The flux linked with a coil of self inductance `2H`, when there is a current of `5.8A` flowing through it is

Options

A. `11.6 Wb`
B. `2.9 Wb`
C. `8.7 Wb`
D. independent of orientation of coil.

Answer

To solve the problem, we need to find the magnetic flux linked with a coil given its self-inductance and the current flowing through it. The formula we will use is:

\[ \Phi = L \cdot I \]

where:
- \(\Phi\) is the magnetic flux linked with the coil,
- \(L\) is the self-inductance of the coil,
- \(I\) is the current flowing through the coil.

### Step-by-Step Solution:

1. **Identify the given values**:
- Self-inductance \(L = 2 \, \text{H}\) (Henry)
- Current \(I = 5.8 \, \text{A}\) (Amperes)

2. **Apply the formula for magnetic flux**:
\[ \Phi = L \cdot I \]

3. **Substitute the values into the formula**:
\[ \Phi = 2 \, \text{H} \cdot 5.8 \, \text{A} \]

4. **Calculate the magnetic flux**:
\[ \Phi = 2 \cdot 5.8 = 11.6 \, \text{Wb} \]

5. **Conclusion**:
The flux linked with the coil is \(11.6 \, \text{Wb}\) (Webers).