A very long solenoid is made out of a wire with n turns per units length. The radius of the cylinder is a and is negligible compared to its length l. The interior of the cylinder is filled with materials such that the linear magnetic permeability varies with the distance r from axis according to `mu(r)={{:(mu_(1)="constant , for "0ltrlt b),(mu_(2)="constant , for "b lt r lt a):}}`
The self - inductance of the solenoid is

Options

A. `pin^(2)l[mu_(1)b^(2)+mu_(2)b^(2)]`
B. `pin^(2)l[mu_(1)+mu_(2)]a^(2)`
C. `pin^(2)l[mu_(1)b^(2)+mu_(2)(a^(2)-b^(2))]`
D. `pin^(2)l[mu_(1)b^(2)+(mu_(1)+mu_(2))a^(2)]`

Answer

To find the self-inductance of the solenoid with varying magnetic permeability, we can follow these steps:

### Step 1: Understand the Configuration
We have a long solenoid with:
- Length \( l \)
- Number of turns per unit length \( n \)
- Radius \( a \)
- Two regions of magnetic permeability:
- \( \mu_1 \) for \( 0 < r < b \)
- \( \mu_2 \) for \( b < r < a \)

### Step 2: Magnetic Field Inside the Solenoid
The magnetic field \( B \) inside the solenoid can be expressed as:
\[ B = \mu H \]
where \( H \) is the magnetic field intensity. For a solenoid, \( H \) is given by:
\[ H = nI \]
Thus, the magnetic field \( B \) can be expressed as:
\[ B = \mu_1 n I \quad \text{for } 0 < r < b \]
\[ B = \mu_2 n I \quad \text{for } b < r < a \]

### Step 3: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the solenoid can be calculated by integrating the magnetic field over the cross-sectional area.

1. **For the inner region (0 to b)**:
\[ \Phi_1 = \int_0^b B \cdot dA = \int_0^b (\mu_1 n I) \cdot (\pi r^2) \, dr \]
\[ \Phi_1 = \mu_1 n I \cdot \pi \int_0^b r^2 \, dr = \mu_1 n I \cdot \pi \left[ \frac{r^3}{3} \right]_0^b = \mu_1 n I \cdot \pi \cdot \frac{b^3}{3} \]

2. **For the outer region (b to a)**:
\[ \Phi_2 = \int_b^a B \cdot dA = \int_b^a (\mu_2 n I) \cdot (\pi r^2) \, dr \]
\[ \Phi_2 = \mu_2 n I \cdot \pi \int_b^a r^2 \, dr = \mu_2 n I \cdot \pi \left[ \frac{r^3}{3} \right]_b^a = \mu_2 n I \cdot \pi \left( \frac{a^3}{3} - \frac{b^3}{3} \right) \]

### Step 4: Total Magnetic Flux
The total magnetic flux \( \Phi \) through the solenoid is:
\[ \Phi = \Phi_1 + \Phi_2 = \mu_1 n I \cdot \pi \cdot \frac{b^3}{3} + \mu_2 n I \cdot \pi \left( \frac{a^3}{3} - \frac{b^3}{3} \right) \]

### Step 5: Self-Inductance Calculation
The self-inductance \( L \) is given by:
\[ L = \frac{N \Phi}{I} \]
where \( N = n \cdot l \) is the total number of turns in the solenoid. Substituting for \( \Phi \):
\[ L = n l \cdot \left( \frac{\mu_1 n I \cdot \pi \cdot \frac{b^3}{3} + \mu_2 n I \cdot \pi \left( \frac{a^3}{3} - \frac{b^3}{3} \right)}{I} \right) \]
This simplifies to:
\[ L = n l \cdot \left( \mu_1 \cdot \frac{\pi b^3}{3} + \mu_2 \left( \frac{\pi a^3}{3} - \frac{\pi b^3}{3} \right) \right) \]
\[ L = \frac{n l \pi}{3} \left( \mu_1 b^3 + \mu_2 (a^3 - b^3) \right) \]

### Final Expression for Self-Inductance
Thus, the self-inductance \( L \) of the solenoid is:
\[ L = \frac{n l \pi}{3} \left( \mu_1 b^3 + \mu_2 (a^3 - b^3) \right) \]