Simple Generator - Calculating "N" Loops


by PeachBanana
Tags: generator, loops, simple
PeachBanana
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#1
Sep26-12, 10:38 PM
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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ehild
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#2
Sep26-12, 11:31 PM
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Quote Quote by PeachBanana View Post
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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