A generator has a square coil consisting of 269 turns. The coil rotates at 113 rad/s in a 0.222-T magnetic field. The peak output of the gen

Question

A generator has a square coil consisting of 269 turns. The coil rotates at 113 rad/s in a 0.222-T magnetic field. The peak output of the generator is 79.2 V. What is the length of one side of the coil?

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Mộc Miên 3 years 2021-08-22T01:09:57+00:00 2 Answers 13 views 0

Answers ( )

  1. minhkhang
    0
    2021-08-22T01:11:00+00:00
    Reply

    Answer:

    Explanation:

    number of turns, n = 269

    angular velocity, ω = 113 rad/s

    magnetic field, B = 0.222 T

    Voltage, V = 79.2 V

    Let the area of the coil is A and the side of the square is s.

    The maximum emf generated by the coil is

    V = n x B x A x ω

    79.2 = 269 x 0.222 x A x 113

    A = 0.01174 m²

    so, s x s = 0.01174

    s = 0.108 m

    Thus, the side of the square coil is 0.108 m.

  2. kimchi2
    0
    2021-08-22T01:11:35+00:00
    Reply

    Answer:

    length of one side of the coil is 0.1083 m

    Explanation:

    given data

    no of turn = 269

    coil rotates = 113 rad/s

    magnetic field = 0.222-T

    peak output = 79.2 V

    solution

    we will apply here formula for maximum emf induced in a coil that is

    εo = N × A × B × ω    ……………………1

    and here Area A = l²

    so

    l = \sqrt{A}

    put here value of A from equation 1

    I = \sqrt{\frac{\epsilon _o}{N\times B\times \omega} }  

    put here value and we will get

    l =  \sqrt{\frac{79.2}{269\times 0.222\times 113}}

    l = 0.1083 m

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