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10 April, 17:46

A coaxial cable carries a current I = 0.02A through a wire of radius Rw=.2mmin one direction and an identical current through the outer sheath of radius Rs=3.4mm in the opposite direction. Determine the magnetic field everywhere.

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  1. 10 April, 17:48
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    a) B = 10⁻¹ r, b) B = 4 10⁻⁹ / r, c) B=0

    Explanation:

    For this exercise let's use Ampere's law

    ∫ B. ds = μ₀ I

    Where I is the current locked in the path. Let's take a closed path as a circle

    ds = 2π dr

    B 2π r = μ₀ I

    B = μ₀ I / 2μ₀ r

    Let's analyze several cases

    a) r
    Since the radius of the circumference is less than that of the wire, the current is less, let's use the concept of current density

    j = I / A

    For this case

    j = I / π Rw² = I'/π r²

    I' = I r² / Rw²

    The magnetic field is

    B = (μ₀ / 2π) r²/Rw² 1 / r

    B = (μ₀ / 2π) r / Rw²

    calculate

    B = 4π 10⁻⁷ / 2π r / 0.002²

    B = 10⁻¹ r

    b) in field between Rw
    In this case the current enclosed in the total current

    I = 0.02 A

    B = μ₀ / 2π I / r

    B = 4π 10⁻⁷ / 2π 0.02 / r

    B = 4 10⁻⁹ / r

    c) the field outside the coaxial Rs
    In this case the waxed current is zero, so

    B = 0
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