Magnetic flux of 20 μWb is linked with a coil when current of 5 mA is flown through it. What is the self-inductance of the coil (in `xx 10^(-3)` H) ?

Answer

To find the self-inductance of the coil, we can use the relationship between magnetic flux (Φ), current (I), and self-inductance (L). The formula we will use is:

\[ \Phi = L \cdot I \]

Where:
- \(\Phi\) is the magnetic flux in webers (Wb)
- \(L\) is the self-inductance in henries (H)
- \(I\) is the current in amperes (A)

### Step 1: Convert the given values to SI units
- The magnetic flux is given as \(20 \, \mu Wb\).
\[ \Phi = 20 \, \mu Wb = 20 \times 10^{-6} \, Wb = 2 \times 10^{-5} \, Wb \]

- The current is given as \(5 \, mA\).
\[ I = 5 \, mA = 5 \times 10^{-3} \, A \]

### Step 2: Substitute the values into the formula
Now, we can substitute the values of \(\Phi\) and \(I\) into the formula to find \(L\):
\[ \Phi = L \cdot I \implies L = \frac{\Phi}{I} \]
Substituting the values we have:
\[ L = \frac{2 \times 10^{-5}}{5 \times 10^{-3}} \]

### Step 3: Simplify the expression
Now, we simplify the expression:
\[ L = \frac{2}{5} \times \frac{10^{-5}}{10^{-3}} = \frac{2}{5} \times 10^{-2} \]

### Step 4: Convert to milliHenries
To express \(L\) in milliHenries (mH), we multiply by \(10^3\):
\[ L = \frac{2}{5} \times 10^{-2} \times 10^3 = \frac{2 \times 10^1}{5} = \frac{20}{5} = 4 \, mH \]

### Final Answer
Thus, the self-inductance of the coil is:
\[ L = 4 \times 10^{-3} \, H \]