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 \( \frac{n_2}{n_1} \).