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8 February, 23:02

A stationary conductive loop with an internal resistance of 0.5 Ω is placed in a time varying magnetic field. When the loop is closed, a current of 5 A flows through it. What will the current be if the loop is opened to create a small gap and a 2-Ω resistor is connected across its open ends?

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  1. 8 February, 23:28
    0
    e = - AdB/dt

    5*0.5 = - AdB/dt

    i (2+0.5) = - AdB/dt

    from above expressions

    2.5 = i2.5

    so i = 1A

    More Answers:

    v=0.5x5 = 2.5

    2.5 / (2+0.5) = 1 A
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