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23 August, 07:34

A 40-turn, 4.0-cm-diameter coil with R=0.40Ω surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The 60 Hz current through the solenoid is I=I0sin (2πft). What is the Io if maximum induced current is. 2 A?

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  1. 23 August, 07:44
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    Magnetic field B is produced when a current I Amphere passes through a solenoid. B is parallel to its axis.

    B=U N/I I. N is number of turns in the solenoid of lm length

    N=200, l = 20cm = 0.2m, I = 1₀sin (2πft) where f is equal to 60Hz

    B = 4π * 10⁻⁷ (200/0.2) l₀ sin (2πft) T

    =1.256 * 10⁻³ l₀ sin (2πft) Tesla

    Area of the coil is πr² = π (1.5cm) ² = 2.25π * 10⁴m²

    magnetic flux which is through the coil is given by

    Ф = B. A = BA cosФ

    ФФ = O since B is in direction of A

    A = 40 * π*2.25 * ⁻⁴m² which is the number of turns being 40.

    Flux Ф through the coil is,

    1.256 * 10⁻³ l₀ sin (2πft) * 9π * * 10⁻³m²

    =35.5 * 10⁻⁶ l₀ sin ₀ (2πft) ab

    Ф is time-varying emf will be generated in the coil

    ∈ = dФ/dt

    ∈ = d/dt [35.5 * 10⁻d l₀ sin (2πft) ab]

    ∈ = 35.5 * 10⁻⁶ l₀ 2πf cos 2πftV

    f = 60Hz

    ∈∈ 13376.4 * 10⁻⁶ l₀ cos 2πftV

    Current I amp shall be induced in the cell of resistance Rohm so

    I = E/R

    I = 13376.4 * 10⁻⁶ l₀ cos 2πft) V/0.4∩

    =33441 * 10⁻⁶ I₀ cos 2πft A

    I = 3344q * 10 ⁻⁶ l₀

    But I = 0.2A

    l₀ = (0.2) (10⁶) / 33441 = 6.0A
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