Question

There are two long co-axial solenoids of the same length \( l \). The inner and outer coils have radii \( r_1 \) and \( r_2 \) and the number of turns per unit length \( n_1 \) and \( n_2 \), respectively. The ratio of mutual inductance to the self-inductance of the inner coil is:

• A. \( \frac{n_1}{n_2} \)
• B. \( \frac{n_2}{n_1} \cdot \frac{r_1}{r_2} \)
• C. \( \frac{n_2}{n_1} \cdot \frac{r_2^2}{r_1^2} \)
• D. \( \frac{n_2}{n_1} \)

Hint:

The mutual inductance depends on the turns per unit length of both solenoids, while self-inductance depends only on the inner solenoid.

Answer 1

The Correct Option is D

Step 1: {Expression for Mutual Inductance}
The mutual inductance \( M \) between the two co-axial solenoids is given by: \[ M = \mu_0 n_1 n_2 \pi r_1^2 l \] Step 2: {Expression for Self-Inductance of the Inner Coil}
The self-inductance \( L \) of the inner solenoid is given by: \[ L = \mu_0 n_1^2 \pi r_1^2 l \] Step 3: {Finding the Ratio \( \frac{M}{L} \)}
\[ \frac{M}{L} = \frac{\mu_0 n_1 n_2 \pi r_1^2 l}{\mu_0 n_1^2 \pi r_1^2 l} = \frac{n_2}{n_1} \] Thus, the correct answer is \( \frac{n_2}{n_1} \).

Answer 2

Step 1: Understand mutual and self-inductance
- Mutual inductance \( M \): Induced emf in one coil due to a change of current in another.
- Self-inductance \( L \): Induced emf in the same coil due to its own changing current.

Step 2: Use the formula for mutual inductance between co-axial solenoids
When the inner solenoid is completely inside the outer solenoid, mutual inductance is:
\[ M = \mu_0 n_1 n_2 A_1 l \]
Where:
- \( \mu_0 \) is the permeability of free space
- \( n_1 \), \( n_2 \) are number of turns per unit length of inner and outer coils
- \( A_1 = \pi r_1^2 \) is the cross-sectional area of the inner coil
- \( l \) is the length of the solenoid

Step 3: Use the formula for self-inductance of the inner solenoid
\[ L_1 = \mu_0 n_1^2 A_1 l \]

Step 4: Find the ratio \( \frac{M}{L_1} \)
\[ \frac{M}{L_1} = \frac{\mu_0 n_1 n_2 A_1 l}{\mu_0 n_1^2 A_1 l} = \frac{n_2}{n_1} \]

Step 5: Final Answer
The ratio of mutual inductance to self-inductance of the inner coil is:
\( \frac{n_2}{n_1} \)