Find the inductance of a coil of 100 turns carrying 5 mA producing a magnetic flux of `10^(-5)` Wb:

Options

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

Explanation

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

\[ \Phi = L \cdot I \]

Where:
- \(\Phi\) is the magnetic flux in Webers (Wb)
- \(L\) is the inductance in Henrys (H)
- \(I\) is the current in Amperes (A)

Given:
- Number of turns, \(N = 100\)
- Current, \(I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A}\)
- Magnetic flux, \(\Phi = 10^{-5} \, \text{Wb}\)

### Step 1: Write the formula for inductance
From the relationship between magnetic flux and inductance, we can rearrange the formula to solve for \(L\):

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

### Step 2: Substitute the values into the formula
Now we can substitute the values of \(\Phi\) and \(I\) into the equation:

\[ L = \frac{10^{-5} \, \text{Wb}}{5 \times 10^{-3} \, \text{A}} \]

### Step 3: Calculate the inductance
Now perform the calculation:

\[ L = \frac{10^{-5}}{5 \times 10^{-3}} = \frac{10^{-5}}{5 \times 10^{-3}} = \frac{10^{-5}}{5 \times 10^{-3}} = \frac{10^{-5}}{5} \times 10^{3} = 2 \times 10^{-3} = 0.2 \, \text{H} \]

### Final Result
Thus, the inductance of the coil is:

\[ L = 0.2 \, \text{H} \]