A piece of copper wire is formed into a single circular loop of radius 13 cm. A magnetic field is oriented parallel to the normal to the loo

Question

A piece of copper wire is formed into a single circular loop of radius 13 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.86 T in a time of 0.67 s. The wire has a resistance per unit length of 4.4 x 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire? Number Units

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Thu Thủy 5 days 2021-07-22T14:51:15+00:00 1 Answers 9 views 0

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    2021-07-22T14:53:00+00:00

    Answer:

    0.07 J

    Explanation:

    Parameters given:

    Radius of loop, r = 13 cm = 0.13 m

    Change in magnetic field, B = 0.86 T

    Time taken, t = 0.67 s

    Resistance per unit length of wire, R/L = 4.4 * 10^{-2} ohms/m

    Resistance in the entire length of wire, R = R/L * L (where L = circumference of wire)

    R = 4.4 * 10^{-2} * 2\pi r = 4.4 * 10^{-2} * 2\pi * 0.13

    R = 0.036 ohms

    Electrical energy is given as the product of Power and Time;

    E = P * t

    Electrical power, P, is given as:

    P = \frac{V^2}{R}

    where V = EMF/Voltage

    The average EMF induced in a coil is given as:

    V = \frac{-BA}{t}

    where A = Area of coil = \pi r^2 = 0.13^2\pi  = 0.0531 m^2

    Therefore, Power becomes:

    P = \frac{(\frac{-BA}{t})^2 }{R}\\ \\\\P = \frac{(BA)^2}{Rt^2}

    Then, Electrical energy becomes:

    E = \frac{(BA)^2}{Rt^2} * t\\\\\\E = \frac{(BA)^2}{Rt}

    E = \frac{(0.86 * 0.053)^2}{(4.4 *10^{-2} * 0.67)}\\ \\\\E = 0.07 J

    The average electrical energy dissipated in the resistance of the wire is 0.07 J

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