A coil of `100` turns carries a current of `5mA` and creates a magnetic flux of `10^(-5)`weber. The inductance is

Options

A. `0.2mH`
B. `2.0mH`
C. `0.02mH`
D. none of these

Explanation

To find the inductance \( L \) of the coil, we can use the formula that relates the magnetic flux \( \Phi \), the number of turns \( N \), the current \( I \), and the inductance \( L \):

\[ N \Phi = L I \]

Where:
- \( N \) = number of turns (100 turns)
- \( \Phi \) = magnetic flux (in Weber, \( 10^{-5} \) Wb)
- \( I \) = current (in Amperes, \( 5 \text{ mA} = 5 \times 10^{-3} \text{ A} \))

### Step 1: Write down the known values
- \( N = 100 \)
- \( \Phi = 10^{-5} \) Wb
- \( I = 5 \times 10^{-3} \) A

### Step 2: Substitute the known values into the formula
Substituting the values into the equation \( N \Phi = L I \):

\[ 100 \times 10^{-5} = L \times 5 \times 10^{-3} \]

### Step 3: Simplify the equation
Now we simplify the left side:

\[ 100 \times 10^{-5} = 1 \times 10^{-3} \]

So the equation becomes:

\[ 1 \times 10^{-3} = L \times 5 \times 10^{-3} \]

### Step 4: Solve for \( L \)
To find \( L \), we can rearrange the equation:

\[ L = \frac{1 \times 10^{-3}}{5 \times 10^{-3}} \]

### Step 5: Perform the division
Calculating the right side:

\[ L = \frac{1}{5} \times \frac{10^{-3}}{10^{-3}} = \frac{1}{5} = 0.2 \]

### Step 6: Convert to millihenry
Since \( 0.2 \) Henry can also be expressed in millihenry:

\[ L = 0.2 \text{ H} = 200 \text{ mH} \]

### Final Answer
Thus, the inductance \( L \) of the coil is:

\[ L = 200 \text{ mH} \]