2 \times 10^{-8} \, \text{T} \).

### Final Answer:
The magnitude of the magnetic vector \( \vec{B} \) is \( 1.2 \times 10^{-8} \, \text{T} \).