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

Answer: (A)

1 : 2

Answer 2

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: LA = 2LB nA = $\frac{1}{2}$nB Let LA and LB be the self-inductances of solenoids A and B respectively.

Then, LA ∝ nA2lA and LB ∝ nB2lB.

Since lA = 2lB and nA = nB/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, LA : LB = 1 : 2