Question

A simple generator is used to generate a peak output voltage of 19.0 V . The square armature consists of windings that are 6.65 cm on a side and rotates in a field of 0.434 T at a rate of 49.8 rev/s . Part A How many loops of wire should be wound on the square armature

Answer 1

One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

$\epsilon = NBA\omega$

Where,

N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

$\omega$ = Angular velocity

Re-arrange to find N,

$N = \frac{\epsilon}{BA\omega}$

Our values are given as,

$\epsilon = 19V$

$B = 0.434T$

$\omega = 49.8\frac{rev}{s} (\frac{2\pi rad}{1 rev}) = 99.6\pi rad/s$

$A = (6.65*10^{-2})^2 m^2$

Replacing at our equation we have:

$N = \frac{\epsilon}{( 0.434)A\omega}$

$N = \frac{19}{( 0.434)((6.65*10^{-2})^2)(99.6\pi)}$

$N = 31.63 \approx 32$

Therefore the number of loops of wire should be wound on the square armature is 32 loops