You are looking down on a single coil in a constant magnetic field B = 1.2 T which points directly into of the screen. The dimensions of the

Question

You are looking down on a single coil in a constant magnetic field B = 1.2 T which points directly into of the screen. The dimensions of the coil go from a = 8 cm and b = 17 cm, to a* = 16 cm and b* = 22 cm in t = 0.04 seconds. If the coil has resistance that remains constant at 1.2 ohms. What would be the magnitude of the induced current in amperes?

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Latifah 4 years 2021-08-06T21:57:21+00:00 1 Answers 17 views 0

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  1. Doris
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    2021-08-06T21:59:14+00:00
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    Answer:

    The  current is I = 0.5425 \ A

    Explanation:

    From the question we are told that

       The  magnetic field is  B = 1.2 \ T

       The first length is  a = 8 \ cm = 0.08 \ m

        The  second length is  a^* = 16 \ cm = 0.16 \ m

        The first width is  b = 17 \ cm = 0.17 \ m

         The second  width is  b^* = 22 \ cm = 0.22 \ m

        The time interval  is  dt = 0.04 \ s

         The resistance is  R = 1.2 \ \Omega

    Generally the first area is

         A = a * b

    =>    A = 0.08 * 0.17

    =>     A = 0.0136 \ m^2

    The second area is  

          A^* = a^* * b^*

    =>   A^* = 0.16 * 0.22

    =>     A^* = 0.0352 \ m^2

    Generally the induced emf is mathematically represented as

           \epsilon = - \frac{ B * [A^* - A]}{dt}

    This negative show that it is moving in the opposite direction of the motion producing it

    =>   |\epsilon | = \frac{ 1.2 * [ 0.0352-0.0135]}{0.04}

    =>    |\epsilon | = 0.651 \ V

    The induced current is

         I = \frac{|\epsilon|}{R}

    =>   I = \frac{ 0.651}{1.2}

    =>   I = 0.5425 \ A

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