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STANDARDPhysics-Electromagnetic Induction

STANDARD Physics Magnetic Flux MCQ Question

Type: MCQ-numerical-Medium-Class 12

The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA is

A

5 × 10⁻⁸ Wb

B

2 × 10⁻⁸ Wb

C

3 × 10⁻⁵ Wb

D

8 × 10⁻⁷ Wb

Correct Answer

Option A

Detailed Explanation

The total magnetic flux linked with the coil is given by Na = LI. Here, N = 400, L = 10 mH = 10 × 10⁻³ H, I = 2 mA = 2 × 10⁻³ A. Therefore, Na = LI = (10 × 10⁻³) × (2 × 10⁻³) = 2 × 10⁻⁵ Wb. Dividing by the number of turns, the flux per turn is 5 × 10⁻⁸ Wb.

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