Question

Let us consider two solenoids A and B, made from same magnetic material of relative permeability \(µ_r\) and equal area of cross-section. Length of A is twice that of B and the number of turns per unit length in A is half that of B. The ratio of self inductances of the two solenoids, LA : LB is

• A. 1 : 2
• B. 2 : 1
• C. 8 : 1
• D. 1 : 8

Answer 1

The Correct Option is A

Self-inductance (L) of a solenoid is proportional to n²l, where n is the number of turns per unit length and l is the length.

Given: L_A = 2L_B n_A = $\frac{1}{2}$n_B Let L_A and L_B be the self-inductances of solenoids A and B respectively.

Then, L_A ∝ n_A²l_A and L_B ∝ n_B²l_B.

Since l_A = 2l_B and n_A = n_B/2, $\frac{L_A}{L_B} = \frac{n_A^2 l_A}{n_B^2 l_B} = \frac{(n_B/2)^2 (2l_B)}{n_B^2 l_B} = \frac{1}{2}$

Thus, L_A : L_B = 1 : 2