Ac Generator
A rectangular armature coil of an AC generator has 100 turns and area 0.05 square metres, rotating at an angular speed of 60 radians per second in a uniform magnetic field of 0.4 tesla. What is the peak emf produced?
Select the correct option:
Solution
120 V
NCERT Class 12, Chapter 6 (Electromagnetic Induction) explains that an AC generator produces a sinusoidal emf as its coil rotates, with peak value ε0=NBAω, arising because the flux through the coil varies as cosωt and its rate of change is maximum when the coil plane is parallel to the field. Substituting: ε0=100×0.4×0.05×60=120 V. The option 60 V halves the result by dropping a factor of two somewhere in the product. The option 240 V doubles it, perhaps by mis-reading the turns. The option 12 V drops a factor of ten in the area or turns. As a sanity check, the units combine as (turns × T × m² × rad/s), and since T\cdotpm2=Wb and Wb/s=V, the result is correctly in volts. A peak emf of 120 V is realistic for such a compact laboratory generator, and the actual output alternates sinusoidally between +120 V and −120 V. It is instructive to see that the peak emf depends linearly on the rotational speed, so spinning the coil twice as fast would double both the peak voltage and the output frequency. The emf is zero at the instant the coil plane is perpendicular to the field, because the flux is then maximum and momentarily unchanging, and it is greatest a quarter-turn later when the flux is changing most rapidly. This continuous conversion of mechanical rotational energy into electrical energy is the fundamental operating principle behind the generators that supply electrical power stations worldwide.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- ac generator
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
120 V
NCERT Class 12, Chapter 6 (Electromagnetic Induction) explains that an AC generator produces a sinusoidal emf as its coil rotates, with peak value ε0=NBAω, arising because the flux through the coil varies as cosωt and its rate of change is maximum when the coil plane is parallel to the field. Substituting: ε0=100×0.4×0.05×60=120 V. The option 60 V halves the result by dropping a factor of two somewhere in the product. The option 240 V doubles it, perhaps by mis-reading the turns. The option 12 V drops a factor of ten in the area or turns. As a sanity check, the units combine as (turns × T × m² × rad/s), and since T\cdotpm2=Wb and Wb/s=V, the result is correctly in volts. A peak emf of 120 V is realistic for such a compact laboratory generator, and the actual output alternates sinusoidally between +120 V and −120 V. It is instructive to see that the peak emf depends linearly on the rotational speed, so spinning the coil twice as fast would double both the peak voltage and the output frequency. The emf is zero at the instant the coil plane is perpendicular to the field, because the flux is then maximum and momentarily unchanging, and it is greatest a quarter-turn later when the flux is changing most rapidly. This continuous conversion of mechanical rotational energy into electrical energy is the fundamental operating principle behind the generators that supply electrical power stations worldwide.
This medium difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of ac generator. It appeared in the 2025 exam.
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