# Simple Generator - Calculating N Loops

by PeachBanana
Tags: generator, loops, simple
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 P: 191 1. The problem statement, all variables and given/known data A simple generator is used to generate a peak output voltage of 26.0 V . The square armature consists of windings that are 6.3 cm on a side and rotates in a field of 0.440T at a rate of 60.0 rev/s. How many loops of wire should be wound on the square armature? Express your answer as an integer. 2. Relevant equations ε / 2Blv sin θ = N 3. The attempt at a solution l = 0.063 m ε = 26.0 V B = 0.440 T v = 1.89 m/s (Is this what is incorrect?) v = ωr v = (60.0 rev/s)(0.0315 m) v = 1.89 m/s (26.0 V) / (2)(0.440 T)(0.063 m)(1.89 m/s) (26.0 V) / (.104 T m^2 / s) = 250 loops ETA : I assumed the angle is 90 degrees.
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P: 10,671
 Quote by PeachBanana 1. The problem statement, all variables and given/known data A simple generator is used to generate a peak output voltage of 26.0 V . The square armature consists of windings that are 6.3 cm on a side and rotates in a field of 0.440T at a rate of 60.0 rev/s. How many loops of wire should be wound on the square armature? Express your answer as an integer. 2. Relevant equations ε / 2Blv sin θ = N 3. The attempt at a solution l = 0.063 m ε = 26.0 V B = 0.440 T v = 1.89 m/s (Is this what is incorrect?) v = ωr v = (60.0 rev/s)(0.0315 m) v = 1.89 m/s (26.0 V) / (2)(0.440 T)(0.063 m)(1.89 m/s) (26.0 V) / (.104 T m^2 / s) = 250 loops ETA : I assumed the angle is 90 degrees.
ω=2pi times revolutions/second

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