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A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4×10⁻³ Wb. The self-inductance of the solenoid is

Asked in NEET 2016 Phase-I · Self-inductance and self-induced EMF

Answer: (2) 1 H

Step-by-step solution

The total flux linkage is the flux per turn times the number of turns:

Nφ=1000×4×10⁻³=4 Wb.

Self-inductance is defined by Nφ=LI, so

L=(Nφ)/I=4/4=1 H.

The common slip is to use the flux per turn alone and forget the factor of N.

Why the other options are wrong

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