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A solenoid has a total of 500 turns, and a current of 2A flows through it. The magnetic flux associated with each turn is 4 × 10⁻³ Wb. Using the relationship Φ = LI or NΦ = LI, calculate the self-inductance L of the solenoid.

1. 500 × 4 × 10⁻³ / 2 henry = 1 H
2. Mutual inductance between two coils M is √(M₁M₂) = √(2mH × 8mH) = 4mH
3. L = 2mH, i = 2e⁻t
4. E = -L (di/dt) = -L (2e⁻t - 2te⁻t)

Correct answer: 500 × 4 × 10⁻³ / 2 henry = 1 H

Solution

The formula for self-inductance is correctly applied as L = NΦ/I. Substituting the given values, L = (500 × 4 × 10⁻³) / 2 = 1 H. Thus, option A is correct.

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