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(a) Suppose you imbedded some free charge in a piece of glass. About how long would it take for the charge to flow to the surface?

(b) Silver is an excellent conductor, but it’s expensive. Suppose you were designing a microwave experiment to operate at a frequency of1010Hz. How thick would you make the silver coatings?

(c) Find the wavelength and propagation speed in copper for radio waves at role="math" localid="1655716459863" 1MHz. Compare the corresponding values in air (or vacuum).

Short Answer

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Answer

(a) The time in which the charge flows to the surface is 20s.

(b) The thickness of the silver coating is role="math" localid="1655716514781" 6.34×10-4mm.

(c) The wavelength and propagation speed in copper for radio waves is 0.4mmand 400msrespectively, and the wavelength for radio waves in air is 300m, and the propagation speed in copper is losses gradually with respect to the penetration depth.

Step by step solution

01

Expression for the characteristic time and the relativity permittivity of the medium:

Write the expression for the characteristic time.

ζ=εσ …… (1)

Here, ζis the characteristic time, εis the permittivity of the medium, and σis the conductivity.

Write the expression for the relativity permittivity.

εr=εε0ε=εrε0 …… (2)

Here, ε0is the Permittivity of the free space.

02

Determine the time in which the charge flow to the surface:

(a)

Substitute the value of equation (2) in equation (1).

ζ=εrε0σ …… (3)

Write the expression for the relation between relative permittivity and refractive index of the medium.

εr=n2

Here, n is the refractive index of glass n=1.5.

Substitute n=1.5, ε0=8.85×10-12C2/N.mand σ=10-12Ω-1·m-1in equation (3).

ζ=n2ε0σζ=1.52×8.85×10-12C2/N·m10-12Ω-1·m-1ζ=19.91s20sζ=20s

Therefore, the time in which the charge flows to the surface is 20s.

03

Determine the thickness of the silver coating:

(b)

Write the expression for the skin depth.

d=1k …… (4)

Here,kis the imaginary constant.

Write the expression for the value ofk.

d=12πfμ1ρ2d=2ρ2πfμd=ρπfμ

For σ>>>ωε0, the above equation becomes,

k=ωμσ2 …… (5)

Substitute the value of equation (5) in equation (4).

d=12πfμ1ρ2d=2ρ2πfμd=ρπfμ

Here, ρis the resistivity of silver 1.59×10-8Ω·m-1and f is the frequency.

Substitute ρ=1.59×10-8Ω·m-1, f=1010Hz, and μ=4π×10-7H/min the above expression.

d=1.59×10-8Ω·m-1π×1010Hz×4π×10-7H/md=6.34×10-7m×103mm1md=6.34×104mm

Therefore, the thickness of the silver coating is 6.34×104mm.

04

Determine the wavelength of radio waves:

(c)

Write the expression for the wavelength of a radio wave.

λ=2πk

Here,kis the wavenumber.

Substitute the value of equation (5) in the above expression.

λ=2πωμσ2λ=2π22πf1ρμ

Substitute f=1MHz, ρ=1.59×10-8Ω·m-1and μ=4π×10-7H/min the above expression.

λ=2π22π×1MHz×106Hz1MHz×11.59×10-8Ω·m-14π×10-7H/mλ=4×10-4m×103mm1mλ=0.4mm

05

Determine the propagation speed for radio waves:

Write the expression for propagation speed.

v=λf

Substitute λ=4×10-4mand f=1MHzin the above expression.

v=4×10-4×1MHz×106Hz1MHzv=400m/s

Calculate the wavelength in a vacuum.

λ=cfλ=3×108ms1MHz×106Hz1MHzλ=300m

The speed of radio waves is much lesser than the speed of light, which means the propagation speed in copper is losses gradually with respect to the penetration depth.

Therefore, the wavelength and propagation speed in copper for radio waves is 0.4mmand 400msrespectively, and the corresponding values in the air is 300m, and the propagation speed in copper is losses gradually with respect to the penetration depth.

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Most popular questions from this chapter

Suppose

E(r,θ,ϕ,t)=Asinθr[cos(krωt)(1/kr)sin(krωt)]ϕ^

(This is, incidentally, the simplest possible spherical wave. For notational convenience, let role="math" localid="1658817164296" (krωt)u in your calculations.)

(a) Show that E obeys all four of Maxwell's equations, in vacuum, and find the associated magnetic field.

(b) Calculate the Poynting vector. Average S over a full cycle to get the intensity vector I. (Does it point in the expected direction? Does it fall off liker2, as it should?)

(c) Integrate role="math" localid="1658817283737" Ida over a spherical surface to determine the total power radiated. [Answer: 4πA2/3μ0c ]

[The naive explanation for the pressure of light offered in Section 9.2.3 has its flaws, as you discovered if you worked Problem 9.11. Here’s another account, due originally to Planck.] A plane wave traveling through vacuum in the z direction encounters a perfect conductor occupying the region z0, and reflects back:

E(z,t)=E0[cos(kz-ωt)-cos(kz+ωt)]x^,(z>0),

(a) Find the accompanying magnetic field (in the region role="math" localid="1657454664985" (z>0).

(b) Assuming inside the conductor, find the current K on the surface z=0, by invoking the appropriate boundary condition.

(c) Find the magnetic force per unit area on the surface, and compare its time average with the expected radiation pressure (Eq. 9.64).

Question: Obtain Eq. 9.20 directly from the wave equation by separation of variables.

Question:According to Snell's law, when light passes from an optically dense medium into a less dense one the propagation vector bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle

θc=sin-(n2n1)

Then , and the transmitted ray just grazes the surface. If exceeds , there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium ; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.26

Figure 9.28

A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with and

kT=kTsinθTx^+cosθTz^

the only change is that

sinθT=n1n2sinθI

is now greater than, and

cosθT=1-sin2θT

is imaginary. (Obviously, can no longer be interpreted as an angle!)

(a) Show that

ET(r,t)=E0Te-kzeI(kx-ωt)

Where

kωc(n1sinθ1)2-n22

This is a wave propagating in the direction (parallel to the interface!), and attenuated in the direction.

(b) Noting that (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate theirreflection coefficient for polarization parallel to the plane of incidence. [Notice that you get reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]

(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).

(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

Er,t=E0e-kzcoskx-ωty^Br,t=E0ωe-kzksinkx-ωtx^+kcoskx-ωtz^

(e) Check that the fields in (d) satisfy all of Maxwell's equations (Eq. 9.67).

(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction.

Work out the theory of TM modes for a rectangular wave guide. In particular, find the longitudinal electric field, the cutoff frequencies, and the wave and group velocities. Find the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency, for a given wave guide. [Caution: What is the lowest TM mode?]

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