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Introduction to Electrodynamics
Found in: Page 432
Introduction to Electrodynamics

Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

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Short Answer


E(r,θ,ϕ,t)=A sinθr[cos(kr-ωt)-1krsin(kr-ωt)] ϕ

(This is, incidentally, the simplest possible spherical wave. For notational convenience, let(kr-ωt)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 . (Does it point in the expected direction? Does it fall off like r-2, as it should?)

(c) Integrate over a spherical surface to determine the total power radiated. [Answer:4πA2/3μ0c]


The value of divergence of electric field of Maxwell’s equation is .E=0 .

The value of curl of electric field is ×E=1r sinθθsinθEϕr-1rθrEϕθ

The value of magnetic field isB=2Acosθωr2sin u+1krcos ur+Asinθωr-kcos u1kr2cos u+1rsin uθ.

The value of Gauss law of magnetism is .B=0.

The value of Ampere’s law.is 1c2Et=×B.


The value of Intensity vector is I=A2sin2θ2μ0cr2rand the pointing vector S over the full

cycle is S=A2sinθμ0ωr22cosθrsinucos u+1krcos2u-sin2u-1k2r2sinucosuθ-sinθ-kcos2u+1kr2cos2u+1rsinucosu+1rsinucosu-1k2r3sinucosu-1kr2sin2ur

(c) The value of total power radiated is4π3A2μ0c.

See the step by step solution

Step by Step Solution

Step 1: Write the given data from the question.

Consider this, incidentally, the simplest possible spherical wave. For notational convenience, let kr-ωtu in your calculations.

Determine the formula of divergence of electric field of Maxwell’s equation, curl of electric field, magnetic field, Gauss law of magnetism, Ampere’s law, Intensity vector and total power radiated.

Write the formula of electric field is,

.E=1r2r(r2Er)+1rsinθθ(sinθEθ)+1rsinθEϕ …… (1)

Here,role="math" localid="1658485458510" E is the electric field component of a spherical wave, role="math" localid="1658485452985" Eris electric field component of a spherical wave, r is radius, Eθelectric field component of a spherical wave andEϕis electric field component of a spherical wave.

Write the formula of curl of electric field.

×E=-Br …… (2)

Here, Bis the magnetic field strength andris radius.

Write the formula of magnetic field.

B=1rsinθθ[Asin2θr cosu-1krsinu]r-1rr[Asinθcosu-1krsinu] θ …… (3)

Here, r is radius,krepresent the wave number and A is constant.

Write the formula of Gauss law of magnetism.

.B=0 …… (4)

Here, is the magnetic field strength.

Write the formula of Ampere’s law.

×B=μσE+1c2Et …… (5)

Here, E is the electric field component of a spherical wave, μ is permeability, cdenotes the speed of light.

Write the formula of intensity vector.

I=(S) …… (6)

Here, S is Poynting vector.

Write the formula of total power radiated.

P=I.da …… (7)

Here,I is intensity vector.

Step 3: (a) Determine the electric field of Maxwell’s equation.

From Gauss’s law,


Here, ρf is the free charge density.

Determine the divergence of electric field is,

Substitute 0 for Er,Eθ andAsinθrcoskr-ωt-1krsinkr-ωtfor Eϕ.E=1r2rr20+1rsin θrsinθ0+1rsin θAsinθrcoskr-ωt-1krsinkr-ωtϕ =1r sin θEϕ =0 .

As there is no free charge density here, therefore .E=0.

Hence, Gauss’s law is obeyed.

According to Faraday’s Law.

Determine the curl of electric field is,

role="math" localid="1658487525637" ×E=1rsinθθsinθEϕ-Eθϕr+1rrrEϕϕ+1rrrEθ-Erθϕ

Therefore, the value of curl of electric field is


Substitute -Btfor ×E.

-Bt=1rsinθθAsin2θrcosu-1krsin ur-1rtAsin θcos u-1kr-1krsin uθ …… (8)

Here, u=(kr-ωt).

Integrate equation (8),

B=1r sinθθAsin2θrcosu-1krsin ur-1rθAsinθcos u-1krsin uθ …… (9)

Substitute role="math" localid="1658488205000" -1ωsin ucos u dt for cos u dt and 1ωcos ufor sin u dt into equation (9).

B=2Acosθωr2sinu+1krcosur+2Acosθωr2-kcos u1kr2cos u+1rsin u)θ

Therefore, the value of magnetic field is

B=2Acosθωr2sin u+1krcos u)r+Asin θωr-kcos u1kr2cos u+1rsin u)θ

Determine the Gauss’s law of magnetism,

Substitute 2Acos θωr2sin u+1krcos u)r+Asin θωr-kcos u1kr2cos u+1rsin uθ for B into equation (4).

.B=1r2rr2B+1rsinθθsinθBθ =1r2r2Acosθωrsin u+1krcos u +1r sinθrAsin2θωr-kcos u+1kr2cos u+1rsin u =1r22Acosθωk cos u-1kr2cos u-1rsin u +1rsinθ2Asinθcosθωr-k cosu+1kr2cos u+1rsin u

Solve further as

.B=2Acosθωr2k cosu-1kr2cos u-1rsin u-k cosu+1rsin u =0

Hence, the Gauss law of magnetism is obeyed.

Determine the Ampere’s law,

Asσ=0, therefore,

×B=1c2Et =1rrrBθ-Bθϕ

Substitute 2Acosθωr2sin u+1krcos ur+Asin θωr-kcos u1kr2cos u+1rsin uB for B .

×B=1rrAsinθω-k cosu+1kr2cos u+1rsin u-θ2Acosθωr2sin u+1krcos uϕ =kωAsinθωk sin u+1rcos uϕ =Asinθωk sin u+1rcos uϕ

Solve the term 1c2Et,

1c2Et=1c2Asinθrωsin u+ωkrcos uϕ =1c2ωkAsinθrksin u+1rcos uϕ =1cAsinθrksin u+1rcos uϕ =×B

Hence, Ampere’s law is obeyed.

Step 4: (b) Determine the Poynting vector and energy per unit time.

Determine the Poynting vector is given by the following equation.

S=1μ0E×B …… (10)

Substitute 2Acos θωr2sin u+1KRcos ur+Asin θωr-kcos u1kr2cos u+1rsin uθ for B and A sin θr cos u-1krsin uϕ for E into above equation (10).

S=1μ0Asinθrcosu-1krsin uϕx2Acosθωr2sin u+1krcos ur+Asinθωr-kcosu+1kr2cos u+1rsin uθ=A2sinθμ0ωr22cosθrsin ucos u+1rcos2u-sin2u-1k2r2sinucosuθ-sin-kcos2u+1rsinucosu+1rsinucosu-1k2r2sinucosu-1kr2sin2ur

Average over a full cycle is,

sin u cos u=0 sin2 u=cos2u=12

Determine the Intensity vector.

Substitute A2sinμ0ωr2k2sinθr for Sinto equation (6).

I=A2sinμ0ωr2k2sinθr =A2sin2θ2μ0cr2r

The intensity fluctuates as 1r2 and faces in the direction of a. A spherical wave is predicted to behave in this way.

Therefore, the intensity vector isA2sin2θ2μ0cr2r.

Step 5: (c) Determine the total power radiated.

Determine the total power radiated is,

Substitute A2sin2θ2μ0cr2r for I.

role="math" localid="1658492229808" P=A22μ0csin2θr2r2sinθdθdϕ =A22μ0c2π0πsin2θdθ =4π3A2μ0c

Therefore, the value of total power radiated is 4π3A2μ0c .

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