Electromagnetic Waves
Questions for Short Answer
1. In a microwave oven, the food is kept in a plastic container and the microwave is directed towards the food. The food is cooked without melting or igniting the plastic container. Explain.
ANSWER: The food we keep in a plastic container in a microwave oven is generally rich in water content. Water molecules are bipolar but not the molecules of the plastic container or air. When bipolar molecules come in the way of microwaves (Which is electromagnetic radiation), they try to align themselves along the electric field by rotating. But the electric field is an oscillating one. So the bipolar molecules are continuously aligning themselves by rotating. In this process, they hit nearby molecules, thus increasing the kinetic energy of the molecules. It results in increasing the temperature of the water/bipolar materials in the food. The plastic container's molecules are unaffected as there is no polarity.
Note: It is not true that the frequency of microwaves matches the frequency of water molecules and resonates.
2. A metal rod is placed along the axis of a solenoid carrying a high-frequency alternating current. It is found that the rod gets heated. Explain why the rod gets heated.
ANSWER: Due to the alternating current in the solenoid, the magnetic flux inside it is everchanging and its direction is also reversing. In such a magnetic field when the metal rod is placed, emf is induced in it, and current flows on its surface. Since there is no guiding path for the current, it seeks its own path and flows in many irregular loops. These are called eddy currents. Also due to the reversing direction of the alternating current, the direction of eddy currents changes. These produce thermal energy and the metal rod gets heated.
3. Can an electromagnetic wave be deflected by an electric field? By a magnetic field?
ANSWER: An electromagnetic wave do not carry a charge or charged particles, hence it can not be deflected by an electric or magnetic field.
4. A wire carries an alternating current i =iₒ sin ωt. Is there an electric field in the vicinity of the wire?
ANSWER: Due to the moving charge in a wire, a magnetic field is created around it. In alternating current, this magnetic field is always varying. The varying magnetic field creates an electric field in the vicinity of the wire.
5. A capacitor is connected to an alternating current source. Is there is a magnetic field between the plates?
ANSWER: Due to the alternating current, the electric field between the plates changes with time. The changing electric field between the plates produces a magnetic field. Hence there will be a magnetic field between the plates.
6. Can an electromagnetic wave be polarized?
ANSWER: Since an electromagnetic wave is a transverse wave, it can be polarized. In an unpolarized electromagnetic wave, the planes of the electric field and the magnetic field of different waves are randomly arranged. It can be polarized by restricting only the waves to pass through one plane and all these have the same planes of electric and magnetic fields.
7. A plane electromagnetic wave is passing through a region. Consider the quantities (a) electric field, (b) magnetic field, (c) electrical energy in a small volume, and (d) magnetic energy in a small volume. Construct pairs of the quantities that oscillate with equal frequencies.
ANSWER: In a plane electromagnetic wave, the electric field is given as
E =Eₒ sin ω(t -x/c)
and the magnetic field is given as
B =Bₒ sin ω(t -x/c).
We see that both the electric and magnetic fields oscillate with the same angular frequency and hence with the same frequency.
The electrical energy stored in a small volume dV is
Ue =½ϵₒ E²dV
=½ϵₒ {Eₒsin ω(t -x/c)}²dV
=½ϵₒ Eₒ²{½(1-cos 2ω(t -x/c))}dV
=¼ϵₒ Eₒ²{1-cos 2ω(t -x/c)}dV
The magnetic energy stored in a small volume dV is,
Ub =(1/2µₒ)B²dV
=(1/2µₒ){Bₒ sin ω(t -x/c)}²dV
=(1/4µₒ)Bₒ²{1 -cos 2ω(t -x/c)}²dV
We see that both Ue and Ub have the same angular frequency =2ω and hence the same frequency =2ω/2π =ω/π.
So the electrical energy and magnetic energy in a small volume oscillate with the same frequency.
OBJECTIVE-I
1. A magnetic field can be produced by
(a) moving charge
(b) a changing electric field
(c) none of them
(d) both of them
ANSWER: (d).
EXPLANATION: A magnetic field can be produced both by a moving charge and a changing electric field. The current by the moving charge is called conduction current and due to the changing electric field is called the displacement current. Option (d) is correct.
2. A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through resistance. The compass needle
(a) does not deflect
(b) deflects for a very short time and then comes back to the original position
(c) deflects and remains deflected as long as the battery is connected
(d) deflects and gradually comes to the original position in a time that is large compared to the time constant.
ANSWER: (d).
EXPLANATION: When the capacitor is connected to a battery, the plates start to be charged. During this charging, there is a changing electric field between the plates. This results in a magnetic field between the plates. So the compass needle deflects. Up to the time constant 𝝉 =CR, the plates are charged up to 63%. For the rest of 37% of charging, the time taken is many times longer than 𝝉. So the rate of charging of the plates (and hence the rate of change of electric field between the plates) up to the time t =𝝉 is comparatively faster than the rest of the charging time. Since the magnetic field between the plates depends upon the rate of change of the electric field (which is highest initially), the magnetic field created initially is maximum and then it decreases. So the compass needle deflects initially to maximum and then gradually comes to the original position (when the plates are fully charged and the rate of the change is almost zero). The time taken to come to the original position is many times more than the time constant, as discussed above. Hence option (d) is correct.
3. Dimensions of 1/(µₒεₒ) is
(a) L/T
(b) T/L
(c) L²/T²
(d) T²/L²
ANSWER: (c).
EXPLANATION: The speed of an electromagnetic wave is given as,
c =1/√(εₒµₒ)
→1/(εₒµₒ) =c²
Since the dimensions of speed c are LT⁻¹, the dimensions of
1/(εₒµₒ) =(LT⁻¹)² =L²/T².
Hence option (c) is correct.
4. Electromagnetic waves are produced by
(a) a static charge
(b) a moving charge
(c) an accelerating charge
(d) chargeless particles.
ANSWER: (c).
EXPLANATION: Electromagnetic waves are produced by an accelerated charge because an accelerated charge produces a changing electric field and a changing magnetic field which constitutes the wave.
Hence option (c) is correct.
5. An electromagnetic wave going through a vacuum is described by
E =Eₒ sin(kx-⍵t); B =Bₒsin(kx-⍵t)
Then
(a) Eₒk =Bₒ⍵
(b) EₒBₒ =⍵k
(c) Eₒ⍵ =Bₒk
(d) none of these
ANSWER: (a).
EXPLANATION: Since Eₒ/Bₒ =c
and ⍵/k =c, where k is the wave number.
Thus, ⍵/k =Eₒ/Bₒ
→Eₒk =Bₒ⍵.
Option (a) is correct.
6. An electric field E and a magnetic field B exist in a region. The fields are not perpendicular to each other.
(a) This is not possible.
(b) No electromagnetic wave is passing through the region.
(c) An electromagnetic wave may be passing through the region.
(d) An electromagnetic wave is certainly passing through the region.
ANSWER: (c).
EXPLANATION: An electric field E and a magnetic field B may exist independently in a region at any angle. Hence option (a) is not correct.
The mutually perpendicular and varying electric and magnetic fields of an electromagnetic wave are linked to each other and they are not affected by the electric and magnetic fields of other sources. Hence an electromagnetic wave can pass unaffected in such external fields. So in the given condition, an electromagnetic wave may be passing through the region. Option (c) is correct. Options (b) and (d) are not binding and hence not correct.
7. Consider the following two statements regarding a linearly polarized, plane electromagnetic wave:
(A) The electric field and the magnetic field have equal average values.
(B) The electric energy and the magnetic energy have equal average values.
(a) Both A and B are true.
(b) A is false but B is true
(c) B is false but A is true
(d) Both A and B are false.
ANSWER: (a).
EXPLANATION: The electric and magnetic fields are sinusoidal waves in an electromagnetic wave. In a single time period, the average value of the sinusoidal wave is zero. If we take the average value over a long time, the value of the average field will be the same Hence both the electric and magnetic fields have an equal average value equal to zero. Statement A is true.
In an electromagnetic wave, the average electric field energy density over a long time is
Ue =¼εₒEₒ²
and average magnetic field energy density over a long time is
Ub ={1/(4µₒ)}*Bₒ²
Since Eₒ =cBₒ, and µₒεₒ =1/c²
Ub =(εₒc²/4)*(Eₒ/c)² =¼εₒEₒ² =Ue
So average electric and magnetic field energy value is equal. Statement B is also true.
Hence option (a) is correct.
8. A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving
(a) along the electric field
(b) along the magnetic field
(c) along the direction of propagation of the wave
(d) In a plane containing the magnetic field and the direction of propagation.
ANSWER: (a).
EXPLANATION: When a free electron is placed in the path of an electromagnetic wave, it will only be affected by the electric field. A charge at rest is unaffected by a magnetic field. So the electron will start moving along the electric field. Option (a) is correct.
9. A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E.
(a) p =0, E ≠0.
(b) p ≠0, E =0.
(c) p ≠0, E ≠0.
(d) p =0, E =0.
ANSWER: (c)
EXPLANATION: The relation between momentum p and energy E is
p =E/c, where c is the speed of light.
Clearly, options (a) and (b) are incorrect.
Also, an electromagnetic wave always carries energy, hence it has some non-zero momentum from the above relation. Thus option (d) is not correct.
Only option (c) is correct.
OBJECTIVE-II
1. An electromagnetic wave going through the vacuum is described by
E =Eₒ sin(kx -⍵t).
Which of the following is/are independent of the wavelength?
(a) k
(b) ⍵
(c) k/⍵
(d) k⍵.
ANSWER: (c).
EXPLANATION: k =2π/λ, So it depends on the wavelength. Option (a) is incorrect.
⍵ =2π𝜈 =2πc/λ
{Since c =𝜈λ}
So ⍵ also depends on the wavelength. Option (b) is incorrect.
k/⍵ =(2π/λ)/(2πc/λ)
=1/c.
It is free of wavelength. So option (c) is correct.
k⍵ =(2π/𝜆)*(2πc/𝜆)
=4π²c/𝜆².
It also depends on wavelength. Option (d) is incorrect.
2. Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor
(a) increases
(b) decreases
(c) does not change
(d) is zero.
ANSWER: (a), (b).
EXPLANATION: Electric field between the plates, E =Q/(εₒA)
The flux of this field through area A,
φE ={Q/(εₒA)}*A =Q/εₒ
Displacement current is,
iᵨ =εₒ.dφE/dt
=εₒ.d(Q/εₒ)/dt
=dQ/dt
So the displacement current is proportional to the rate of change of the charge accumulated on the capacitor plate. If dQ/dt = zero, then the displacement current is also zero. So options (c) and (d) are incorrect. Options (a) and (b) are correct.
3. The speed of electromagnetic waves is the same
(a) for all wavelengths
(b) in all media
(c) for all intensities
(d) for all frequencies.
ANSWER: (c).
EXPLANATION: Though the speed of the electromagnetic waves in a vacuum is independent of their wavelength and frequency, it depends on the vacuum wavelength and frequency of the wave in a medium. Hence options (a), (b), and (d) are incorrect.
Since the intensity of a wave is defined as the energy crossing per unit area per unit time perpendicular to the direction of propagation of the wave, the speed of the wave does not depend on it. Option (c) is correct.
4. Which of the following has zero average value in a plane electromagnetic wave?
(a) electric field
(b) magnetic field
(c) electric energy
(d) magnetic energy.
ANSWER: (a), (b).
EXPLANATION: Since the electric field and the magnetic field in a plane electromagnetic wave are sinusoidal waves, their average value will be zero. Options (a) and (b) are correct.
The electric energy density and the magnetic energy density are equal and given as
uₐᵥ =½εₒEₒ² =(1/2µₒ)Bₒ²
Here εₒ and µₒ are non-zero constants, also Eₒ² and Bₒ² have positive values, hence average electric energy and magnetic energy are not zero. Options (c) and (d) are incorrect.
5. The energy contained in a small volume through which an electromagnetic wave is passing oscillates with
(a) zero frequency
(b) the frequency of the wave
(c) half the frequency of the wave
(d) double the frequency of the wave.
ANSWER: (d).
EXPLANATION: Total energy per unit volume of the wave,
u = ½εₒE²+B²/2µₒ
=½εₒEₒ²sin²(⍵t-kx)+(1/2µₒ)Bₒ²sin²(⍵t-kx)
=½εₒEₒ²{1-cos2(⍵t-kx)}+(Bₒ²/2µₒ){1-cos2(⍵t-kx)}
So the oscillation of energy has an angular frequency of 2⍵ i.e. double the frequency of the wave. So option (d) is correct and the rest incorrect.
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 dimension 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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