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Question

Calculate the peak output voltage of a simple generator whose square armature windings are 5.50 cm5.50 \mathrm{~cm} on a side; the armature contains 125125 loops and rotates in a field of 0.200 T0.200 \mathrm{~T} at a rate of 120rev/s120 \mathrm{rev} / \mathrm{s}.

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Givens:\color{#4257b2}\text{Givens:}

We are given a generator of a square armature rotating in a magnetic field,

  • l=5.5  cml= 5.5\;\mathrm{cm}: is the length if the square armature of the generator.

  • N=125  turnsN= 125\;\mathrm{turns}: is the number of loops of the generator coil.

  • B=0.2  TB= 0.2\;\mathrm{T}: is the magnetic field in which the generator rotates.

  • f=120  rev/sf= 120\;\mathrm{rev/s}: is the frequency of rotation.

We are required to evaluate the peak value of the output voltage ϵo\epsilon_o of the generator.

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