Q4. Assertion: the emf of an AC generator is zero at the instant the plane of its coil is perpendicular to the field. Reason: at that instant the flux through the coil is largest and momentarily not changing. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: ε = NBAω sin ωt is zero at ωt = 0, where ΦB = BA cos ωt is at its peak (NCERT Eq. (6.19), page 171).
Q5. A generator coil of area 0.025 m2 turns at 50 revolutions per second in a field of 0.20 T. How many turns are needed for a peak emf of 300 V?
ε0 = NBA(2πν), so N = 300/(0.20 × 0.025 × 2π × 50) ≈ 190.99, so 191 turns (190 turns would give only 298 V) (NCERT Eqs. (6.19)–(6.21), page 171).
Q6. The coil of a generator is turned twice as fast. What happens to the peak emf and to the frequency of the output?
Both double: ε0 = NBAω and the frequency is ω/2π (NCERT Eqs. (6.19)–(6.21), page 171).
Q7. For the generator of Example 2, find the emf after its normal turns through 30° from the direction of B, with the positive rotation and emf convention of the lesson.
ε = ε0 sin 30° ≈ 126 × 0.5 ≈ 63 V (NCERT Eq. (6.20), page 171).