The self inductane of coil of 400 turns is 8 mH. If current of 5 mA flows in it, then flux associated with the coil is

Options

A. `(mu_(0)//4pi)`
B. `mu_(0)`
C. `(100mu_(0))/(pi)`
D. `(4pi//mu_(0))`

Explanation

To find the magnetic flux associated with a coil given its self-inductance, the number of turns, and the current flowing through it, we can use the formula:

\[ \Phi = L \cdot I \]

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns, \(N = 400\) turns (not directly needed for this calculation).
- Self-inductance, \(L = 8 \, \text{mH} = 8 \times 10^{-3} \, \text{H}\).
- Current, \(I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A}\).

2. **Substitute the values into the formula:**
\[ \Phi = L \cdot I = (8 \times 10^{-3} \, \text{H}) \cdot (5 \times 10^{-3} \, \text{A}) \]

3. **Calculate the product:**
\[ \Phi = 8 \times 5 \times 10^{-3} \times 10^{-3} = 40 \times 10^{-6} \, \text{Wb} \]

4. **Convert to standard form:**
\[ \Phi = 4.0 \times 10^{-5} \, \text{Wb} \]

### Final Answer:
The magnetic flux associated with the coil is \(4.0 \times 10^{-5} \, \text{Wb}\).

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