A solenoid having 100 turns with an area of cross section `10 cm^(2)` is 12 cm long. When iron core is introduced in it, the self-inductance increases to 1000 times. Calculate the relative permeability of the core.

Answer

To solve the problem, we need to calculate the relative permeability of the iron core introduced into the solenoid. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Number of turns (N) = 100
- Area of cross-section (A) = 10 cm² = \(10 \times 10^{-4} \, m^2\) (converting to square meters)
- Length of the solenoid (L) = 12 cm = 0.12 m
- The self-inductance increases to 1000 times when the iron core is introduced.

### Step 2: Write the formula for self-inductance
The self-inductance \(L\) of a solenoid is given by the formula:
\[ L = \frac{\mu_0 N^2 A}{L} \]
Where:
- \(\mu_0\) is the permeability of free space.
- \(N\) is the number of turns.
- \(A\) is the cross-sectional area.
- \(L\) is the length of the solenoid.

### Step 3: Calculate the initial self-inductance \(L_1\)
For the initial solenoid (without the iron core), we have:
\[ L_1 = \frac{\mu_0 N^2 A}{L} \]

### Step 4: Write the formula for self-inductance with the iron core
When the iron core is introduced, the self-inductance becomes \(L_2\):
\[ L_2 = \frac{\mu_0 \mu_r N^2 A}{L} \]
Where \(\mu_r\) is the relative permeability of the iron core.

### Step 5: Relate \(L_2\) to \(L_1\)
According to the problem, \(L_2 = 1000 L_1\). Therefore, we can write:
\[ \frac{\mu_0 \mu_r N^2 A}{L} = 1000 \left(\frac{\mu_0 N^2 A}{L}\right) \]

### Step 6: Simplify the equation
Dividing both sides by \(\frac{\mu_0 N^2 A}{L}\) (assuming it is not zero), we get:
\[ \mu_r = 1000 \]

### Step 7: Conclusion
The relative permeability of the iron core is:
\[ \mu_r = 1000 \]