A magnetic flux of `5xx10^(-1)wb` is associated with every 10 turns of a 500 turns coil. The electric current flowing through the wire is 5A. The self inductance of coil will be

Options

A. `0.5H`
B. `5xx10^(-3)H`
C. `5H`
D. `5xx10^(-2)H`

Explanation

To find the self-inductance of the coil, we can follow these steps:

### Step 1: Identify the given values
- Magnetic flux (Φ) = \(5 \times 10^{-1} \, \text{Wb} = 0.5 \, \text{Wb}\)
- Total number of turns in the coil (N) = 500 turns
- Current (I) = 5 A

### Step 2: Calculate the effective number of turns
The magnetic flux is associated with every 10 turns of the coil. Therefore, we need to calculate the effective number of turns (N_eff):

\[ N_{\text{eff}} = \frac{N}{10} = \frac{500}{10} = 50 \, \text{turns} \]

### Step 3: Use the formula for self-inductance
The self-inductance (L) of a coil can be calculated using the formula:

\[ L = \frac{N_{\text{eff}} \cdot \Phi}{I} \]

### Step 4: Substitute the values into the formula
Now, substituting the values we have:

\[ L = \frac{50 \cdot 0.5}{5} \]

### Step 5: Simplify the expression
Calculating the numerator:

\[ 50 \cdot 0.5 = 25 \]

Now substituting back into the equation:

\[ L = \frac{25}{5} = 5 \, \text{H} \]

### Final Answer
The self-inductance of the coil is \(5 \, \text{H}\).

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