What will be the self inductance of a coil of 100 turns, if current of 2 ampere gives rise to a magnetic flux of `5 xx 10 ^(-5)` Weber through the coil :

Options

A. 2.5 Henry
B. 2.5 milli Henry
C. `2.5 xx 10 ^(-4)` Henry
D. `0.25` Henry

Explanation

To find the self-inductance \( L \) of the coil, we can use the formula that relates self-inductance, magnetic flux, and current:

\[ L = \frac{\Phi}{I} \]

where:
- \( L \) is the self-inductance in henries (H),
- \( \Phi \) is the total magnetic flux through the coil in webers (Wb),
- \( I \) is the current in amperes (A).

### Step 1: Identify the given values
From the problem statement, we have:
- Number of turns \( N = 100 \) (though this is not directly needed for calculating self-inductance),
- Current \( I = 2 \, \text{A} \),
- Magnetic flux \( \Phi = 5 \times 10^{-5} \, \text{Wb} \).

### Step 2: Substitute the values into the formula
Now, we can substitute the values of \( \Phi \) and \( I \) into the formula for self-inductance:

\[ L = \frac{5 \times 10^{-5} \, \text{Wb}}{2 \, \text{A}} \]

### Step 3: Perform the calculation
Calculating the above expression:

\[ L = \frac{5 \times 10^{-5}}{2} = 2.5 \times 10^{-5} \, \text{H} \]

### Step 4: Convert to a more standard unit
Since \( 1 \, \text{H} = 10^6 \, \mu\text{H} \), we can convert:

\[ L = 2.5 \times 10^{-5} \, \text{H} = 25 \, \mu\text{H} \]

### Final Answer
Thus, the self-inductance of the coil is:

\[ L = 25 \, \mu\text{H} \]

---