Class 12 Physics Chapter 6: Electromagnetic Induction case study question with answers
One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
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Answer: (c) 236 V
ε₀ = NBAω = 120 × 0.25 × 0.050 × 2π × 25 = 236 V. 37.5 V leaves out 2π; 471 V doubles it; 1.96 V leaves out N.
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Answer: (d) parallel to the field
ε = ε₀ sin ωt is greatest at θ = 90° and 270°, when the plane of the coil is parallel to B: the flux is zero but changing fastest.
Answer and marking scheme
ωt = 2π × 25 × (1/300) = π/6, so ε = ε₀ sin 30° = 236 × 0.5 = 118 V. The emf is zero twice in each revolution, so successive zeros are half a period apart: T/2 = 1/(2 × 25) = 0.020 s.
- ε = ε₀ sin(π/6) = 118 V (1)
- T/2 = 0.020 s (1)
Answer and marking scheme
ε₀ = NBAω is proportional to ω, so it halves to about 118 V; the frequency equals the rate of revolution, 12.5 Hz. The kinetic energy of the flowing water, through the turbine's mechanical energy, is converted into electrical energy.
- peak emf about 118 V and frequency 12.5 Hz (1)
- mechanical (water) energy to electrical energy (1)
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