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

The self-inductance (L) of a solenoid is directly proportional to the square of the number of turns per unit length (n²) and the length (l).

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

Therefore, L_A ∝ n_A²l_A and L_B ∝ n_B²l_B.

With l_A = 2l_B and n_A = n_B/2, the ratio of inductances is $\frac{L_A}{L_B} = \frac{n_A^2 l_A}{n_B^2 l_B} = \frac{(n_B/2)^2 (2l_B)}{n_B^2 l_B} = \frac{1}{2}$.

Consequently, the ratio L_A : L_B is 1 : 2.