CLS Aipmt 16 17 XII Phy Study Package 6 SET 1 Chapter 8

Chapter

8

Electromagnetic Waves Solutions SECTION - A School / Board Exam. Type Questions Very Short Answer Type Questions : 1.

A capacitor of capacitance C, is being charged up by connecting it across a d.c. voltage source of voltage V. How do the conduction and displacement currents, in this set up compare with each other (i)

During the charging up process?

(ii)

After the capacitor gets fully charged?

Sol. (i) (ii) 2.

During the charging Ic = Id After charging Ic = Id = 0

The velocity of propagation (in vacuum) and the frequency of (i) x-rays and (ii) radio waves are denoted by (vx, fx) and (vR, fR) respectively. How are these values related to each other? (i)

vx and vR

(ii)

fx and fR

Compare with each other. Sol. (i) (ii) 3.

vX = vR fX > fR

Let the wavelengths of electromagnetic waves used quite often for (i)

Killing germs in household water purifiers.

(ii)

Remote sensing.

(iii) Treatment of cancer. be labelled as 1, 2 and 3. Arrange 1, 2 and 3 in increasing order. Sol. 2 > 1 > 3 4.

Name the part of electromagnetic spectrum of wavelength 102 m and mention its one application.

Sol. Radio wave. It is used for communication purpose. 5.

Name the E.M. waves used for studying crystal structure of solids. What is its frequency range?

Sol. X-ray, 1016 Hz to 1020 Hz Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

74 6.

Electromagnetic Waves

Solution of Assignment (Set-1)

 If you find closed loop of B in a region in space, does it necessarily mean that actual charges are flowing across the area bounded by the loops?

Sol. Not, necessarily. 7.

Give difference between displacement current and conduction current.

Sol. Conduction current is due to flow of electrons through a cross-section but displacement current is due to changing of electric flux through a cross-section with time. 8.

Is displacement current like conduction current, a source of magnetic field?

Sol. Yes 9.

The charging current for a capacitor is 0.2 A. What is the displacement current?

Sol. Id = Ic = 0.2 A 10. What modification was made by Maxwell in Ampere’s Circuital Law? Sol. Maxwell modifies Ampere circuital law as  

∫ B.d l   I 0

C

d ⎤ ⎡  Id    0 ⎢IC   0 E ⎥ dt ⎦ ⎣

Short Answer Type Questions :

 11. Show that average energy density of the electric field E equals the average energy density of the magnetic  field B . Sol. Average energy density of E UE 

1 0E02 4

Average energy density of B

1 B02 4 0

UB 

∵ E0 = cB0 then,

UE 

1 1 0 (cB0 )2  0c 2B02 4 4

1 1 2 = 4 0 (  )  B0 0 0

1 B02 = 4 0 UE = UB 12. A variable frequency a.c. source is connected to a capacitor. Will the displacement current increase or decrease with increase in frequency. Give reason for your answer.

1 ⎤ ⎡ Sol. The increase in frequency of a.c., will decrease the reactance of the capacitor ⎢ XC  and hence will C ⎥⎦ ⎣ increase the conduction current. Since the displacement current is equal to the conduction current, therefore displacement current will increase. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

Electromagnetic Waves

75

13. How can you conclude that electromagnetic wave is transverse in nature? Draw a diagram showing the propagation of an electromagnetic wave along the x-axis, indicating clearly the directions of the oscillating electric and magnetic fields associated with it. Sol. Electromagnetic waves can be polarised. Only transverse wave can be polarised hence it is transverse in nature. y

E

B0 C Direction propagation

z

B

x

E0

14. Define the intensity of electromagnetic wave? Write its formula in terms of r.m.s. value of electric field. Sol. Intensity of electromagnetic wave is defined as the amount of energy crossing per unit time per unit area 2 c . perpendicular to the direction of propagation of electromagnetic waves. I   0 E rms

15. Two students A and B prepare the following table about the electromagnetic waves. Rewrite this table in its corrected form. Student

Direction of Electric field

Sol.

Magnetic field

Peak Value of Propagation

Electric field

Magnetic field

A

Along x-axis

Along z-axis

Along y-axis

E0

B0 = cE0

B

Along y-axis

Along x-axis

Along x-axis

E0 = cB0

B0

Student

Direction of

Peak Value of

Electric field

Magnetic Field

Propagation

Electric Field

A

Along x-axis

Along z-axis

Along – ve y-axis

E

B

Along y-axis

Along x-axis

Along – ve z-axis

E = cB

Magnetic Field B=

E c

B

16. The electric field of a plane electromagnetic wave in vacuum is represented by ⎡ x ⎤ Ey = 0.5 cos ⎢ 2  108 ⎛⎜ t – ⎞⎟ ⎥ c ⎝ ⎠⎦ ⎣

(i)

What is the direction of propagation of electromagnetic wave?

(ii)

Determine the wavelength of the wave.

(iii) Calculate the component of associated magnetic field. Sol. (i) (ii)

Along x-axis k =

2 

2  108 2  c 



=

c 3  108  = 3m 8 10 108

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Electromagnetic Waves

(iii) B0 =

Solution of Assignment (Set-1)

E0 0.5  = 1.7 × 10–9 T c 3  108

Bz = 1.7 × 10–9 cos[(2 × 108 (t –

x )] c

17. Calculate displacement current between the square plates of side 1 cm of a capacitor of electric field between the plates is changing at the rate of 3 × 106 V/ms. Sol. Id = 0

d E dE  0 A dt dt

= 8.85  10 –12  10 –4  3  106 = 26.55 × 10–10 A

x⎞ ⎛ 18. A beam of light travelling along x-axis is described by the magnetic field Bz = 5 × 10–9sin ⎜ t – ⎟ T. Calculate c⎠ ⎝ the maximum electric and magnetic force on a charge, i.e. -particle moving along y-axis with speed of 3 × 107 ms–1. Sol. E0 = B0 c = 5 × 10–9 × 3 × 108 = 1.5 Vm–1 Maximum electric force F = qE0 = 2 × 1.6 × 10–19 × 1.5 = 4.8 × 10–19 N Maximum magnetic force F = qVB0 = 2 × 1.6 × 10–19 × 3 × 107 × 5 × 10–9 = 4.8 × 10–20 N 19. The r.m.s. value of electric field of the light coming from the sun is 720 NC–1. Calculate the average total energy density of the electromagnetic wave. Sol. Average total energy density 2 u = 0 Erms

20. What is microwave oven? Explain its principle of working. Sol. Water molecules in food are polar molecules. They act as electric dipole. These dipoles align in the direction of electric field. When electric field is made to oscillate the water molecules also oscillate. When the frequency of electromagnetic oscillations matches the natural frequency of water molecules, resonance takes place. The oscillating water molecules impart thermal motion to the surrounding molecules of food or vegetables, thus they are heated and hence cooked. 21. What does an electromagnetic wave consist of ? On what factors does its velocity in vacuum depend? Sol. Electromagnetic wave consists of time varying electric and magnetic fields, which are perpendicular to each other and also perpendicular to the direction of propagation of wave. Its velocity depends on electric and magnetic properties of medium. 22. Write the order of frequency range and one use of each of the following electromagnetic radiation. (i)

Radiowaves

(ii)

Ultraviolet rays

(iii) Gamma rays Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

Sol. (i)

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77

Radio wave :

(ii)



Frequency – 500 kHz to 1000 MHz



Used in Radio and Television communication.

Ultraviolet Rays : 

Frequency – 1014 Hz to 1016 Hz.



Used in LASIK eye surgery.



UV lamps are used to kill germs in water purifiers.

(iii) Gamma Rays : 

Frequency 1018 Hz to 1022 Hz



Used for cancer therapy and to study about nuclear structure.

23. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 NC–1 and its frequency is 50 MHz. (i) Determine B0, , k and 

  (ii) Write expression for E and B . B0 =

Sol. (i)

E0 120   4  10–7 T c 3  108

 = 2f = 2 × 50 × 106

(ii)

= 3.14 × 108 Hz (iii) k =

 3.14  108   1.04 m–1 c 3  108

c 3  108   0.6  101 = 6 m f 50  106

(iv)

=

(v)

x⎞ ⎛ x Ey = E 0 sin  ⎛⎜ t – ⎞⎟ = 120 sin × 108 ⎜⎝ t – ⎟⎠ c ⎝ c⎠

x⎞ ⎛ Bz = 4 × 10–7sin × 108 ⎜⎝ t – ⎟⎠ c 24. Find the energy stored in a 90 cm length of a laser beam operating at 10 mW. Sol.

c 90 cm

Time taken by wave t 

l 90  10 –2   3  10 –9 s c 3  108

Energy passing through any cross-section in 3 × 10–9 s is U = (10 mW) (3 × 10–9 s) = (10 × 10–3 J/s) (3 × 10–9 s) = 30 × 10–12 J Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Electromagnetic Waves

Solution of Assignment (Set-1)

25. Experimental observations have shown that x-rays (i)

Travel in vacuum with a speed of 3 × 108 ms–1

(ii)

Shows the phenomena of diffraction and can be polarised.

What conclusion is drawn about the nature of X-rays from each of these observations? Sol. X-ray is an electromagnetic wave and transverse in nature. 26. The following table gives the wavelength range of some constituents of the electromagnetic spectrum. Select the wavelength range and name the electromagnetic waves that are (i)

Widely used in the remote switches of household electronic devices.

(ii)

Produced in nuclear reactions. S. No.

Sol. (i)

Wavelength

1

1 mm to 700 nm

2.

400 nm to 1 nm

3.

1 mm to 10 –3 nm

4.

< 10–3 nm

The wavelength range 1 mm to 700 nm is of infrared rays. This is used in remote switches. Wavelength range < 10–3 nm are produced in nuclear reaction. These waves are known as -rays.

(ii)

27. Give the frequency range and source of production of X-rays. Also write two uses of X-rays. Sol. Frequency range of X-ray 1016 – 1020 Hz. It is produced in coolidge tube when high speed electron strike target material. If it used for (a) Cancer treatment (b) To study crystal structure. 28. How are amplitudes of electric and magnetic fields related in E.M. waves? Name the electromagnetic radiation used for viewing object through fog. What is their wavelength range? Sol. The ratio of amplitude of electric field to that of magnetic field gives speed of E.M. waves. Infrared radiation are used for viewing objects through haze and fog. Its wavelength lies between 1mm – 700 nm. 29. A plane electromagnetic wave travelling along the x-direction has a wavelength of 3 mm. The variation in the electric field occurs in the Y-direction with an amplitude 66 Vm–1. Write the equations of electric and magnetic fields as a function of x and t respectively. Sol. Amplitude of magnetic field B0 =

E0 66   22  10 –8 T = 2.2 × 10–7 T c 3  108

 = ck = 3 × 108 ×

2 3  10–3

= 2 × 1011 Hz

x⎤ ⎡ E y = 66 sin2 × 1011 ⎢t – ⎥ c ⎣ ⎦ x⎤ ⎡ Bz = 2.2 × 10–7 sin 2 × 1011 ⎢t – ⎥ c ⎣ ⎦ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

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79

30. Calculate the speed of electromagnetic wave in a medium of dielectric constant 2.25 and relative permeability 4.

Sol. Use the formula V =

1 0  r 0 0



c r r



3  108 4  2.25

 1  108 m/s

Long Answer Type Questions : 31. (a)

Write the order of frequency range and one use of each of the following electromagnetic wave. (i)

Microwave

(ii) UV rays (iii) -rays (b) Sol. (a)

Find wavelength of electromagnetic wave of frequency 5 × 1019 Hz in free space. Give two applications of this wave. (i)

Microwaves : Frequency range – 109 to 1011 Hz Used in – Radar system.

(ii) UV rays Frequency range – 1014 to 1016 Hz Used in food preservatives. (iii) -rays frequency range – 1018 – 1022 Hz Used in – Radiotherapy (b)

c=f× =

c f

 =

3  108 m 5  1019

 = 6 × 10–12 m This wavelength belongs to -rays Applications – (a) Radiotherapy (b) They are used to get information regarding the structure of atomic nuclei. 32. Write the name of different parts of atmosphere. Discuss the uses of electromagnetic waves. Sol. The envelop of gases surrounding the earth’s surface is known as atmosphere which is divided into following parts: (a)

Troposphere : Just above earth surface.

(b)

Stratosphere : It lies between troposphere and mesosphere. Ozone layer lies in it.

(c)

Mesosphere : Above ozone layer and below ionosphere.

(d)

Ionosphere : It is the top most layer. It has kennelly heaviside layer and appleton layer. It plays vital role Radio communication.

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Electromagnetic Waves

Solution of Assignment (Set-1)

Uses of Electromagnetic Waves: (i)

Radiowave and micro-waves : Used in TV and other communications, in microwave oven, radar.

(ii)

Infrared rays : 

For night vision because they can pass through haze, fog and mist.



In secret writings, in green house warming.

(iii) U.V. rays : 

To preserve food, killing bacteria in drinking water.



In finger prints, detecting invisible writings.



To study structure of molecules.

X-rays : (i)

In medical diagnosis like fracture.

(ii)

In radiotherapy

(iii) For locating faults, cracks, flaws in finished metallic materials. (iv)

To study crystal structure.

-rays : (i)

To get information regarding structure of atomic nuclei.

33. Explain the greenhouse effect. Sol. Due to reflection of long wavelength (Infrared radiation) by low altitude clouds and carbon dioxide back in the earth atmosphere there is warming of the atmosphere of the earth. This phenomenon is known as green house effect. The radiations from the sun enter the earth’s atmosphere and heat the objects on the earth. These heated objects emit radiations which are called as heat (infrared) radiations. These heat radiations are long wavelength waves and are reflected back to earth surface by the layer of carbon dioxide and clouds. Therefore these radiations are trapped in the earth’s atmosphere which results into the increase in temperature of the atmosphere near earth surface. It means CO2 and clouds form blanket of atmosphere. Since due to heavy use of automobile and industry the amount of CO2 in the atmosphere is increasing. So more and more heat radiations are entrapped in the atmosphere which results into increase in temperature of earth’s atmosphere day by day. Due to increase in temperature of atmosphere there is melting of glacier, increasing sea level which may cause damage to the cities on sea shore. 34. (a)

Which of the following, if any, can act as a source of electromagnetic waves? Explain each one. (i)

A charge moving with constant velocity

(ii) An electron moving in orbit in H-atom (iii) A charge at rest. (b)

Show that displacement current is equal to conduction current when a capacitor is charged.

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Solution of Assignment (Set-1)

Sol. (a)

(i)

Electromagnetic Waves

81

A charge moving with constant velocity has zero acceleration and according to classical theory of electromagnetism only accelerated charge can produce electromagnetic wave. Therefore a charge moving with constant velocity cannot produce electromagnetic wave.

(ii) When electron in H-atom moves in circular orbit has centripetal acceleration. According to Maxwell’s classical electromagnetism theory, this should radiate electromagnetic wave. But Bohr postulated that classical electromagnetic wave radiation theory dial not hold for atomic size system. Electron in atom are confined to non-radiating orbits called stationary orbit. (iii) A charge in rest has only static electric field which cannot be the source of electromagnetic wave. (b)

Let at an instant charge on the plates of capacitor be Q. Area of plate is A. Electric field between the plates of capacitor

E

Q A0

...(i)

Flux of this field passing through the surface between the plates is E = E × A

–Q

Q = A  A 0

E 

I

Q 0

...(ii)

Displacement current Id is

Id  0 

A

Q A

+ + + + + +

I EMF Source

d  dt

d ⎛Q⎞ = 0 dt ⎜  ⎟ ⎝ 0⎠ 1 dQ = 0   dt 0 Id 

dQ dt

...(iii)

dQ is the rate at which charge is reaching to positive plate of capacitor through conducting wire dt

therefore Ic 

dQ dt

...(iv)

From equations (iii) and (iv) Id = Ic 35. A capacitor made of two circular plates each of radius 15 cm and separated by 5.0 mm. Capacitor is charged by a constant current of 0.23A. Obtain displacement current across the plates given

dV  1.87  109 Vs –1 . dt

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Electromagnetic Waves

Sol. Id =  0

Solution of Assignment (Set-1)

d E d  EA  dE = 0 = 0 A dt dt dt

E=

V d

Id =

 0 A dV d dt

0 = 8.85 × 10–12 C2N–1m–2 r = 15 cm = 15 × 10–2 m d = 5.0 mm = 5 × 10–3 m dV = 1.87 × 109 Vs–1 dt

Id =



8.85  10 –12  3.14  15  10 –2 5  10 –3



2

 1.87  109

= 0.234 A Id  0.23 A 36. (a)

How does a charge q oscillating at certain frequency produce electromagnetic wave? Sketch a schematic   diagram depicting E and B of an electromagnetic wave propagating along x-axis.

(b)

The amplitude of magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is amplitude of electric field part of the wave?

Sol. (a)

A charge q oscillating at certain frequency produces an oscillating electric field in the space which in turn produces an oscillating magnetic field in the same space. This oscillating magnetic field is source of an oscillating electric field. This process is repeated as oscillating electric and magnetic field regenerate each other. As a result an electromagnetic wave propagates through space. The frequency of this E.M. wave is equal to the frequency of oscillation of the charge q.

Y E

E0

C B Z (b)

c

E0 = B 0

0

= cB0

X

B0

E0 = 3 × 108 × 510 × 10–9 NC–1 = 153 NC–1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

37.

Electromagnetic Waves

83

(a) When can a charge act as source of electromagnetic wave? How are the directions of electric and magnetic field vectors in E.M. wave related to each other and to the direction of propagation of wave. Which physical quantity, if any, has the same value for waves belonging to different parts of electromagnetic spectrum? (b)

Calculate ratio of wavelength and speed of X-ray of wavelength 10–10 m and red light of wavelength 6800 Å.

Sol. (a)

A charge particle acts as a source of electromagnetic wave only when it is accelerated e.g. in L.C. oscillation. The direction of electric field and magnetic field vector are perpendicular to each other as well as to the direction of propagation. The speed of all waves has same value.

(b)

Wavelength of X-ray x = 10–10m Wavelength of Red ray R = 6.8 × 103 × 10–10 x 10 –10 1   10 –3  R 6.8  103  10 –10 = 6.8

Cx 1 Speed of x-ray and red light will be same. Therefore C  1 R

38. The four Maxwell’s Equations and Lorentz Force Law (which together constituted the foundation of all classical electromagnetism) are listed below.  

q

(i)

∫ E .d s  

(ii)

∫ B.ds  0

(iii)

  d ∫c E.dl  – dt

(iv)

∫

s

0

 

s

c

 

 ∫ B.ds  s

  d B.dl  0I  0 0 dt

 

 ∫ E.ds  and Lorentz Force F s

   = q E v B





(a)

Give the name associated with the four equations above.

(b)

If magnetic monopole existed, which of the equations would be modified? Suggest how they are modified.

(c)

Which of the four equations shows that electrostatic field cannot form closed loops?

(d)

Which of the four equations shows that magnetic field lines cannot start from a point nor end at a point?

Sol. (a)

(i)

 Gauss’s law in electrostatics

(ii)  Gauss’s law in magnetism (iii)  Faraday’s law of electromagnetic induction (iv)  Modified ampere’s circuital law. (b)

Gauss law in magnetism would be modified, this will be

(c)

Equation (1) i.e., Gauss law in electrostatics.

 

q

∫ E.ds   (d)

 

∫ B.ds  0.

0

Equation (ii) i.e. Gauss law in magnetism.

 

∫ B.ds  0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Electromagnetic Waves

Solution of Assignment (Set-1)

39. Suppose that the electric field part of an electromagnetic wave in vacuum is

 E = 30 cos 1.8 y  5.4  108 t  iˆ in SI units. Calculate (a)

What is direction of motion of Electromagnetic waves?

(b)

  What is amplitude of E and B ?

(c)

What is frequency of electromagnetic waves?

(d)

Write the equation of magnetic field part of the electromagnetic waves.

Sol. (a) (b)

Electromagnetic wave is moving along – ve y-axis. Amplitude of E0 = 30 NC –1 Amplitude of magnetic field B

B0 = (c)

E0 30   10 –7 T c 3  108

 = 5.4 × 108  = 2f f=

5.4  108 2

f=

2.7  108 Hz 

= 0.859 × 108 Hz f = 8.6 × 107 Hz. (d)

 B  10 –7 cos 1.8 y  5.4  108 t kˆ Tesla.





40. Explain each part in brief (i)

On what principle is working of microwave oven based?

(ii)

How are microwaves generated?

(iii) How are X-rays generated? (iv)

How are -rays produced?

(v)

How are radiowaves produced?

(vi)

Radiowaves, Infrared radiation and visible light are parts of electromagnetic radiations. How do they differ from each other?

(vii) In what respect are all parts of electromagnetic spectrum same? (viii) Why are microwaves used in satellite? (ix) How is ozone layer helpful for human being on earth? (x)

Clouds, CO2 are responsible for green house effect. How?

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Solution of Assignment (Set-1)

Sol. (i) (ii)

Electromagnetic Waves

85

It is based on principle of resonance. It is generated by spark discharge.

(iii) By bombarding high energy electrons on a metal target of high atomic no in Coolidge tube. (iv)

During Nuclear deexcitation

(v)

By accelerated charge particle.

(vi)

They have different wavelength and frequency.

(vii) They have same speed 3 × 108 m/s in vacuum. (viii) Microwaves have short wavelength and propagates in straight line. (ix) It absorbs U.V. radiations which is harmful for human being. (x)

They trap heat radiations, the longer wavelength reflected back from earth surface.

41. The terminology of the different parts of electromagnetic spectrum is given in the text. Use formula E = h (for energy of a quantum of radiation : Photon) and obtain the photon energy in unit of eV for (i)

Radio wave (  1 m)

(ii)

Visible light ( = 500 nm)

(iii) X-rays (10–10 m) (iv)

-rays ( = 10–12 m)

What difference do you observe in the energy levels of visible light and -rays? Sol. Energy of photon E = h E= (i)

c⎞ hc ⎛ ⎜∵    ⎟  ⎝ ⎠

Radiowave ( = 1m)

E=

6.63  10 –34  3  108 eV 1  1.6  10 –19

= 1.24 × 10–6 eV (ii)

Visible light ( = 500 nm) E =

=

hc 1  eV  e

6.63  10 –34  3  108 500  10 –9  1.6  10 –19

= 2.5 eV (iii) X-ray  = 10–10 m E =

hc 1  eV  e

6.63  10 –34  3  108 = 10 –10  1.6  10 –19

= 1.24 × 104 eV Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Electromagnetic Waves

(iv)

Solution of Assignment (Set-1)

-rays E=

=

hc 1  eV  e

6.63  10 –34  3  108 10 –12  1.6  10 –19

= 1.24 × 106 eV In visible light energy, levels are spaced by few eV while in case of -rays. This spacing is about 1 MeV. 42. (a)

Show that average energy density of electric field is equal to average energy density of magnetic field in electromagnetic radiation.

(b)

Suppose that the electric field amplitude of an electromagnetic wave is E0 = 150 NC–1 and its frequency is 50 MHz. (i)

Determine B0 (amplitude of magnetic field),  (angular frequency) and  wavelength.

  (ii) Write expression for E and B .

Sol. (a)

E

=

1  0 E 02 4

⎧ E0 ⎫ ⎨∵ c  ⎬ B0 ⎭ ⎩

=

1  0 c2B2 2

{∵ E0 = cB0}

1 ⎛ 1  ⎜ = 2 0⎜   ⎝ 0 0

=

1 B02 4 0

2

⎞ 2 ⎟ B ⎟ ⎠

(This is average energy density due to magnetic field.)

E  B

(b)

(i) B0 =

150 E0 = = 50 × 10–8 = 5 × 10–7 T 3  108 c

 = 2f = 2 × 3.14 × 50 × 106 = 3.14 × 108 rad/sec =

c 3  10 8 3 300    10 2  6m 6 f 50 50 50  10

(ii) Let wave propagates along x-axis

 E is along y-axis and  B is along z-axis  E = E0sin(t – kx) ˆj = 150 sin(3.14 × 108t – 1.05x) ˆj   B = B 0 sin  t – kx  kˆ = 5 × 10–7sin (3.14 × 108t – 1.05x) kˆ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

43. (a) (b) Sol. (a)

Electromagnetic Waves

87

A radio can be tuned to any station in the 6 MHz to 15 MHz frequency band. What is the corresponding wavelength band? Derive expression for displacement current. f1 = 6 MHz = 6 × 106 Hz f2 = 15 MHz = 15 × 106 Hz 1 =

2 =

c 3  108   50m f1 6  106 c 3  108   20 m f2 15  106

Wavelength band is  = 50m – 20m = 30 m (b)

Let at an instant charge on capacitor plate is Q, then electric field between the plate is E

Q A0

Electric flux through the plates  = E × A Q  =  0

...(i)

Charge on capacitor plate changes with time i=

dQ dt

...(ii)

1 dQ d =  dt dt 0 dQ d  0 dt dt Id   0

44. (a)

d dt

Write the relations among the followings. (i)

  Direction of propagation and direction of oscillations of E and B in electromagnetic wave.

(ii) Velocity of electromagnetic wave in vacuum and the permeability and permittivity of free space. (iii) Amplitude of oscillating electric field and magnetic field vector with velocity of electromagnetic wave. (b)

 An electromagnetic wave is travelling in a medium with velocity v  viˆ . The electric field oscillations are along y-axis then (i)

Identify the direction in which magnetic field oscillations are taking place.

(ii) Depict them in diagram. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

88

Electromagnetic Waves

Sol. (a)

Solution of Assignment (Set-1)

  The direction of propagation of electromagnetic wave is in the direction of E  B .



(i)



1 (ii) c =

0 0

E0 (iii) c = B 0

(b)

  E  B = along x-axis

(i)

E0 jˆ  B0Bˆ = v 0 iˆ

Bˆ  kˆ

 It means B must oscillate along z-axis. y

(ii)

E

C

x

B z 45. Sea water at frequency f = 4 × 108 Hz has permittivity  = 80 0, permeability  = 0 and resistivity  = 0.25 -m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage Vt = V0sin2ft. What fraction of the conduction current density is the displacement current density? Sol. Let separation between the plates of the capacitor is d. Then the electric field E=

V d

E=

V0 sin 2ft d

The conduction current density is given by Jcon =

1 v0 sin2ft  d

=

1 v0 sin2ft  d

=

v0 sin2ft d

Jcon = J0(con) sin2ft Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

Electromagnetic Waves

89

The displacement current density is given by Jd = 

dE dt

= 

 ⎛ v0 ⎞ ⎜ sin 2 ft ⎟ t ⎝ d ⎠

 2 fv 0 cos 2 ft d

=

Jd = d 0 d  cos 2 ft

J0  d  J0 con



2f V0 d = 2f  d v0

= 2800 f  0.25 = 40f0

=

f  10 4  108  10 4 = = 9 9 9  10 9  10 9

SECTION - B Model Test Paper Very Short Answer Type Questions : 1.

Name the electromagnetic radiation to which following wavelengths belong: (i)

1Å

(ii)

4800 Å

Sol. (i) (ii) 2.

X-ray Visible light

From the following identify the electromagnetic waves having the (i)

Maximum frequency

(ii)

Minimum frequency

Sol. (i) (ii) 3.

[1 Mark]

Gamma rays have maximum frequency Radiowaves have minimum frequency.

Experimental observations have shown that x-rays travel in vacuum with speed of light. What conclusion can be drawn about X-rays from this observation?

Sol. X-rays are electromagnetic wave. 4.

State the condition under which a microwave oven heats up a food item containing water molecules most efficiently.

Sol. When frequency of microwave matches resonant frequency of water molecules. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

90 5.

Electromagnetic Waves

Solution of Assignment (Set-1)

Name the part of electromagnetic spectrum which is used (i)

in T.V. communication.

(ii)

to maintain the warmth on earth.

Sol. (i) Radiowaves (ii) Infrared radiations. 6.

Is displacement current a source of magnetic field?

Sol. Yes displacement current like conduction current is a source of magnetic field. 7.

Name the physical quantity(s) that oscillate in electromagnetic wave.

Sol. Electric field and magnetic field vectors. 8.

Name for scientists related with history of electromagnetic wave.

Sol. (a) Maxwell (b) Hertz (c) Bose (d) Marconi Short Answer Type Questions : 9.

[2 Marks]

(i)

State two applications of UV rays.

(ii)

Name the electromagnetic radiations used for viewing the objects through haze and fog.

Sol. (i) (ii) 10. (i) (ii)

Sol. (i) (ii) 11. (i) (ii)

Sol. (i) (ii)

(a) Sterilizing the surgical instruments, (b) To preserve the food stuff. Infrared radiations. When -rays and X-rays have same energy, how can they be distinguished? During charging of a capacitor the conduction current is 0.50 A. What is displacement current between the plates of the capacitor? On the basis of mode of production. 0.5 A, because conduction current is equal to displacement current. What is the ratio of speed of UV rays and infrared rays in vacuum? What is harmful effect of UV rays? Which part of electromagnetic spectrum has largest penetrating power? Write its, frequency range, and one use. Both travels with same speed in vacuum. UV rays can cause skin cancer. -rays, frequency – 1018 Hz to 1022 Hz, used to get information regarding nuclear structure.

12. The magnetic field of a beam emerging from a filter kept in front of light source is given as B = 12 × 10–8 sin(1.20 × 107y – 3.6 × 1015 t) T. What is the average intensity of the beam?

Sol. Iav





2

–8 1 B02  3  108 watt 1 12  10  c = = = 1.71 2 0 m2 2 1.26  10 –6

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Solution of Assignment (Set-1)

Electromagnetic Waves

91

13. A variable frequency a.c. source is connected to a capacitor. How will the displacement current change with decrease in frequency?

1 will increase which will lead to decrease in conduction c current. In this case I0 = Ic hence displacement current will decrease.

Sol. On decreasing the frequency, reactance Xc =

14. Magnetic field part of an electromagnetic wave in vacuum is given by

 B = 2 × 10–8 T cos (1.2 y + 3.6 × 108t) ˆj (i)

What is frequency of electromagnetic wave?

(ii)

Write equation for electric field part of electromagnetic wave.

Sol. (i)

 = 3.6 × 108 2f = 3.6 × 108 f=

3.6  108 Hz 2

f = 5.7 × 107 Hz (ii)

E0 = cB0 = 3 × 108 × 2 × 10–8 =6

N C

 N E  6 cos 1.2y  3.6  108 iˆ C





Short Answer Type Questions :

[3 Marks]

15. Write any four characteristics of electromagnetic waves. Give two uses of each of (i)

Radio waves

(ii)

Microwaves

Sol. Four characteristics properties of E.M. waves. (a)

They are transverse in nature.

(b)

They travel through vacuum with speed 3 × 108 m/s.

(c)

They do not require medium for its propagation.

(d)

Electric and magnetic field oscillate perpendicular to each other. (i)

Uses of Radio waves 

Radio Transmission



Radio astronomy

(ii) Uses of Microwaves 

Microwave ovens



Radar system

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Electromagnetic Waves

Solution of Assignment (Set-1)

16. Give reasons for the followings: (i)

Long distance radio broadcast use shortwave bands.

(ii)

The small ozone layer on the top of the stratosphere is crucial for human survival.

(iii) Satellites are used for long distance T.V. transmission. Sol. (i)

Long distance Radio broadcasts use shortwave bands because shortwave bands are easily reflected back to earth by the ionosphere.

(ii)

The small ozone layer on the top of stratosphere is crucial for human survival because ozone layer absorbs the UV rays from the sun and prevents it from reaching the earth and causing damage to life.

(iii) T.V. signals cannot be properly reflected back by ionosphere. Therefore, their reflections is done by satellites. 17. What is the range of frequencies used in satellite communication? What is common between these waves and light waves? Sol. The waves used for satellite communication lie in the following two frequency ranges: (a)

3.7 – 4.2 GHz for downlink

(b)

5.9 – 6.4 GHz for uplink.

The waves used for satellite communication and light waves are electromagnetic waves and they travel in straight line. 18. How does a charge q oscillating at certain frequency produce electromagnets wave? Sketch a schematic diagram depicting electric and magnetic field for an electromagnetic wave propagating along z-axis. Sol. An oscillating charge is an example of accelerated charge. According Maxwell’s theory accelerated charge produces electromagnetic wave. It is produced because oscillating charge produces time varying magnetic field which produces time varying self sustained electric field. This process goes on, producing electromagnetic wave.

y

E

C z

x

B

19. The oscillating magnetic field in a plane electromagnetic wave is given by By = 8 × 10–6 cos(2 × 1011 t + 300x) T (i)

Calculate the wavelength of E.M. wave.

(ii)

Write down the expression for oscillating electric field.

(iii) Calculate frequency of the wave. Sol. (i)

2  300   =

1 m 150

 = 0.0067 m Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

(ii)

Electromagnetic Waves

93

E0 = cB0 = 3 × 108 × 8 × 10–6 = 2400

V m

E2 = 2400 cos(2 × 1011t + 300 x)

V m

(iii) 2 × 1011 = 2f f=

1  1011 

f = 0.3183 × 1011 f = 3.183 × 1010 Hz 20. You are provided with capacitors of 4 F. How would you establish an instantaneous displacement current of 2 mA in the space between them? Sol. IC = C

dV dt

–6 2 × 10–3 = 4  10 

dV dt

dv 2 10 –3   dt 4 10 –6

= 0.5 × 103 = 500 volt/sec By applying a time varying voltage of 500

volt across the capacitor. s

Long Answer Type Questions :

[5 Marks]

x⎞ ⎛ 21. A beam of light travelling along x-axis is described by the magnetic field Bz = 5 × 10–9sin ⎜⎝ t – ⎟⎠ T. Calculate c

the maximum electric and magnetic force on a charge i.e. -particle moving along y-axis with velocity 3 × 107 m/s. Sol. E0 = cB0 = 3 × 108 × 5 × 10–9 = 1.5

v m

Fe = qE0 = 2 × 1.6 × 10–19 × 1.5 = 4.8 × 10–19 N Maximum magnetic force F = qvB0 = 2 × 1.6 × 10–19 × 3 × 107 × 5 × 10–9 = 4.8 × 10–20 N Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Electromagnetic Waves

Solution of Assignment (Set-1)

22. Find wavelength of electromagnetic waves of frequency 6 × 1012 Hz in vacuum. Give its three applications. Sol. c = f ×  =

c f

3  108 = 6  1012

= 0.5 × 10–4 = 5 × 10–5 m These are infrared radiations. Uses :

(i) It keeps earth warm due to green house effect. (ii) It is used to treat muscular strains. (iii) It is used to take photograph during fog and smoke.

23. The relative permittivity and relative permeability of a medium are 4 and 9 respectively. Find speed of electromagnetic wave in this medium.

1  0 r 0 r

Sol. V =

= c

1 0 0  r r

1 r r

c =

=

r r 3  108 4 9

= 0.5 × 108 m/s = 5 × 107 m/s 24. Find the wavelength of electromagnetic wave of frequency 5 × 1019 Hz in free space. Give its two applications. Sol.  =

c f

3  108 = m 5  1019

= 0.6 × 10–11 = 6 × 10–12 m This wavelength corresponds to X-ray. Uses : (i) In radiotherapy to cure untraceable skin diseases and malignant growth. (ii) In engineering, for detecting faults, cracks, flaws and holes in metal products. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

Electromagnetic Waves

95

25. About 2.5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of the radiation? (i)

at a distance of 1 m from the bulb.

(ii)

at a distance of 10 m from the bulb.

Assume that the radiation is emitted isotropically and neglect reflections. Sol. Power of visible radiation P =

2.5  200  5W 100

(i) Intensity at distance of 1m I=

=

P 4 r 2

5 4  3.14  12

= 0.398 × 10–3

W m2

(ii) Intensity at 10 m I=

P 4r 2

5 = 4  3.14  10 2   = 0.00398 = 3.98 × 10–3

W m2

26. The terminology of different part of the electromagnetic spectrum is given in the text. Use formula E = h and obtain the photon energy in units of eV for different parts. In what way are they different scales of photon energies that you obtain related to the source of electromagnetic radiation? Sol. Energy of photon E =

=

6.6  10 –34  3  108 



19.8  10–26 J 

=

19.8  10 –26 eV   1.6  10 –19

=

12.375  10 –7 eV 

hC 

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Electromagnetic Waves

Solution of Assignment (Set-1)

The given table list photon energy of different parts (m)

10

E(eV)

12.375 × 10

3

1

–10

12.375 × 10

10

–7

–3

10

12.375 × 10

–4

–6

12.375 × 10

10

–1

–8

12.375 × 10

10

–10

12.375 × 10

10

3

–12

12.375 × 10

5

The photon energies for different parts of the spectrum of a source indicate the spacing of the relevant energy level of the source.







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