Find the radiation resistance of the wire joining the two ends of the dipole. (This is the resistance that would give the same average power loss—to heat—as the oscillating dipole in fact puts out in the form of radiation.) Show thatR=790(dλ)2Ω , whereλ is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm ), should you worry about the radiative contribution to the total resistance?

Short Answer

Expert verified

The radiation resistance of the wire joining the two ends of the dipole is R=790dλ2Ω, and as the resistance is larger than the ohmic resistance of the wire in the radio, we do not need to worry about the radiative contribution to the total resistance.

Step by step solution

01

Expression for the power dissipated across a wire and the average power radiated by an oscillating electric dipole:

Write the expression for the power dissipated across a wire at time t.

P=I(t)2R …… (1)

Here, I is the current through the wire, and R is the resistance.

Write the expression for the average power radiated by an oscillating electric dipole.

<P>=μ0p02ω412Πc …… (2)

Here,μ0 is the magnetic constant,p0 is the maximum dipole moment of the dipole,ω is the angular frequency, and c is the speed of light.

02

Determine the power dissipated across a wire at time t :

Write the expression for the current through a wire at time t.

It=dqtdt …… (3)

Here, qt is the charge at time t, which is given as:

qt=q0cosωt

Substitute qt=q0cosωtin equation (3).

It=ddtq0cosωtIt=-q0ωsinωt

Substitute It=-q0ωsinωtin equation (1).

P=-q0ωsinωt2RP=q02ω2Rsin2ωt

The average of sin2ωtis given as 12.

Hence, the above equation becomes,

P=q02ω2R12P=q02ω2R2........(4)

03

Determine the expression for the radiation resistance:

Write the expression for the maximum dipole moment of the dipole.

p0=q0d

Substitute p0=q0din equation (2).

P=μ0q0d2ω412ΠcP=μ0q02d2ω412Πc.........(5)

Equate equations (4) and (5).

q02ω2R2=μ0q02d2ω412ΠcR=μ0d2ω26πc.......(6)

Write the relation between angular frequency and wavelength.

ω=2πcλ

Substitute ω=2πcλin equation (6).

R=μ0d22πcλ26πcR=μ0d24π2c26πcλ2R=2μ0πc3dλ2........(7)

04

Determine the radiation resistance of the wire joining the two ends of the dipole:

Substitute μ0=4π×10-7H/mand c=3×108m/sin equation (7).

role="math" R=24π×10-7H/mπ×3×10-8m/s3dλ2R=790dλ2Ω.........(7)

05

Check the contribution of the radiation resistance to the total resistance:

Write the expression for the wavelength in terms of speed of light and frequency.

λ=cf

Here, f is the frequency of the FM wavef=100MHz .

Substitutec=3×108m/s andf=100MHz in the above expression.

λ=3×108m/s100MHz×106Hz1MHzλ=3m

Substitute d=5cmand λ=3min equation (7).

R=7905cm×10-2m1cm3m2ΩR=7900.0532ΩR=0.219Ω

The resistance is small, althought it is larger than the ohmic resistance of the wire present in the radio. This suggest that the raditiona effects the final resistance.

Therefore, the radiation resistance of the wire joining the two ends of the dipole is R=790dλ2Ω, and as the resistance is larger than the ohmic resistance of the wire in the radio, we do not need to worry about the radiative contribution to the total resistance.

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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?

Deduce Eq. 11.100 from Eq. 11.99. Here are three methods:

(a) Use the Abraham-Lorentz formula to determine the radiation reaction on each end of the dumbbell; add this to the interaction term (Eq. 11.99).

(b) Method (a) has the defect that it uses the Abraham-Lorentz formula—the very thing that we were trying to derive. To avoid this,F(q) let be the total d-independent part of the self-force on a charge q. Then

F(q)=Fint(q)+2F(q2)

WhereFint is the interaction part (Eq. 11.99), andF(q2) is the self-force on each end. Now,F(q) must be proportional to q2, since the field is proportional to q and the force is qE. SoF(q2)=(14)F(q) Take it from there.

(c) Smear out the charge along a strip of length L oriented perpendicular to the motion (the charge density, then, is λ=qL); find the cumulative interaction force for all pairs of segments, using Eq. 11.99 (with the correspondence q2λdy, at one end and at the other). Make sure you don’t count the same pair twice.

Use the “duality” transformation of Prob. 7.64, together with the fields of an oscillating electric dipole (Eqs. 11.18 and 11.19), to determine the fields that would be produced by an oscillating “Gilbert” magnetic dipole (composed of equal and opposite magnetic charges, instead of an electric current loop). Compare Eqs. 11.36 and 11.37, and comment on the result.

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.

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!]

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