If the self inductance of 500 turn coil is 125 mH, then the self inductance of similar coil of 800 turns is

Options

(a) 48.4 mH
(b) 200 mH
(c) 187.5 mH
(d) 320 mH

Correct Answer:

320 mH

Explanation:

Self inductance of a coil, L = μᵣμ₀n² Al

Where μᵣ is the relative permittivity of the coil, n is the number of terms per unit length, A is the area of cross-section, l is the length of the solenoid

L ∝ n²

Self inductance of 500 turns coil = 125 mH Therefore, L for the coil of 800 turns = [125 / (500)²] × 800² L = 320 mH