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6 July, 08:35

The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoid? (b) What is the induced emf in the solenoid if its current drops from 3.2 A to 0 in 55 ms?

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  1. 6 July, 12:31
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    Answer: 0.43 V

    Explanation:

    L = [μ (0) * N² * A] / l

    Where

    L = Inductance of the solenoid

    N = the number of turns in the solenoid

    A = cross sectional area of the solenoid

    l = length of the solenoid

    7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

    1.752*10^-3 = 4π*10^-7 * 202500 * A

    1.752*10^-3 = 0.255 * A

    A = 1.752*10^-3 / 0.255

    A = 0.00687 m²

    A = 6.87*10^-3 m²

    emf = - N (ΔΦ/Δt) ... 1

    L = N (ΔΦ/ΔI) so that,

    N*ΔΦ = ΔI*L

    Substituting this in eqn 1, we have

    emf = - ΔI*L / Δt

    emf = - [ (0 - 3.2) * 7.3*10^-3] / 55*10^-3

    emf = 0.0234 / 0.055

    emf = 0.43 V
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