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Electromagnetic Induction · Class 12 · CBSE · Science Physics

The self-inductance of a coil of 500 turns is 50 mH. When a current of 2 A flows through it, the total magnetic flux linked with each turn of the coi…

Practice question from Electromagnetic Induction (Class 12 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

easy · MCQ · Q-74A6A3

The self-inductance of a coil of 500 turns is 50 mH. When a current of 2 A flows through it, the total magnetic flux linked with each turn of the coil is:

  1. A.2×10−4 Wb2 \times 10^{-4}\text{ Wb}Correct answer
  2. B.10−3 Wb10^{-3}\text{ Wb}
  3. C.5×10−3 Wb5 \times 10^{-3}\text{ Wb}
  4. D.2.5×10−4 Wb2.5 \times 10^{-4}\text{ Wb}

Solution

Total magnetic flux linkage NΦ=LIN\Phi = LI. Given N=500N = 500, L=50×10−3 HL = 50 \times 10^{-3}\text{ H}, I=2 AI = 2\text{ A}. Total flux NΦ=50×10−3×2=0.1 WbN\Phi = 50 \times 10^{-3} \times 2 = 0.1\text{ Wb}. The flux linked with each turn is Φ=0.1500=2×10−4 Wb\Phi = \frac{0.1}{500} = 2 \times 10^{-4}\text{ Wb}.

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