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

Find the angle θmax at which the maximum radiation is emitted, in Ex. 11.3 (Fig. 11.13). Show that for ultra relativistic speeds ( υclose toc), θmax(1β)/2. What is the intensity of the radiation in this maximal direction (in the ultra relativistic case), in proportion to the same quantity for a particle instantaneously at rest? Give your answer in terms ofγ.

Apply Eqs. 11.59 and 11.60 to the rotating dipole of Prob. 11.4. Explain any apparent discrepancies with your previous answer

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?

Assuming you exclude the runaway solution in Prob. 11.19, calculate

(a) The work done by the external force,

(b) The final kinetic energy (assume the initial kinetic energy was zero),

(c) The total energy radiated.

Check that energy is conserved in this process.

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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