5 single loop of copper wire, lying flat in a plane, has an area of 8.60cm2 and a resistance of 3.00. A uniform magnetic field points perpen

Question

5 single loop of copper wire, lying flat in a plane, has an area of 8.60cm2 and a resistance of 3.00. A uniform magnetic field points perpendicular to the plane of the loop. The field initially has a magnitude of .5 T, and the magnitude increases linearly to 3. 0 in a time of 1.10 s, what is the induced current in the loop of wire over this time? .652A

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Lệ Thu 2 days 2021-07-21T23:16:39+00:00 1 Answers 0 views 0

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    2021-07-21T23:18:16+00:00

    Answer:

    0.652 mA

    Explanation:

    According to Faraday’s Law :

    Emf = -  \frac{d \phi}{dt}

    = \frac{- \delta \phi }{\delta \ t}

    |E| = \frac{A ( \delta B)}{\delta t}

    where ;

    A = 8.60*10^{-4}

    \delta B = 3.00 T - 0.5 T = 2.5 T

    |E| = \frac{8.60*10^{-4}*2.5}{1.10}

    |E| = 1.96*10^{-3}

    Induced current  I = \frac{|E|}{R}

    = \frac{1.96*10^{-3}}{3.0}

    = 6.52*10^{-4} A

    = 0.652 mA

    Thus, the induced current in the loop of wire over this time = 0.652 A

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