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Standard XII

Physics

Electric Generator

The number of...

Question

The number of turns in the coil of an AC generator is 5000 and the area of the coil is $0.25 m^{2} .$ The coil is rotated at the rate of 100 cycles /sec in the magnetic field of $0.2 W / m^{2}$ . then the peak value will be :

786 kV

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178 kV

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157 kV

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123 kV

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Solution

The correct option is C 157 kV
We know that
$ϵ_{0} = ω N B A ϵ_{0} = (2 π v) N B A ϵ_{0} = 2 \times 3.14 \times 1000 \times 5000 \times 0.2 \times 0.25 ϵ_{0} = 15 k V$
Thus, the peak value of emt $= 157 k V$

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The number of turns in the coil of an AC generator is 5000 and the area of the coil is0.25 m2. The coil is rotated at the rate of 100 cycles /sec in the magnetic field of 0.2W/m2. then the peak value will be :

The correct option is C 157 kVWe know thatϵ0=ωNBAϵ0=(2πv)NBAϵ0=2×3.14×1000×5000×0.2×0.25ϵ0=15kVThus, the peak value of emt=157kV

The number of turns in the coil of an AC generator is 5000 and the area of the coil is $0.25 m^{2} .$ The coil is rotated at the rate of 100 cycles /sec in the magnetic field of $0.2 W / m^{2}$ . then the peak value will be :

The correct option is C 157 kV
We know that
$ϵ_{0} = ω N B A ϵ_{0} = (2 π v) N B A ϵ_{0} = 2 \times 3.14 \times 1000 \times 5000 \times 0.2 \times 0.25 ϵ_{0} = 15 k V$
Thus, the peak value of emt $= 157 k V$