A 16.0 cm × 16.0 cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis. The resistance of the loop is 0.500 Ω . A m

Question

A 16.0 cm × 16.0 cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis. The resistance of the loop is 0.500 Ω . A magnetic field parallel to the z-axis is given by B = 0.750 y2t, where B is in tesla, y is in meters, and t is in seconds. What is the size of the induced current in the loop at t = 0.490 s ? Express your answer with the appropriate units.

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Tryphena 5 years 2021-07-23T19:09:05+00:00 1 Answers 20 views 0

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  1. diemkieu
    0
    2021-07-23T19:11:04+00:00
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    Answer:

    the induced current is I = 3.2768*10^{-4} \ \ A

    Explanation:

    Given that :

    side of the square (l) = 16 cm = 0.16 m

    Magnetic field  B = 0.750 y² t

    Resistance R = 0.500 Ω

    Time t = 0.490 s

    Let consider a small rectangular; whose length is l and breath is dy

    Hence; to determine the magnetic flux through it small rectangular; we have:

    d \phi = B.dA \\ \\ d \phi = B (i * dy) \\ \\ d \phi = (0.750 \ y^2 t ) *(l *dy) \\ \\ d \phi = ( 0.75*t*l) y^2 dy \\ \\ d\phi = (0.75 tl) y^2 dy

    Let calculate the total flux in the square loop

    \phi = \int\limits^{y=l}_{y=0} B.dA \\ \\ \phi = \int\limits^{y=l}_{y=0} (0.75 \ t \ l ) y^2dy \\ \\ \phi = (0.75 \ t \ l )\frac{l^3}{3} \\ \\ \phi = (0.75 \ t )\frac{l^4}{3}

    Thus the total flux in the square loop is \phi = (0.75 \ t )\frac{l^4}{3}

    Now; going to the induced emf; let consider the Faraday’s Law

    V = \frac{d \phi}{dt}

    V = \frac{d }{dt}(\frac{0.75*l^4}{3})*t

    V = (\frac{0.75*0.16^4}{3})

    V = 1.6384*10^{-4} \ V

    Finally ; the induced current I is given by the expression;

    I = \frac{V}{R} \\ \\ I = (\frac{1.6384*10^{-4}}{0.5})

    I = 3.2768*10^{-4} \ \ A

    Therefore; the induced current is I = 3.2768*10^{-4} \ \ A

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