AIPMT PRELIMS2008Physics-Electromagnetic Induction

AIPMT PRELIMS 2008 Physics Self-Inductance MCQ Question

Type: MCQ-numerical-Medium-Class 12

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

1.0 henry

B

4.0 henry

C

2.5 henry

D

2.0 henry

Correct Answer

Option A

Detailed Explanation

The self-inductance L is given by L = NΦ/I, where N is the number of turns, Φ is the magnetic flux, and I is the current. Substituting the given values, L = 500 × 4 × 10⁻³ / 2 = 1.0 henry.

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