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Use the result of Prob. 10.34 to determine the power radiated by an ideal electric dipole, , at the origin. Check that your answer is consistent with Eq. 11.22, in the case of sinusoidal time dependence, and with Prob. 11.26, in the case of quadratic time dependence.

Short Answer

Expert verified

The power radiated by an ideal electric dipole at the origin isP=0p02412c, and the obtained result is consistent with equation 11.22.

Step by step solution

01

Expression for the power radiated by an ideal electric dipole:

Write the expression for the power radiated by an ideal electric dipole.

P=Srda 鈥︹ (1)

Here, Sris the Poynting vector of an ideal dipole which is given as:

Sr=10ErBr 鈥︹ (2)

Here,Sr is the electric field of the ideal dipole and is the magnetic field of the ideal dipole.

02

Determine the Poynting vector of an ideal dipole:

Write the expression for a non-static ideal dipole in an electric field.

Er,t=-04p,,-r^r^-p,,r+C2p+rcp,r^p+rcp,r

Here, 0is the permeability of free space, p is the dipole moment, r is the distance of the source, and c is the speed of light.

Write the expression for a non-static ideal dipole in the magnetic field.

Br,t=-04r^p,+rcp,,r2

Determine the total power radiated over the sphere with the radius r is,

P=Sda=002432c2sin2r2sindd=002412c

Write the radiation for the fixed time at the origin t=t-rcfrom the electric and magnetic dipole moment.

Er=-04p-r^,,r^p,,r2Br=-04r^p,,rc

Substitute the value of Erand Brin equation (2).

Sr=10-04p,,-r^r^p,,r-04r^p,,rcSr=0162cr2p,,-r^r^p,,r^p,,Sr=0162cr2p,,-r^r^p,,-r^p,,r^r^p,,Sr=0162cr2p,,2-r^p,,r^

03

Determine the power radiated by an ideal electric dipole: 

Substitute the value ofSrin equation (1).

P=0162cr2p2-r^p,,,,2r^-daP=0162cr2p2-r^p,,,,2r^-r2sindd

Here,p2-r^p,,,,2=p2sin2,,

Hence, the equation becomes,

P=0162cr2p..1r2sin2r2sinddP=0162cr2p2..40sin3dP=0pp2..6c 鈥︹ (3)

Write the expression for the electric dipole moment in the case of sinusoidal time dependence.

p=p0cost

Take the double differentiation of the above equation.

p.=-p0sintp..=-2p0cost

Squaring on both sides,

p2..=-4p20cos2t

Take the time average of .

p2..=4p20cos2tp2..=4p20cos2tcos2t=12p2..=124p02

Substitute the value of p..2in equation (3).

P=0124p026cP=0p02412c

Hence, the result is consistent with equation 11.22.

Therefore, the power radiated by an ideal electric dipole at the origin is, P=0p02412c0and the obtained result is consistent with equation 11.22.

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Most popular questions from this chapter

we calculated the energy per unit time radiated by a (non-relativistic) point charge- the Larmor formula. In the same spirit:

(a) Calculate the momentum per unit time radiated.

(b) Calculate the angular momentum per unit time radiated.

A point charge q, of mass m, is attached to a spring of constant k.Y2<<0Attimet=0it is given a kick, so its initial energy is U0=12mv02. Now it oscillates, gradually radiating away this energy.

(a) Confirm that the total energy radiated is equal to U0. Assume the radiation damping is small, so you can write the equation of motion as and the solution as

role="math" localid="1658840767865" x+y+x+02x=0,

and the solution as

x(t)=v00e-yt/2sin(0t)

with 0k/m,Y=02T, and Y2<<0 (drop Y2in comparison to 02, and when you average over a complete cycle, ignore the change in e-y).

(b) Suppose now we have two such oscillators, and we start them off with identical kicks. Regardless of their relative positions and orientations, the total energy radiated must be 2U0. But what if they are right on top of each other, so it's equivalent to a single oscillator with twice the charge; the Larmor formula says that the power radiated is four times as great, suggesting that the total will be 4U0. Find the error in this reasoning, and show that the total is actually2U0, as it should be.

(a) Does a particle in hyperbolic motion (Eq. 10.52) radiate? (Use the exact formula (Eq. 11.75) to calculate the power radiated.)

(b) Does a particle in hyperbolic motion experience a radiation reaction? (Use the exact formula (Prob. 11.33) to determine the reaction force.)

[Comment: These famous questions carry important implications for the principle of equivalence.]

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鈥攖he 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 =qL); find the cumulative interaction force for all pairs of segments, using Eq. 11.99 (with the correspondence q2dy, at one end and at the other). Make sure you don鈥檛 count the same pair twice.

(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:.(t)=Acos(t)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.

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