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2 January, 20:49

A solenoid has a cross-sectional area of 6.0 3 1024 m2, consists of 400 turns per meter, and carries a current of 0.40 A. A 10-turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a 1.5-V resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.050 s. Find the average current induced in the coil.

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  1. 2 January, 21:03
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    Complete Question:

    A solenoid has a cross-sectional area of 6.0*10-4m2 consists of 400 turns per meter, and carries a current of 0.40 A. A 10-turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a 1.5-Ω resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.050 s. Find the average current induced in the coil.

    Answer:

    I = 1.65 * 10⁻⁵A

    Explanation:

    I = Absolute value[-N (d∅/dt) / R]

    N = 10 turns

    ∅ = BA cos Θ

    B = μηI

    B = (4π * 10⁻⁷) (400) (0.4)

    B = 0.0002 T

    A = 6.03 * 10⁻⁴m²

    ∅ = 0.0002 * 6.03 * 10⁻⁴

    ∅ = 12.12 * 10⁻⁸

    d∅/dt = (12.12 * 10⁻⁸) / 0.05

    d∅/dt = 24.24 * 10⁻⁷

    I = Abslute value of [-10 * (24.24 * 10⁻⁷) / 1.5]

    I = 1.6 * 10⁻⁵A
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