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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