Solenoid `S_1` has N turns, radius `R_1` and length l. It is so long that its magnetic field is uniform nearly everywhere inside it and is nearly zero outside, Solenoid `S_2` has `N_2` turns, radius `R_2ltR_1`.and the same length as `S_1` It lies inside `S_1` with their axes prallel. (a) Assume `S_1` carries variable current i. Compute the mutual inductance characterizing the emf induced is `S_2`. (b) Now assume `S_2` carries current i. Compute the mutual inductance to which the emf in `S_1` is proportional. (c) State how we results of parts (a) and (b) compare with each other.

Answer

To solve the problem step by step, we will break down each part of the question regarding the mutual inductance between the two solenoids, \( S_1 \) and \( S_2 \).

### Part (a): Mutual Inductance Characterizing the EMF Induced in \( S_2 \)

1. **Identify the Magnetic Field in \( S_1 \)**:
The magnetic field inside a long solenoid \( S_1 \) is given by:
\[ B_1 = \mu_0 \frac{N_1}{L} I \]
where \( \mu_0 \) is the permeability of free space, \( N_1 \) is the number of turns in \( S_1 \), \( L \) is the length of the solenoid, and \( I \) is the current flowing through \( S_1 \).

2. **Calculate the Magnetic Flux through \( S_2 \)**:
The magnetic flux \( \Phi \) through one turn of solenoid \( S_2 \) is:
\[ \Phi = B_1 \cdot A = B_1 \cdot \pi R_2^2 \]
where \( A = \pi R_2^2 \) is the cross-sectional area of \( S_2 \).

Substituting for \( B_1 \):
\[ \Phi = \left( \mu_0 \frac{N_1}{L} I \right) \cdot \pi R_2^2 \]

3. **Total Flux Linked with \( S_2 \)**:
The total flux linked with \( S_2 \) (which has \( N_2 \) turns) is:
\[ \Phi_{\text{total}} = N_2 \Phi = N_2 \left( \mu_0 \frac{N_1}{L} I \cdot \pi R_2^2 \right) \]

4. **Calculate the Induced EMF in \( S_2 \)**:
The induced EMF \( E_2 \) in \( S_2 \) is given by Faraday's law:
\[ E_2 = -\frac{d\Phi_{\text{total}}}{dt} = -N_2 \left( \mu_0 \frac{N_1}{L} \pi R_2^2 \right) \frac{dI}{dt} \]

5. **Define the Mutual Inductance \( M_{12} \)**:
The mutual inductance \( M_{12} \) is defined as:
\[ M_{12} = -\frac{E_2}{\frac{dI}{dt}} = \mu_0 \frac{N_1 N_2 \pi R_2^2}{L} \]

### Part (b): Mutual Inductance Characterizing the EMF Induced in \( S_1 \)

1. **Identify the Magnetic Field in \( S_2 \)**:
The magnetic field inside solenoid \( S_2 \) when it carries current \( I_2 \) is:
\[ B_2 = \mu_0 \frac{N_2}{L} I_2 \]

2. **Calculate the Magnetic Flux through \( S_1 \)**:
The magnetic flux \( \Phi_1 \) through one turn of solenoid \( S_1 \) is:
\[ \Phi_1 = B_2 \cdot A_1 = B_2 \cdot \pi R_1^2 \]
where \( A_1 = \pi R_1^2 \).

Substituting for \( B_2 \):
\[ \Phi_1 = \left( \mu_0 \frac{N_2}{L} I_2 \right) \cdot \pi R_1^2 \]

3. **Total Flux Linked with \( S_1 \)**:
The total flux linked with \( S_1 \) (which has \( N_1 \) turns) is:
\[ \Phi_{\text{total}} = N_1 \Phi_1 = N_1 \left( \mu_0 \frac{N_2}{L} I_2 \cdot \pi R_1^2 \right) \]

4. **Calculate the Induced EMF in \( S_1 \)**:
The induced EMF \( E_1 \) in \( S_1 \) is given by:
\[ E_1 = -\frac{d\Phi_{\text{total}}}{dt} = -N_1 \left( \mu_0 \frac{N_2}{L} \pi R_1^2 \right) \frac{dI_2}{dt} \]

5. **Define the Mutual Inductance \( M_{21} \)**:
The mutual inductance \( M_{21} \) is defined as:
\[ M_{21} = -\frac{E_1}{\frac{dI_2}{dt}} = \mu_0 \frac{N_1 N_2 \pi R_1^2}{L} \]

### Part (c): Comparison of Results from Parts (a) and (b)

- From part (a), we found:
\[ M_{12} = \mu_0 \frac{N_1 N_2 \pi R_2^2}{L} \]
- From part (b), we found:
\[ M_{21} = \mu_0 \frac{N_1 N_2 \pi R_1^2}{L} \]

- **Comparison**:
The mutual inductance \( M_{12} \) and \( M_{21} \) are similar in form but differ in the radius term:
\[ M_{12} \propto R_2^2 \quad \text{and} \quad M_{21} \propto R_1^2 \]
Since \( R_2 < R_1 \), it follows that \( M_{12} < M_{21} \).