A square loop of wire 10cm on a side with 100 turns of wire rotates in a magnetic field, |B| = 0.01T, twenty times each second. At t = 0, th

Question

A square loop of wire 10cm on a side with 100 turns of wire rotates in a magnetic field, |B| = 0.01T, twenty times each second. At t = 0, the normal of the loop aligns with the magnetic field. Write a symbolic expression for the voltage as a function of time starting from Faraday’s Law. Define each symbol in your expression and give its numerical value. Calculate the peak output voltage of the generator.

in progress 0
Khải Quang 3 years 2021-07-29T01:38:21+00:00 1 Answers 20 views 0

Answers ( )

  1. Xavia
    0
    2021-07-29T01:40:03+00:00
    Reply

    Answer with Explanation:

    We are given that

    Side, l=10 cm=10\times 10^{-2} m

    1 m=100 cm

    Magnetic field,B=0.01 T

    Number of turns ,N=100

    \omega=20rev/s=20\times 2\pi=40\pi rad/s

    Magnetic flux,\phi=NBAcos\omega t=NBAcos\omega t

    By Faraday’s law

    V=\mid- \frac{d\phi}{dt}\mid =\omega NBAsin\omega t=\omega NBA sin\omega t

    Where \omega=Angular frequency

    N=Number of turns

    B=Magnitude of magnetic field

    A=Cross section area  of wire

    For peak voltage

    \omega t=\frac{\pi}{2}

    V_{max}=NBA\omega sin\frac{\pi}{2}=NBA\omega=100\times 0.01\times(10\times 10^{-2})^2\times 40\pi=

    V_{max}=1.26 V

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )