Q. A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 × 10⁻³ Wb. The self-inductance of the solenoid is
A) 4.0 henry
B) 2.5 henry
C) 2.0 henry
D) 1.0 henry
Explanation
[Solution from source] Total flux linkage Nφ = LI, so L = 500 × 4 × 10⁻³/2 = 1.0 H. 2.0 H would forget to divide by the current. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.