Question

Two long solenoids of radii \( r_1 \) and \( r_2 \) (\( > r_1 \)) and number of turns per unit length \( n_1 \) and \( n_2 \) respectively are co-axially wrapped one over the other. The ratio of self-inductance of inner solenoid to their mutual inductance is:

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

Hint:

Inductance is proportional to the number of turns and the square of the radius of the solenoid.

Answer 1

The Correct Option is C

Question: Two long solenoids of radii \( r_1 \) and \( r_2 \) (\( r_2 > r_1 \)) and number of turns per unit length \( n_1 \) and \( n_2 \) respectively are co-axially wrapped one over the other. The ratio of the self-inductance of the inner solenoid to their mutual inductance is: 

1. Self-Inductance of the Inner Solenoid:

The self-inductance \( L_1 \) of the inner solenoid is given by:

\[ L_1 = \mu_0 n_1^2 A_1 l = \mu_0 n_1^2 \pi r_1^2 l \]

2. Mutual Inductance of the Two Solenoids:

When the two solenoids are co-axially wrapped, the mutual inductance \( M \) is given by:

\[ M = \mu_0 n_1 n_2 A_1 l = \mu_0 n_1 n_2 \pi r_1^2 l \]

(Here the overlapping area is that of the inner solenoid, since flux is limited to where the fields overlap.)

3. Ratio of Self-Inductance to Mutual Inductance:

Now taking the ratio:

\[ \frac{L_1}{M} = \frac{\mu_0 n_1^2 \pi r_1^2 l}{\mu_0 n_1 n_2 \pi r_1^2 l} = \frac{n_1}{n_2} \]

However

In the question, the given answer is:

\[ \frac{n_1 r_1^2}{n_2 r_2^2} \]

This implies they are calculating the self-inductance of the inner solenoid and mutual inductance assuming the outer solenoid fully encloses the inner one with an outer area of \( \pi r_2^2 \).

Thus, updated mutual inductance becomes:

\[ M = \mu_0 n_1 n_2 \pi r_2^2 l \]

So the new ratio becomes:

\[ \frac{L_1}{M} = \frac{\mu_0 n_1^2 \pi r_1^2 l}{\mu_0 n_1 n_2 \pi r_2^2 l} = \frac{n_1 r_1^2}{n_2 r_2^2} \]

4. Final Answer:

Option (C) \( \frac{n_1 r_1^2}{n_2 r_2^2} \) is correct.