Question

The ratio of the number of turns per unit length of two solenoids A and B is \(1:3\) and the lengths of A and B are in the ratio \(1:2\). If the two solenoids have the same cross-sectional area, the ratio of the self-inductances of the solenoids A and B is:

• A. \(1:12\)
• B. \(1:6\)
• C. \(1:18\)
• D. \(1:9\)

Hint:

In problems involving ratios of physical quantities, always ensure consistency in units and directly relate the quantities using their defining equations.

Answer 1

The Correct Option is C

Step 1: The self-inductance \(L\) for a solenoid is given by: \[ L = \mu_0 \frac{N^2}{l} A \] where \(N\) is the number of turns, \(l\) is the length, and \(A\) is the cross-sectional area. 
Step 2: Given the number of turns per unit length ratio for A and B as \(1:3\) and length ratio as \(1:2\), the number of turns \(N\) for A and B can be represented as \(N_A = n_A \times l_A\) and \(N_B = n_B \times l_B\). 
Step 3: Using \(n_A:l_A = 1:2\) and \(n_B:l_B = 3:1\), the inductance ratio becomes: \[ \frac{L_A}{L_B} = \frac{\mu_0 \frac{(n_A l_A)^2}{l_A} A}{\mu_0 \frac{(n_B l_B)^2}{l_B} A} = \frac{(n_A l_A)^2}{(n_B l_B)^2} = \left(\frac{n_A l_A}{n_B l_B}\right)^2 \] Step 4: Calculate the ratio \( \frac{n_A l_A}{n_B l_B} \) where \(n_A = 1\), \(n_B = 3\), \(l_A = 1\), \(l_B = 2\). Thus, \[ \left(\frac{1 \times 1}{3 \times 2}\right)^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36} \] So, the correct ratio of the self-inductances is \(1:18\).