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Q9.20P

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Introduction to Electrodynamics
Found in: Page 415
Introduction to Electrodynamics

Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

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Short Answer

(a) Show that the skin depth in a poor conductor σ<<ωε is (εσ)2μ (independent of frequency). Find the skin depth (in meters) for (pure) water. (Use the static values of ε , μ and σ; your answers will be valid, then, only at relatively low frequencies.)

(b) Show that the skin depth in a good conductor (σ<<ωε) is λ2π (where λ is the wavelength in the conductor). Find the skin depth (in nanometers) for a typical metal (σ>>Ωm107-1) in the visible range (ω1015/s) , assuming ε=ε0 and μμ0. Why are metals opaque?

(c) Show that in a good conductor the magnetic field lags the electric field by 45°, and find the ratio of their amplitudes. For a numerical example, use the “typical metal” in part (b).

Answer

(a) The skin depth in a poor conductor is d=2σεμ and the skin depth for pure water is 1.19×104 m.

(b) The skin depth in a god conductor is d=λ2π and the skin depth for a typical metal is 12.5 nm.

(c) The ratio of the real amplitude is 1.12×10-7S/m.

See the step by step solution

Step by Step Solution

Step 1: Expression for the wavenumber and relative permittivity:

Write the expression for the wavenumber for the poor conductor.

k=k+iK …… (1)

Here, K is the imaginary part.

Write the expression for the relativity permittivity.

εr=εε0ε=εrε0 …… (2)

Here, is the Permittivity of the free space.

Step 2: Determine the skin depth in a poor conductor and for pure water:

(a)

Write the expression for imaginary part K.

K=ωεμ21+σεω212

Solve the above expression for the poor conductor.

K=ωεμ21+12σεω2-112K=ωεμ212σεωK=σ2μμ

Write the expression for the skin depth.

d=1K

Substitute K=σ2με in the above expression.

d=1σ2μεd=2σεμ

Substitute the value of equation (2) in the above expression.

d=2σε0εrμ

Take the value of εr from table 4.2 for pure water.

Substitute σ=2.5×105Ωm-1 , role="math" localid="1655714742974" ε0=8.85×10-12 C2/Nm2 , εr=80.1 and μ=4π×10-7 H/m in the above expression.

d=212.5×105Ωm-18.85×10-12C2/Nm280.14π×10-7 H/md=1.19×104 m

Therefore, the skin depth in a poor conductor is d=2σεμ and the skin depth for pure water is 1.19×104 m.

Step 3: Determine the skin depth in a good conductor and for a typical metal:

(b)

Write the expression for the wavenumber for a good conductor.

k=K

Here, K is the imaginary part.

Hence, the wavelength of the conductors will be,

λ=2πk2πK

Calculate imaginary part K for a good conductor.

K=ωεμ21+σεω2-112K=ωεμ2σεωK=ωμσ2

Hence, the skip depth in a good conductor will be,

d=1ωμσ2 …… (3)

Substitute ω=1015 s-1, μ=4π×10-7 H/m and σ=107 Ωm-1 in equation (3).

d=11015 s-14π×10-7 H/m107 Ωm-12d=1.25×10-8 m×109 nm1md=12.5 nm

As the fields do not penetrate far into the metal, they are opaque in nature.

Therefore, the skin depth in a god conductor is d=λ2π and the skin depth for a typical metal is 12.5 nm.

Step 4: Show that in a good conductor, the magnetic field lags the electric field by 45°:

(c)

For a good conductor:

k=K

Calculate the phase angle between the magnetic field and the dielectric field.

ϕ=tan-1Kkϕ=tan-11ϕ=45°

Take the ratio of the real amplitude of E and B.

B0E0=KωB0E0=εμ1+σεω2B0E0=εμσεωB0E0=σμω

Substitute σ=107 Ωm-1, μ=4π×10-7 H/m and ω=1015 s-1 in the above expression.

B0E0=107 Ωm-14π×10-7 H/m1015 s-1B0E0=1.12×10-7 S/m

Therefore, the ratio of the real amplitude is 1.12×10-7 S/m.

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