Equation 11.14 can be expressed in “coordinate-free” form by writing p0cosθ=p0·r^. Do so, and likewise for Eqs. 11.17, 11.18. 11.19, and 11.21.

Short Answer

Expert verified

The equation 11.17, 11.18, 11.19, and 11.21 in the coordinate free form are found as:

Vr,t=-ω4πε0cp0·r^rsinωt-rcAr,t=-μ0ω4πp0rsinωt-rcE=-μ0ω24πr^×p0×r^rcosωt-rcB=-μ0ω24πcp0×r^rcosωt-rc

Also,

S=μ0ω432π2cp0×r^2r2r^

Step by step solution

01

Given information:

Given data:

The equation is given as p0cosθ=p0·r^.

02

Express equation 11.14 in coordinate free form:

Write the expression for equation 11.14.

V(r,θ,t)=-p0ω4Πε0c(cosθr)sin[ωt-rc]

Here, r is the radial distance, t is the time,θ is the angle, c is the speed of lightε0 is the susceptibility in free space, andω is the angular frequency.

Substitutep0cosθ=p0·r^ in the above expression.

Vr,t=-ω4πε0cp0·r^rsinωt-rc

03

Express equation 11.17 in coordinate free form:

Write the expression for equation 11.17.

Ar,θ,t=-μ0p0ω4πrsinωt-rcz^

Substitutep0cosθ=p0·r^ in the above expression.

Ar,t=-μ0ω4πp0rsinωt-rc

04

Express equation 11.18 in coordinate free form:

Write the expression for equation 11.18.

E=-V-AtE=-μ0p0ω24πsinθrcosωt-rcθ^

Substitute p0cosθ=p0·r^in the above expression.

E=-μ0ω24πr^×p0×r^rcosωt-rc

05

Express equation 11.19 in coordinate free form:

Write the expression for equation 11.18.

B=×AB=-μ0p0ω24πsinθrcosωt-rcϕ^

Substitutep0cosθ=p0·r^ in the above expression.

B=-μ0ω24πcp0×r^rcosωt-rc

06

Express equation 11.21 in coordinate free form:

Write the expression for equation 11.21.

S=μ0p02ω432π2csin2θr2r^

Substitutep0cosθ=p0·r^in the above expression.

S=μ0ω432π2cp0×r^2r2r^

Therefore, the equations 11.17, 11.18, 11.19, and 11.21 in the coordinate free form are found as:

Vr,t=-ω4πε0cp0·r^rsinωt-rcAr,t=-μ0ω4πp0rsinωt-rcE=-μ0ω24πr^×p0×r^rcosωt-rcB=-μ0ω24πcp0×r^rcosωt-rc

Also,

S=μ0ω432π2cp0×r^2r2r^

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Most popular questions from this chapter

A radio tower rises to height h above flat horizontal ground. At the top is a magnetic dipole antenna, of radius b, with its axis vertical. FM station KRUD broadcasts from this antenna at (angular) frequency ω, with a total radiated power P (that’s averaged, of course, over a full cycle). Neighbors have complained about problems they attribute to excessive radiation from the tower—interference with their stereo systems, mechanical garage doors opening and closing mysteriously, and a variety of suspicious medical problems. But the city engineer who measured the radiation level at the base of the tower found it to be well below the accepted standard. You have been hired by the Neighborhood Association to assess the engineer’s report.

(a) In terms of the variables given (not all of which may be relevant), find the formula for the intensity of the radiation at ground level, a distance R from the base of the tower. You may assume that bc/ωh. [Note: We are interested only in the magnitude of the radiation, not in its direction—when measurements are taken, the detector will be aimed directly at the antenna.]

(b) How far from the base of the tower should the engineer have made the measurement? What is the formula for the intensity at this location?

(c) KRUD’s actual power output is 35 kilowatts, its frequency is 90 MHz, the antenna’s radius is 6 cm, and the height of the tower is 200 m. The city’s radio-emission limit is 200 microwatts/cm2. Is KRUD in compliance?

A charged particle, traveling in from along the x axis, encounters a rectangular potential energy barrier

U(x)={U0,if0<x<L,0,otherwise}

Show that, because of the radiation reaction, it is possible for the particle to tunnel through the barrier-that is, even if the incident kinetic energy is less thanU0, the particle can pass through 26.[Hint: Your task is to solve the equation

a=τa˙+Fm

Subject to the force

F(x)=U0[δ(x)+δ(xL)]

Refer to Probs. 11.19 and 11.31, but notice that this time the force is a specified function ofx, nott. There are three regions to consider: (i)x<0, (ii) 0<x<L, (iii)x>L. Find the general solution fora(t), υ(t), andx(t)in each region, exclude the runaway in region (iii), and impose the appropriate boundary conditions at x=0andx=L. Show that the final velocity (υf)is related to the time spent traversing the barrier by the equation

,L=υfTU0mυf(τeT/τ+Tτ)

and the initial velocity (atx=) is

υi=υfU0mυf[111+υ0mvf2(eT/τ1)]

To simplify these results (since all we're looking for is a specific example), suppose the final kinetic energy is half the barrier height. Show that in this case

υi=υf1(L/υfτ)

In particular, if you choose L=υfτ/4 , then υi=(4/3)υf, the initial kinetic energy is (8/9)U0, and the particle makes it through, even though it didn't have sufficient energy to get over the barrier!]

(a) A particle of charge qmoves in a circle of radiusRat a constant speedv. To sustain the motion, you must, of course, provide a centripetal forcemv2Rwhat additional force (Fe) must you exert, in order to counteract the radiation reaction? [It's easiest to express the answer in terms of the instantaneous velocityv.] What power (Pe) does this extra force deliver? ComparePewith the power radiated (use the Larmor formula).

(b) Repeat part (a) for a particle in simple harmonic motion with amplitudeand angular frequency:ω.ω(t)=Acos(ωt)z Explain the discrepancy.

(c) Consider the case of a particle in free fall (constant accelerationg). What is the radiation reaction force? What is the power radiated? Comment on these results.

An ideal electric dipole is situated at the origin; its dipole moment points in the z direction and is quadratic in time:

p(t)=12p¨0t2z^    (<t<)

wherep¨0is a constant.

  1. Use the method of Section 11.1.2 to determine the (exact) electric and magnetic fields for all r > 0 (there's also a delta-function term at the origin, but we're not concerned with that).
  2. Calculate the power, P(r,t), passing through a sphere of radius r.
  3. Find the total power radiated (Eq. 11.2), and check that your answer is consistent with Eq. 11.60.21

A particle of mass m and charge q is attached to a spring with force constant k, hanging from the ceiling (Fig. 11.18). Its equilibrium position is a distance h above the floor. It is pulled down a distance d below equilibrium and released, at timet=0.

(a) Under the usual assumptions (dλh), calculate the intensity of the radiation hitting the floor, as a function of the distance R from the point directly below q. [Note: The intensity here is the average power per unit area of floor.]

FIGURE 11.18

At whatR is the radiation most intense? Neglect the radiative damping of the oscillator.

(b) As a check on your formula, assume the floor is of infinite extent, and calculate the average energy per unit time striking the entire floor. Is it what you’d expect?

(c) Because it is losing energy in the form of radiation, the amplitude of the oscillation will gradually decrease. After what timebhas the amplitude been reduced to d/e? (Assume the fraction of the total energy lost in one cycle is very small.)

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