Answered

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A simple generator is used to generate a peak output voltage of 23.0 V . The square armature consists of windings that are 5.1 cm on a side and rotates in a field of 0.500 T at a rate of 55.0 rev/s .
How many loops of wire should be wound on the square armature?

Sagot :

Answer: 51

Explanation:

Given

Output is 23 V

The square armature side is [tex]a=5.1\ cm[/tex]

Magnetic field [tex]B=0.5\T[/tex]

Rate of revolution [tex]n=55\ rev/s[/tex]

Angular speed

[tex]\omega =2\pi n\\\omega=2\pi \times 55=110\pi\ rad/s[/tex]

Peak voltage is given by

[tex]E_{peak}=NB\omega A\quad [\text{N=Number of windings; A=area of cross-section}]\\\\N=\dfrac{E_{peak}}{B\omega A}\\\\N=\dfrac{23}{0.5\times 110\pi\times (0.051)^2}\approx 51[/tex]

So, there are approximate 51 loops

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