The number of turns in the coil of an AC generator is $$5000$$ and the area of the coil is $$0.25$$ $$m2$$. The coil is rotated at the rate of $$100$$ $$cycles/$$$$sec$$ in a magnetic field of $$0.2$$ T. The peak value of the emf (in$$ $$kV) generated is[Take π$$=3.14$$]

ϵ$$0=$$ωNBAϵ$$0=2$$×$$3.14$$×$$100$$×$$5000$$×$$0.2$$×$$0.25$$ϵ$$0=157$$ kV

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The number of turns in the coil of an AC generator is 5000 and the area of the coil is 0.25 m². The coil is rotated at the rate of 100 cycles/sec in a magnetic field of 0.2 T. The peak value of the emf (in kV) generated is
[Take $π = 3.14$ ]

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detailed solution

157

$ϵ_{0} = ωNBA$

$ϵ_{0} = 2 \times 3.14 \times 100 \times 5000 \times 0.2 \times 0.25$

$ϵ_{0} = 157 kV$

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The number of turns in the coil of an AC generator is 5000 and the area of the coil is 0.25 m2. The coil is rotated at the rate of 100 cycles/sec in a magnetic field of 0.2 T. The peak value of the emf (in kV) generated is[Take π=3.14]

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The number of turns in the coil of an AC generator is 5000 and the area of the coil is 0.25 m². The coil is rotated at the rate of 100 cycles/sec in a magnetic field of 0.2 T. The peak value of the emf (in kV) generated is
[Take $π = 3.14$ ]