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 setup
We have a long solenoid with:
- Length \( L \)
- Radius \( a \)
- Number of turns per unit length \( n \)
- 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 a solenoid is given by:
\[ B = \mu n I \]
where \( I \) is the current flowing through the solenoid. We will consider the two regions separately.

### Step 3: Calculate the magnetic field in each region
1. For the region \( 0 < r < b \):
\[ B_1 = \mu_1 n I \]
2. For the region \( b < r < a \):
\[ B_2 = \mu_2 n I \]

### Step 4: Calculate the flux through each region
The magnetic flux \( \Phi \) through one turn of the solenoid can be calculated as:
\[ \Phi = B \cdot A \]
where \( A \) is the area.

1. For the region \( 0 < r < b \):
- Area \( A_1 = \pi b^2 \)
- Flux \( \Phi_1 = B_1 \cdot A_1 = \mu_1 n I \cdot \pi b^2 \)

2. For the region \( b < r < a \):
- Area \( A_2 = \pi (a^2 - b^2) \)
- Flux \( \Phi_2 = B_2 \cdot A_2 = \mu_2 n I \cdot \pi (a^2 - b^2) \)

### Step 5: Total flux for all turns
The total flux \( \Phi_{total} \) through the solenoid considering all \( N \) turns (where \( N = nL \)):
\[ \Phi_{total} = N \cdot (\Phi_1 + \Phi_2) = nL \left( \mu_1 n I \cdot \pi b^2 + \mu_2 n I \cdot \pi (a^2 - b^2) \right) \]

### Step 6: Simplify the expression
\[ \Phi_{total} = nL \cdot nI \cdot \pi \left( \mu_1 b^2 + \mu_2 (a^2 - b^2) \right) \]

### Step 7: Relate total flux to self-inductance
The self-inductance \( L_s \) is defined as:
\[ L_s = \frac{\Phi_{total}}{I} \]
Substituting the expression for \( \Phi_{total} \):
\[ L_s = n^2 \pi L \left( \mu_1 b^2 + \mu_2 (a^2 - b^2) \right) \]

### Final Expression for Self-Inductance
Thus, the self-inductance of the solenoid is:
\[ L_s = n^2 \pi L \left( \mu_1 b^2 + \mu_2 (a^2 - b^2) \right) \]