# 9 - Electromagnetic wave - Theory & Solved Exam..Module-4pdf

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### 9 - Electromagnetic wave - Theory & Solved Exam..Module-4pdf

• 1. ELECTROMAGNETIC WAVES Total number of questions in this chapter are : (i) In chapter Examples ....................... 11 (ii) Solved Examples ....................... 06 Total no. of questions ....................... 17
• 2. 1. BASIC EQUATIONS OF ELECTRICITY AND MAGNETISM The whole concept of electricity and magnetism can be explained by the four basic equations we have deal so far. (1)  ds . E = 0 Q  (Gauss law for electrostatic) (2)  ds . B = 0 (Gauss law for magnetism) (3)  dl . B = µ0i (Ampere’s law for Magnetism) (4)  dl . E = 0 (Ampere’s law for electrostaties) The above stated equation are true for non-time varying fields 2. FARADAYS LAW FOR TIME VARYING MAGNETIC FIELD To understand the concept of faradays law we consider a circular conducting loop placed in a region where time dependent magnetic field is present x x x x x x x x x x x E E E E     e e e e Conducting loop Time dependent magnetic field is switched on at t = 0 From the earlier concept we know that an induced emf will be produced in the conducting loop due to which current will flow in the loop. For current to flow a force must act on the electron which will move then from static state. This force cannot be due to magnetic field (since magnetic force does not act on stationary charge). Hence this force must be due to an electric field which has been generated due to changing Magnetic field. Note :- This electric field is non conservative in nature. Faraday stated this fact in his equation  dl . E = –        dt d B 3. CONCEPT OF DISPLACEMENT CURRENT (MODIFIED AMPERE’S LAW) Maxwell tried to generalis the concept of faradays law that if changing magnetic field can produce changing electric field then the reverse should also be true i.e. changing electric field must produce magnetic fied. To understand the concept of displacement current let us try to understand this experiment when the switch was closed at t = 0 both the needles deflected. V t = 0 Magnetic Magnetic Needle (2) Parallel plate capacitor Needle (1) • Deflection of needle (1) is under stood as M.F. is produced due to current flowing in the wire. But why did needle 2 deflect? It is lying in between the two plates of capacitor where there is no current. This magnetic field between the plates is due to the changing electric field between the plates (During charging of capacitor). Hence maxwell conducted that changing electric field produces a magnetic field For Needle (1) Amper’s law  dl . B = µ0ic ..... (1) For needle (2) Amper’s law  dl . B = µ00 dt d E  ..... (2) Hence there are two methods of producing M.F. (a) Due to flow of electron which is known as conduction current (b) Due to changing electric field combining eq. (1) and eq. (2)  dl . B = µ0                dt d i E 0 C Modifield ampere’s law Note : dt d E 0   is known as displacement current)
• 3. Displacement current Ex. 1 A parallel plate capacitor with plate area A and seperation between the plated d, is charged by a constant current i, consider a plane surface of area A/4 parallel to the plates and drawn symetrically between the plates what is the displacement current through this area. (A) i (B) 2i (C) i/4 (D) i/2 Sol.(C) Electric field between the plates of the capacitor is given by E = 0   or 0 A q  Flux through the area considered  = 0 A q  × 4 A = 0 4 q  Displacement corrent id = 0 dt d E  = 0 × dt d         0 4 q = 4 i Ex. 2 The charging current for a capacitor is 1 A then what is the displacement current ? (A) 1 A (B) 0.5 A (C) 0 (D) 2 A Sol. (A) Electric field between the plates is E = A Q 0   E = E. A or A Q 0  × A A  id = 0 dt d E  or 0 dt d         0 Q  id = dt dQ = i (charging current) Hence id = 1A Ans. 4. FINAL FORM OF MAXWELL’S EQUATION (a)  ds . E = 0 q  (b)  ds . B = 0 (c)  dl . E = – dt d B  (d)  dl . B = µ0          dt d E 0 I The above equation is known as maxwell’s equation for time varying form. Howover for free space there are no charges and no conduction current the equations that are significant.         dt d µ dl . B dt d dl . E E 0 0 B Solving these two differential equation the equation of electric field and magnetic field that satisfies these differential equations are obtained Ey = E0 sin (t – x/c) Bz = B0 sin (t – x/c) Here 0 0 B E = C and 0 0 µ 1  = C 5. TRANSEVERSE NATURE OF ELECTRO MAGNETIC WAVE AND ITS PROPERTIES 5.1 Electromagnetic waves : The idea of electromagnetic waves was given by Maxwell and experimental verification was provided by Hertz and other scientists. A brief history of electromagnetic waves is as follows: On the basis of experimental study of electromagnetic induction, Faraday concluded that a magnetic field changing with time at a point produces a time varying electric field at that point. Maxwell in 1864 pointed out an electric field changing with time at a point also produces a time varying magnetic field. The two fields are mutually perpendicular to each other. This idea led Maxwell to conclude that the mutually perpendicular time varying electric and magnetic fields produce electromagnetic disturbances in space. These disturbances have the properties of wave which are called as electromagnetic waves. According to Maxwell, the electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vectors at right angle to each other as well as right angles to the direction of wave propagation. An electromagnetic wave is shown in fig.
• 4. Z Y X Direction of propagation E B E B E B Envelope of magnetic induction vector Envelope of electric intensity vector The velocity of electromagnetic wave in free space is given by c = ) µ ( 1 0 0  where µ0 (1 = 1.257 × 10–6 T mA–1) and 0 (= 8.854 × 10–12 C2N–1m–2) are permeability and permittivity of free space respectively. The velocity of electromagentic waves in free space is equal to the velocity of light. Therefore, light is electromagnetic waves. The electromagnetic waves are transverse in nature. 5.2 Transverse nature of electromagnetic waves: We have seen that electromagnetic waves consists of a sinusoidally verying electric and magnetic field. These fields act right angles to each other as well as right angles to the direction of propagation of waves. These fields are represented by E = E0 sin (t – x/c) and B = B0 sin (t – x/c) respectively. The two fields combine to constitute electromagnetic wave. The electromagnetic wave propagates in space in a direction perpendicular to the directions of both fields as shown in fig. The electric field vectors (E) is along Y-axis and magnetic field vector (B) along Z-axis while the wave propagation direction is along X-axis. As both the fields are perpendicular to the direction of propagation of electromagnetic wave and hence the electromagnetic waves are transverse in nature. 6. POYNTING VECTOR (DERIVATION NOT REQUIRED) Poynting vector is a vector that describes the magnitude and direction of energy flow rate. 0 µ B E S       S  Poynting vector The magnitude of poynting vector represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propogation SI unit J/sm2 or w/m2 Properties of E.M. wave Ex.3 If E and B be the electric and magnetic field of E.M. wave then the direction of propogation of E.M. wave is along the direction. (A) E (B) B (C) E × B (D) B × E Sol. (C) Correct option is (C) Ex.4 Which of the following pairs of space and time varying E and B fields would generate a plane electromagnetic wave travelling in (–Z) direction- (A) Ex, By (B) Ey, Bx (C) Ex, By (D) Ey, Bx Sol. (B) Since the direction of propogation of EM wave is given by E × B hence correct option is (b) ( ^ j × ^ i = – ^ k ) Ex.5 Choose the wrong statement for E.M. wave. They- (A) Are Transverse (B) Travels in free space with the speed of light (C) Are produced by accelerated charges (D) Travels with same speed in all medium Sol. (D) Speed of E.M. wave = r r 0 µ µ 1   in medium hence it will travell with different speed in different medium.
• 5. 7. ENERGY DENSITY AND INTENSITY We know that electric and magnetic field have energy and since EM wave have both these components hence it carries energy with it. We know that energy density associated with E.F. = 2 1 0E2 We know that energy density associated with M.F. = 0 2 µ 2 B Thus total energy of EM wave is given by U = UE + UB or 2 1 0E2 + 0 2 µ 2 B putting the values of E.F. and M.F. U = 2 1 0E0 2 sin2 (t – x/c) + 0 2 2 0 µ 2 ) c / x t ( sin B   If we take the average value over a long period of time Av 2 c x t sin               = 2 1  2 0 0 2 0 av B 4µ 1 E 4 1 U    The above equation can also be written as Uav = 2 E2 0 0  or Uav = 0 2 0 µ 2 B Intensity : Energy crossing per unit area per unit time perpendicular to the direction of propogation is called intensity of wave. Energy contianed in the volume : x C T  A E.M. wave U = Uav × vol = 2 1 0E2 (ACT) intensity = T A U  = 2 C E2 0   2 0 0 2 0 B 2µ 1 or CE 2 1 0 I   Energy density and intensity Ex.6 The intensity of light from a source is 500/W/m2. Find the amplitude of electric field in this wave- (A) 3 × 102 N/C (B) 2 3 × 102 N/C (C) 2 3 × 102 N/C (D) 2 3 ×101 N/C Sol. (B) I = 2 1 0CE0 2 E0 = C I 2 0  or 3 10 36 10 500 2 8 9          E0 = 2 3 × 102 N/C Hence correct Ans is (B) Ex.7 A point source of 2 watts is radiating uniformly in all direction in vaccum. Find the amplitude of electric field at a distance 2m from it- (A) 3 × 10–4 (B) 3 × 10–2 (C) 3 × 10–4 (D) 3 × 10–2 Sol. (C) I = A P or 4 4 2   =  8 1 W/m2 I = 2 1 0 E0 2 c E0 = C I 2 0  or 8 10 3 8 36 1 2       3 × 10–4 N/C Hence correct ans in (C) Ex.8 In a EM wave the amplitude of electric field is 10 V/m. The frequency of wave is 5 × 104 Hz. The wave is propogating along Z- axis. Then the average energy density of Magnetic field is- (A) 2.21 × 10–10 J/m3 (B) 2.21 × 10–8 J/m3 (C) 2 × 10–8 J/m3 (D) 2 × 10–10 J/m3 Sol. (A) UB = 0 2 0 µ 4 B Also 0 0 B E = C  B0 = C E0
• 6. Hence B0 = 0 0 0 µ 1 E    UB = 0 0 0 2 0 µ 4 µ E  or 4 10 84 . 8 100 12    UB = 2.21 × 10–10 J/m3 Hence correct ans is (A) 8. ELECTROMAGNETIC SPECTRUM Orderly arrangement of electromagnetic radiations according to their wavelength or frequency is called as electromagnetic spectrum. The electromagnetic spectrum encompasses radiowaves, microwaves, infrared rays, visible light, ultraviolet rays, X-rays and gamma rays fig. shown the most commonly enecountered. 102 107 10 108 01 109 10–1 1010 10–2 1011 10–3 750 700 650 600 550 500 450 400 nm 1012 10–4 1013 10–5 1014 10–6 1015 10–7 1016 10–8 1017 10–9 1018 10–10 1019 10–11 1020 10–12 1021 10–13 V.H. Radio wave Radio wave Micro wave Infra red Ultra violet Y rays X rays Red Orange Yellow Green Blue Violet Visible light Frequency in Hz Wavelength in nm Name Frequency range (Hz) Wavelength range (m) -ray 5 × 1020 – 3 × 1019 6 × 10–13 – 1 × 10–10 X-ray 3 × 1019 – 1 × 1016 1 × 10–10 – 3 × 10–8 Ultraviolet 1 ×1016 – 8 × 1014 3 × 10–8 – 4 × 10–7 Visible light 8 × 1014 – 4 × 1014 4 × 10–7 – 8 × 10–7 Infra-red 4 × 1014 – 1 × 1013 8 × 10–7 – 3 × 10–5 Micro-waves 3 × 1011 – 1 × 109 1 × 10–3 – 3 × 10–1 Ultra high radio frequencies 3 × 109 – 3 × 108 1 × 10–1 – 1 Very high Radio frequencies 3 × 109 – 3 × 107 1 – 10 Radio frequencies 3 × 107 – 3 × 104 10 – 104
• 7. VISIBLE SPECTRUM Colour Wavelength (m) Violet 4 × 10–7 – 4.5 × 10–7 Blue 4.5 × 10–7 – 5 × 10–7 Green 5.5 × 10–7 – 5.7 × 10–7 Yellow 5.7 × 107 – 5.9 × 10–7 Orange 5.9 × 10–7 – 6.2 × 10–7 Red 6.2 × 10–7 – 7.5 × 10–7 9. PROPOGATION OF EM WAVE IN ATMOSPHERE Electromagnetic wave of frequency ranging from a few KHz to a few MHz are called radio waves the various frequency ranges used in radiowave of microwave communication system are (i) Medium Freq. Band (MF) 300 to 3000 KHz (ii) High Freq. Band (HF) 3 to 30 MHz (iii) Very High Freq. Band (VHF) 30 to 300 MHz (iv) Ultra High Freq. Band (UHF) 300 to 3000 MHz Radiowaves emitted from a transmilter antenna can reach to the receiver antenna by any one of the following modes of propogation (a) Ground wave propogation (b) Sky wave propogation (c) Space wave progation (a) Ground wave propogation : The radiowaves which progress along the surface of the earth are known as ground wave or surface wave ground wave can have frequency upto 1500 KHz. Not suitable for long distance communication (b) Sky wave propogation : The propogation of radio waves from one place to another place via reflection from ionosphere is called sky wave propogation. Sky waves are suitable for progation of signals having freq. band from 1500 KHz to 40 MHz. Suitable for long distance communication up to 4000 km. (c) Space wave communication : The space wave communication can travell through atmosphere from transmitter antenna to receiver antena either directly or by reflection from earth’s troposphere space ware propogation is utilized in very high frequency band (30 MHz to 300 MHz), Ultra high frequency band etc. Space wave communication is utilized in TV and radar communication * Relation between height of antenna and distance up to which signal can be recieved directly Let PQ be a TV transmitter antenna of hight h located on earth surface. The signal transmitted by this antenna can be recieved within a circle of radius QS or TQ (As hight of antenna << Radius of earth) Antenna Earth R T h R R S P Q O From right angletriangle OTQ We have (OQ)2 = (OT)2 + (QT)2 ..... (1) Here, QT = QS = d, PQ = h  OQ = OP + PQ = (R + h) Substituting in equation (1) (R + h)2 = R2 + d2 Or d2 = h2 + 2Rh d2 2RH (R >> h) d = Rh 2
• 8. The area covered for TV transmission = d2 or  (2hR) Population covered = Population density × Area covered. Electromagnetic spectrum Ex.9 The height of transmitting antenna for transmitting a TV telecast so that it covers a radius of 128 kM is (Radius of earth, R = 6.4 × 106 m) (A) 128 m (B) 1028 m (C) 1280 m (D) 1208 m Sol. (C) The height of transmitting anterna is given by h = R 2 d2 = 6 2 3 10 4 . 6 2 ) 10 128 (    = 1280 m Hence correct ans is C Ex.10 Tv transmission is possible by- (A) Ground wave (B) Sky wave (C) space wave (D) None Sol. (C) Ex.11 A TV tower has a hight of 100 m. The average population density arround the tower is 1000 person/ Km2 and radius of earth is 6.4 × 106. the population covered is- (A) 4 × 106 (B) 6 × 104 (C) 8 × 106 (D) 16 × 104 Sol. (A) The distance ‘d’ upto which TV transmission can be viewed is given by d = hR 2 Area is which transmission is viewed A = d2 =  × 2hR = 2hR A = 2 × 3.14 × 100 × 6.4 × 106  Population = (P.D) × Area = 4 × 106 person Hence correct ans is (A)
• 9. SOLVED EXAMPLES Ex.1 Elecromagnetic waves travel in a medium with a speed of 2 × 108 m/s. The relative permeability of the medium is 1. What is the relative permittivity of the medium- Sol. The speed of electromagnetic waves and in a medium is given by  = ) µ ( 1  Where µ and are absolute permeability and absolute permittivity of the medium. We know that, µ = µ0µr and = 0r. Hence =   r 0 r 0 . µ µ 1   =   0 0 µ 1  ×   r r µ 1  or v =   r r µ c  or r = ) µ ( c r 2 2   r = 1 ) 10 2 ( ) 10 3 ( 2 8 2 8    = 2.25 Ex.2 Fig. shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference between the plates (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule valid at each plate of the capacitor ? Given 0 = 8.85 × 10–12 C2 N–1 m–2 Sol. Here, R = 12 cm = 0.12 m, d = 5.0 mm = 5 × 10–3 m., I = 0.15 A, 0 = 8.85 × 10–12 C2N–1m–2 Area, A = R2 = 3.14 × (0.12)2m2 ] (a) We know that capacity of a parallel plate capacitor is given by C = d A 0  = 3 2 12 10 5 ) 12 . 0 ( 14 . 3 10 85 . 8       = 80.1 × 10–12 F = 80.1 pF Now q = CV or dt dq = C × dt dV or I = C dt dV (I = dt dq ) or dt dV = C I = 12 10 1 . 80 15 . 0   = 1.87 × 109 Vs–1 (b) Displacement current is equal to conduction current i.e. 0.15 A. (c) Yes, Kirchhoff’s first rule is valid at each plate of the capacitor provided. We take the current to be the sum of the conduction and displacement currents. Ex.3 A magnetic field in a plane electromagnetic wave is given by By = 2 × 10–7 sin(0.5 × 103x + 1.5 × 1011t) tesla. (a) What is wavelength and frequency of the wave ? (b) Write down an expression for the electric field. (x is in metre and t in second.) Sol. Given By = 2 × 10–7 sin (0.5 × 103x + 1.5 × 1011) (a) Comparing it with a standard equation for a progresive wave travelling along the negative direction of x-axis is y = r sin   2 (x + vt) = r sin            vt 2 x 2 = r sin           vt 2 x 2 We have,  x 2 = 0.5 × 103 x or  = 3 10 5 . 0 2   = 12.56 × 10–3x = 12.56 mm 2 = 1.5 × 1011
• 10.  =   2 10 5 . 1 11 = 23.9 × 109 Hz = 23.9 Hz (b) E0 = cB0 = (3 × 108) ×(2 × 10–7) = 60 Vm–1  Ez = E0 sin   2 (x + vt) = E0sin           T t 2 x 2 or Ez = 60 sin(0.5 × 103 x + 1.5 × 1011 t) V/m. Ex.4 The electric field of a plane electromagnetic wave in vacuum is represented by Ex = 0, Ey = 0.5 cos[2 × 108 (t – x/c)] and Ez = 0 (a) What is the direction of propagation of electromagnetic wave ? (b) Determine the wavelength of the wave. (c) Compute the component of associated magnetic field Sol. (a) The given equation Ey = 0.5 cos[2 × 108 (t – x/c)] ..... (1) indicates that the electromagnetic waves are propagating along the positive direction of X-axis. (b) The standard equation of electromagnetic wave is given by Ey = E0 cos(t – x/c) ..... (2) Comparing the given eq. (1) with the standard eq. (2), we get  = 2 × 108 or 2 = 2 × 108   = 108 per second Now,  =  c = 8 8 10 10 3 = 3 m (c) Here, the wave is propagating along X-axis and electric field is along Y-axis and hence the magnetic field must be along Z-axis. So the components of associated magnetic field are : Bx = 0, By = 0 and Bz = B0 cos                c x t 10 2 8 or Bz = c E0 cos                c x t 10 2 8 or Bz = 8 10 3 5 . 0  cos                c x t 10 2 8 Ex.5 A light beam travelling in the X-direction is described by the electric field Ey = (300 V/m) sin(t – x/c). An electron is constrained to move along the Y-direction with a speed of 2.0 × 107 m/s. Find the maximum electric force and the maximum magnetic force on the electron. Sol. The maximum value of magnetic field (B0) is given by B0 = c E0 = 8 10 3 300  = 10–6 tesla The magnetic field will be along Z-axis The maximum magnetic force on the electron is given by Fe = qE0 = (1.6 × 10–19) × 300 = 4.8×10–17 N The maximum magnetic force on teh electron is Fb = |q (v × B)| = q  B0 = (1.6 × 10–19) × (2.0 × 107) × (10–6) = 3.2 × 10–18 N Ex.6 The TV transmission tower at a particular staion has a height of 160 m. (a) What is its coverage rang ? (b) How much population is covered by transmission, if the average population density around the tower is 1200/km2. (c) By how much the height of tower be increased to double its coverage range ? Sol. (a) We know that d = ) hR 2 ( = ) 10 4 . 6 ( 160 2 6    = 45255 m (b) Population covered = d2 × population density = 3.14 × (45.255)2 × 1200 = 77.29 lakh (c) In this case, d’ = ) R h 2 (  = 2d  ) R h 2 (  = 2 × ) R h 2 (  h = 4h = 4 × 160 = 640 m Increase in height of tower = h– h = 640 – 160 = 480 m.