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23 January, 21:25

Consider a uniformly wound solenoid having N=210 turns, length l=0.18 m, and cross-sectional area A = 4.00 cm2. Assume l is much longer than the radius of the windings and the core of the solenoid is air. (A) Calculate the inductance of the solenoid. (B) Calculate the self-induced emf in the solenoid if the current it carries decreases at the rate of 120 A/s. Inductor

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  1. 23 January, 21:35
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    Number of turns

    N = 210turns

    Length of solenoid

    l = 0.18m

    Cross sectional area

    A = 4cm² = 4 * 10^-4m²

    A. Inductance L?

    Inductance can be determined using

    L = N²μA/l

    Where

    μ is a constant of permeability of the core

    μ = 4π * 10^-7 Tm/A

    A is cross sectional area

    l is length of coil

    L is inductance

    Therefore

    L = N²μA / l

    L=210² * 4π * 10^-7 * 4 * 10^-4 / 0.18

    L = 1.23 * 10^-4 H

    L = 0.123 mH

    B. Self induce EMF ε?

    EMF is given as

    ε = - Ldi/dt

    Since rate of decrease of current is 120 A/s

    Then, di/dt = - 120A/s, since the current is decreasing

    Then,

    ε = - Ldi/dt

    ε = - 1.23 * 10^-4 * - 120

    ε = 0.01478 V

    ε ≈ 0.015 V
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