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Distinguish between the phase and group velocity. Assuming that a 100 KHz plane wave is used to communicate a submerged submarine for sea water, σ = 4 s/m, ε = 80ε₀, µ = µ₀, compute the attenuation, phase velocity and skin depth of the wave.

Physics

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Answer

b

1.257 Np/m1.257 \text{ Np/m}

5×105 m/s5 \times 10^5 \text{ m/s}

0.796 m0.796 \text{ m}

Step-by-step solution

5 steps

here are the answers to question 4:

Part a

Distinguish between phase and group velocity

  • Phase velocity (vpv_p): This is the speed at which a point of constant phase on a single, monochromatic wave propagates through a medium. It describes how fast the crests or troughs of a wave move. For a wave with angular frequency ω\omega and wave number β\beta, the phase velocity is given by vp=ωβv_p = \frac{\omega}{\beta}.
  • Group velocity (vgv_g): This is the speed at which the overall shape of a wave's amplitude envelope (the "group" of waves) propagates through a medium. It represents the speed at which energy or information is transmitted by the wave. For a wave in a dispersive medium, the group velocity is given by vg=dωdβv_g = \frac{d\omega}{d\beta}. In non-dispersive media, phase and group velocities are equal.

Part b

1.257 Np/m1.257 \text{ Np/m}

Compute the attenuation, phase velocity, and skin depth for a 100 KHz plane wave in sea water.

Given:

  • Frequency f=100 KHz=105 Hzf = 100 \text{ KHz} = 10^5 \text{ Hz}
  • Conductivity σ=4 S/m\sigma = 4 \text{ S/m}
  • Relative permittivity ϵr=80\epsilon_r = 80
  • Relative permeability μr=1\mu_r = 1 (since μ=μ0\mu = \mu_0)
  • Permittivity of free space ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}
  • Permeability of free space μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \text{ H/m}
  1. Calculate angular frequency ω\omega and absolute permittivity ϵ\epsilon.

    ω=2πf=2π(105) rad/s\omega = 2\pi f = 2\pi (10^5) \text{ rad/s} ϵ=ϵrϵ0=80×(8.85×10−12)=7.08×10−10 F/m\epsilon = \epsilon_r \epsilon_0 = 80 \times (8.85 \times 10^{-12}) = 7.08 \times 10^{-10} \text{ F/m}
  2. Determine if sea water is a good conductor or a good dielectric at this frequency.

    We compare the conduction current to the displacement current by calculating the ratio σωϵ\frac{\sigma}{\omega\epsilon}:

    σωϵ=4(2π×105)×(7.08×10−10)=44.4488×10−4≈8991.1\frac{\sigma}{\omega\epsilon} = \frac{4}{(2\pi \times 10^5) \times (7.08 \times 10^{-10})} = \frac{4}{4.4488 \times 10^{-4}} \approx 8991.1

    Since σωϵ≫1\frac{\sigma}{\omega\epsilon} \gg 1, sea water is a good conductor at 100 KHz100 \text{ KHz}.

  3. Calculate the attenuation constant α\alpha.

    For a good conductor, the attenuation constant is given by:

    α=ωμσ2\alpha = \sqrt{\frac{\omega\mu\sigma}{2}}

    Substitute the values:

    α=(2π×105)×(4π×10−7)×42\alpha = \sqrt{\frac{(2\pi \times 10^5) \times (4\pi \times 10^{-7}) \times 4}{2}} α=32π2×10−22=16π2×10−2=0.16π2\alpha = \sqrt{\frac{32\pi^2 \times 10^{-2}}{2}} = \sqrt{16\pi^2 \times 10^{-2}} = \sqrt{0.16\pi^2} α=0.4π≈0.4×3.14159\alpha = 0.4\pi \approx 0.4 \times 3.14159 α≈1.257 Np/m\alpha \approx \boxed{1.257 \text{ Np/m}}
  4. Calculate the phase velocity vpv_p.

    For a good conductor, the phase constant β\beta is approximately equal to the attenuation constant α\alpha.

    β≈α=0.4π rad/m\beta \approx \alpha = 0.4\pi \text{ rad/m}

    The phase velocity is given by:

    vp=ωβv_p = \frac{\omega}{\beta} vp=2π×1050.4π=2×1050.4v_p = \frac{2\pi \times 10^5}{0.4\pi} = \frac{2 \times 10^5}{0.4} vp=5×105 m/sv_p = \boxed{5 \times 10^5 \text{ m/s}}
  5. Calculate the skin depth δ\delta.

    The skin depth is the reciprocal of the attenuation constant:

    δ=1α\delta = \frac{1}{\alpha} δ=10.4π=11.2566\delta = \frac{1}{0.4\pi} = \frac{1}{1.2566} δ≈0.796 m\delta \approx \boxed{0.796 \text{ m}}
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