An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the

Question

An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)

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Cherry 3 years 2021-08-13T12:31:52+00:00 1 Answers 8 views 0

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  1. thugiang
    0
    2021-08-13T12:33:40+00:00
    Reply

    Answer:

    The induced  emf is  \epsilon =7.68 \ V

    Explanation:

    From the question we are told that

         The number of turns is  N = 1300 \ turns

        The diameter is  d = 2.2 \ cm = 2.2*10^{-2}

         The initial magnetic field is  B_i = 0.14 \ T

          The final magnetic field is  B_f = 0 \ T

          The  time taken is  dt = 9.0ms = 9.0*10^{-3} \ s

     

    The radius is mathematically evaluated as

          r = \frac{d}{2 }

    substituting values

         r = \frac{2.2 *10^{-2}}{2 }

         r = 1.1*10^{-2} \ m

    The induced emf is mathematically represented as

        \epsilon =- N * \frac{d\phi }{dt }

    Where  d\phi is the change in magnetic field which is mathematically represented as

            d\phi = dB * A * cos\theta

    =>   d\phi = [B_f - B_i ] * A * cos\theta

    Here  \theta = 0 given that the axis of the coil is parallel to the field

    Also A is the cross-sectional area which is mathematically represented as

           A = \pi r^2

    substituting values

          A = 3.142 * [1.1*10^{-2}]^2

           A = 3.8 *10^{-4] \ m^2

    So

        d\phi = [0 - 0.14 ] * 3.8*10^{-4}

        d\phi = -5.32*10^{-5} \ weber

    So  

         \epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }

        \epsilon =7.68 \ V

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