Electromagnetic Waves
EXERCISES, Q1 to Q9
1. Show that the dimensions of the displacement current εₒdⲫE/dt are that of an electric current.
ANSWER: Dimensions of εₒ (electric permittivity of vacuum):-
εₒ =q₁q₂/4πd²Fₑ
[εₒ] =[AT]²/[L²][MLT⁻²]
=[M⁻¹L⁻³T⁴A²]
Dimensions of electric flux:-
φ =EA
[φ] ={[MLT⁻²]/[AT]}[L²]
=[ML³T⁻³A⁻¹]
Dimensions of time t =[T]
Hence the dimensions of displacement current:--
[εₒdⲫE/dt] =[M⁻¹L⁻³T⁴A²][ML³T⁻³A⁻¹]/[T]
=[A]
Which is the dimensions of electric current.
2. A point charge is moving along a straight line with a constant velocity v. Consider a small area A perpendicular to the direction of motion of the charge (figure 40-E1). Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large so the electric field at any instant is essentially given by Coulomb's law. 
The figure for Q-2
ANSWER: Electric field at small area due to the point charge q,
E =(1/4πεₒ)q/x².
Electric flux φ =EA
Hence the displacement current through the area
Id =|εₒdφ/dt|
=|εₒ d(1/4πεₒ)q/x²}A/dt|
=|(qA/4π)(dx⁻²/dt)|
=|(qA/4π){(dx⁻²/dx)(dx/dt)}|
=|(qA/4π){-2x⁻³*v}|
=qAv/2πx³.
3. A parallel-plate capacitor having plate-area A and plate separation d is joined to a battery of emf Ɛ and internal resistance R at t =0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.
ANSWER: Electric field in a parallel plate capacitor,
E =Q/εₒA
Hence the electric flux linked to the given area,
φ =Electric field*area =E*A/2
=(Q/εₒA)*A/2
=Q/2εₒ
Charge on the capacitor at any time t, during charging,
Q =ƐC(1 -e-t/RC)
Hence,
φ =ƐC(1-e-t/RC)/2εₒ
Hence the displacement current at time t,
Id =εₒdφ/dt
=½ƐC{-e-t/RC (-1/RC)}
=Ɛ/2Re-t/RC
But C = Aεₒ/d, hence,
Id =(Ɛ/2R)e-td/εₒAR.
4. Consider the situation of the previous problem. Define displacement resistance Rd =V/id of the space between the plates where V is the potential difference between the plates and id is the displacement current. Show that Rd varies with time as
Rd = R(et/𝜏 -1).
ANSWER: Electric flux between the plates,
φ =EA =(Q/εₒA)*A
=Q/εₒ
Displacement current,
Id =εₒdφ/dt
=εₒd(Q/εₒ)dt
=dQ/dt
Also at time t, Q =CƐ(1-e-t/RC),
→Id =CƐe-t/RC *(1/RC)
=(Ɛ/R)e-t/RC
Displacement resistance,
Rd =Ɛ/Id -R
=Ɛ/{(Ɛ/R)e-t/RC} -R
=Ret/RC -R
=R(et/𝜏 -1).
5. Using B = µₒH find the ratio Eₒ/Hₒ for a plane electromagnetic wave propagating through a vacuum. Show that it has the dimensions of electric resistance. The ratio is a universal constant called the impedance of free space.
ANSWER: From the given relation, in a vacuum, Bₒ =µₒHₒ
The relation between magnetic and electric fields is,
Bₒ =µₒεₒcEₒ
→µₒHₒ =µₒεₒcEₒ
→Eₒ/Hₒ =1/εₒc
=1/(8.85x10⁻¹² C²/Nm²*3x10⁸ m/s)
=(10000/26.55) Nms/C²
=377 Ω.
{Nms/C² =(Nm/C)/(C/s)
=Volt/Ampere =Ohm}
Dimensions of this ratio:--
=Dimensions of 1/εₒc
Unit of εₒ =C²/Nm² =A²s²/Nm²
Its dimensions =[A²T²]/[MLT⁻²L²]
=[M⁻¹L⁻³T⁴A²]
Dimensions of c =[LT⁻¹]
Hence dimensions of the ratio Eₒ/Hₒ =1/[M⁻¹L⁻³T⁴A²][LT⁻¹]
=[ML²T⁻³A⁻²]
Unit of resistance,
Ohm =Nms/C²
=Nm/A²s
=[MLT⁻²L]/[A²T]
=[ML²T⁻³A⁻²]
Hence the ratio Eₒ/Hₒ has the same dimensions as that of the electric resistance.
6. The sunlight reaching the earth has a maximum electric field of 810 V/m. What is the maximum magnetic field in this light?
ANSWER: Given that Eₒ =810 V/m.
The maximum magnetic field,
Bₒ =µₒεₒcEₒ
=4πx10⁻⁷*8.85x10⁻¹²*3x10⁸*810 T
=2.70x10⁻⁶ T
=2.70 µT.
7. The magnetic field in a plane electromagnetic wave is given by
B =(200 µT)sin[(4.0x10¹⁵ s⁻¹)(t -x/c)]
Find the maximum electric field and the average energy density corresponding to the electric field.
ANSWER: From the given wave description, the Maximum value of the magnetic field,
Bₒ =200 µT =2x10⁻⁴ T.
Hence the maximum value of the electric field,
Eₒ =cBₒ
=3x10⁸*2x10⁻⁴ N/C
=6x10⁴ N/c.
Average energy density,
Uₐᵥ =½εₒEₒ²
=½*(8.85x10⁻¹²)*(6x10⁴)² J/m³
=160x10⁻⁴ J/m³
=0.016 J/m³.
8. A laser beam has an intensity of 2.5x10¹⁴ W/m². Find the amplitude of electric and magnetic fields in the beam.
ANSWER: The relation between intensity and maximum electric field is,
I =½εₒEₒ²C.
→Eₒ² =2I/cεₒ
=2*2.5x10¹⁴/(3x10⁸*8.85x10⁻¹²) (N/C)²
=0.188x10¹⁸ (N/C)²
→Eₒ =0.433x10⁹ N/C
=4.33x10⁸ N/C.
Amplitude of the magnetic field,
Bₒ =Eₒ/c
=4.33x10⁸/3x10⁸ T
=1.44 T.
9. The intensity of sunlight reaching the earth is 1380 W/m². Assume this light to be a plane, monochromatic wave. Find the amplitudes of electric and magnetic fields in this wave.
ANSWER: Since I =½εₒEₒ²c
→Eₒ² =2I/εₒc
=2*1380/(8.85x10⁻¹²*3x10⁸) (N/c)²
=104x10⁴ (N/C)²
Eₒ =10.2x10² N/C
=1.02x10³ N/C.
And Bₒ =Eₒ/c
=1.02x10³/3x10⁸ T
=0.34x10⁻⁵ T
=3.40x10⁻⁶ T.
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Links to the Chapters
Links to the Chapters
CHAPTER- 40- Electromagnetic WavesCHAPTER- 39- Alternating CurrentCHAPTER- 38- Electromagnetic Induction
CHAPTER- 37- Magnetic Properties of MatterCHAPTER- 36- Permanent Magnets
CHAPTER- 35- Magnetic Field due to a Current
CHAPTER- 34- Magnetic Field
CHAPTER- 33- Thermal and Chemical Effects of Electric Current
CHAPTER- 40- Electromagnetic Waves
CHAPTER- 39- Alternating Current
CHAPTER- 38- Electromagnetic Induction
CHAPTER- 37- Magnetic Properties of Matter
CHAPTER- 36- Permanent Magnets
CHAPTER- 35- Magnetic Field due to a Current
CHAPTER- 34- Magnetic Field
CHAPTER- 33- Thermal and Chemical Effects of Electric Current
CHAPTER- 32- Electric Current in ConductorsCHAPTER- 31- CapacitorsCHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
OBJECTIVE -I
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-11 to Q-18
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 32- Electric Current in Conductors
CHAPTER- 31- Capacitors
CHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for→ Newton's laws of motion - Objective - I
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
