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21 December, 09:58

A planar loop consisting of seven turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field that increases uniformly in magnitude from 14 * 10-3 T to 38 * 10-3 T in a time of 8.0 * 10-3 s. What is the resulting induced current in the coil if the resistance of the coil is 5.0 Ω?

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  1. 21 December, 10:15
    0
    Given Information:

    Area of loop = A = 200 cm² = 0.0200 m²

    Change in time = Δt = 8x10⁻³ seconds

    Change in magnetic field = ΔB = (38x10⁻³ - 14x10⁻³) T

    Number of turns = N = 7

    Resistance of coil = R = 5 Ω

    Required Information:

    Induced current = I = ?

    Answer:

    Induced current = 0.084 A

    Explanation:

    From the Faraday's law the induced EMF ξ in the coil is given by

    ξ = - NΔΦ/Δt

    Where ΔΦ is the change in flux

    ΔΦ = ΔBA

    Where A is the area of the planer loop

    ΔΦ = (38x10⁻³ - 14x10⁻³) * 0.0200

    ΔΦ = 0.00048

    So the induced emf becomes

    ξ = - NΔΦ/Δt

    ξ = (-7*0.00048) / 8x10⁻³

    ξ = - 0.42 V

    The negative sign indicates that the induced emf opposes the change that produced it in the first place.

    Finally, we can now find the induced current using Ohm's law

    I = ξ/R

    I = 0.42/5

    I = 0.084 A

    Therefore, the resulting induced current in the coil is 0.084 A
  2. 21 December, 10:28
    0
    The induced current is 0.084 A

    Explanation:

    the area given by the exercise is

    A = 200 cm^2 = 200x10^-4 m^2

    R = 5 Ω

    N = 7 turns

    The formula of the emf induced according to Faraday's law is equal to:

    ε = (-N * dφ) / dt = (N * (b2-b1) * A) / dt

    Replacing values:

    ε = (7 * (38 - 14) * (200x10^-4)) / 8x10^-3 = 0.42 V

    the induced current is equal to:

    I = ε / R = 0.42/5 = 0.084 A
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