A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The

Question

A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday’s Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.

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Thu Thảo 1 year 2021-08-31T14:50:59+00:00 1 Answers 5 views 0

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    2021-08-31T14:51:59+00:00

    Answer:

    The average  emf induce is   [tex]V = 2.625 * 10^{-5} \ V[/tex]

    Explanation:

    From the question we are told that

      The radius of the coil is  [tex]r = 0.30 \ m[/tex]

       The number of turns is  [tex]N = 420 \ turns[/tex]

        The frequency of the transition radio wave is  [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]

         The magnetic field is  [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]

         The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]

         

    The period of the transmitted radio wave is  [tex]T = \frac{1}{f}[/tex]

        So  

                  [tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]

     The potential difference can be mathematically represented as

                   [tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]

               [tex]V = NA ([B_{max} – B_{min} ] * 4f)[/tex]

    Where  [tex]B_{min} = 0T[/tex]

    substituting values

                       [tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]

                      [tex]V = 2.625 * 10^{-5} \ V[/tex]

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