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24 March, 00:28

A single piece of wire is bent into the shape of Texas, with a total area of 13.3 cm2. This Texas shaped loop is perpendicular to a magnetic field which increases uniformly in magnitude from 0.728 T to 2.27 T in a time of 3 s. The wire has a total resistance of 4 Ω. What is the current? Answer in units of mA.

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Answers (2)
  1. 24 March, 01:05
    0
    E = induced emf = - 6.82*10-⁴V

    Explanation:

    Given

    Area = 13.3cm³ = 13.3*10-⁴m²

    ΔB = 2.27T - 0.728 = 1.54T

    Δt = 3s

    So

    ΔB/Δt = 1.54/3 = 0.513T/s

    ΔФ/Δt = ΔB/Δt*Area = 0.513*13.3*10-⁴ = 6.82*10-⁴Wb/s

    By Faraday's law of electromagnetic induction,

    E = - ΔФ/Δt = - (6.82*10-⁴) = - 6.82*10-⁴V
  2. 24 March, 01:08
    0
    Given Information:

    Time = Δt = 3 s

    Area of wire = A = 13.3 cm² = 0.00133 m²

    Change in magnetic field = ΔB = (0.728 - 2.27) T

    Resistance of wire = R = 4 Ω

    Required Information:

    Current = I = ?

    Answer:

    Current = 0.1708 mA

    Step-by-step explanation:

    We know that current is given by

    I = ξ/R

    Where R is the resistance and ξ is induced emf given by

    ξ = - ΔΦ/Δt

    Where ΔΦ is the change in flux and is given by

    ΔΦ = ΔBA

    ΔΦ = (0.728 - 2.27) * 0.00133

    ΔΦ = - 0.00205 T. m²

    ξ = - ΔΦ/Δt

    ξ = - (-0.00205) / 3

    ξ = 0.0006833 V

    I = ξ/R

    I = 0.0006833/4

    I = 0.0001708 A

    I = 0.1708 mA
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