Physics-27.21-Electromagnetic waves

UNIT-21 ELECTROMAGNETIC WAVES 1. DISPLACEMENT CURRENT (i) Let us consider a circuit containing the capacitor as shown in fig. 21.1 (ii) According to Ampere's circuital law, the line integral of magnetic field along any closed path is 0 times the total current enclosed by the closed path. Mathematically In this law it is assumed that conductiona current flows through the connecting wires charging the condenser plates but no current flows in the space in between the plates. Actually it is not true. Illustration 1: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency 2 × 1010 Hz and amplitude 48 V/m. The amplitude of oscillating magnetic field will be: (A) Wb/m2 (B) 16 × 10–8 Wb/m2 (C) 12 × 10–7 Wb/m2 (D) Wb/m2 Solution: (B) Oscillating magnetic field B = = 16 × 10–8 Wb/m2 (iii) When the circuit is closed, conduction current flows from the plate P of the capacitor to the other plate Q through the conducting wires. Maxwell suggested that due to time varying electric field between the plates, an electric current, called displacement current (ID), also flows across the space between the plates of the capacitor. (iv) Thus, there is a continuous flow of current in a capacitive circuit alos, through the conducting wire there is flow of conduction current IC and through the space across the plates of capacitor, there is flow of displacement current ID. (v) Maxwell pointed out that in Ampere's circuital law, the current I should be treated as total current i.e., the sum of the conduction current IC and displacement current ID and modified the law as It is called the Ampere-Maxwell's circuital law. (vi) The displacement current is defined as ID = where is the electric flux linked between the plates of the capacitor at any instant. Therefore Ampere-Maxwell circuital law may be expressed as Illustration 2: A parallel plate capacitor of plate separation 2 mm is conncected in an electric circuit having source voltage 400V. If the plate area is 60 cm2, then the value of displacement current for 10–6 sec. will be: (A) 1.062A (B) 1.062 × 10–2A (C) 1.062 × 10–3A (D) 1.062 × 10–4A Solution: (D) ID = or ID = = 1.062 × 10–4A Illustration 3: In an electric circuit, there is a capacitor of reactance 100 connected across the source of 220V. The displacement current will be: (A) 2.2A (B) 0.22A (C) 4.2A (D) 2.4A Solution: (A) Displacement current and conduction current are equal. ∴ 2.2A (vii) The conduction current and the displacement current are always equal, i.e., IC = ID (viii) Like conduction current, the displacement current is also the source of magnetic field. 2. MAXWELL'S EQUATIONS AND LORENTZ FORCE The existence of electro-magnetic waves that propagate through the space in the form of varying electric and magnetic fields has been predicted by the four basic laws of electromagnetism which are called Maxwell's equations. (i) Gauss's law in electrostatics% It states that the total electric flux through any closed surface is equal to times the net charge enclosed by Mathematically, This equation is called Maxwell's first equation. (ii) Gauss'law in magnetism% It states that the net magnetic flux crossing any closing surface is always zero. Mathematically, This equation is called Maxwell's second equation. A direct consequence of this equation is that the magnetic monpoles do not exist. (iii) Faradays's law of electromagnetic induction% It states that the induced emf produced in a circuit is numerically equal to the rate of change of magnetic flux through it. Mathematically, Butq Line integral of electric field. ∴ This equation is called Maxwell's third equation. The negative sign in this equation indicates that the induced emf produced opposes the rate of change of magnetic flux. Illustration 4: A point source of electromagnetic radiation has an average power output of 800W. The maximum value of electric field at a distance 3.5m from the source will be: (A) 56.7 V/m (B) 62.6 V/m (C) 39.3 V/m (D) 47.5 V/m Solution: (B) Intensity of electromagnetic wave given is by Em = = = 62.6 V/m Illustration 5: In the above problem, the maximum value of magnetic field will be: (A) 2.09 × 10–5 T (B) 2.09 × 10–6 T (C) 2.09 × 10–7 T (D) 2.09 × 10–7 T Solustration: (C) The maximum value of magnetic field is given by = 2.09 × 10–7 T (iv) Maxwell-Ampere circuital law% It states that the line integral of magnetic field along a closed path is equal to μ0 times the total current (i.e., sum of conduction and displacement currents threading the surface bounded by that closed path) Mathematically, This equation is called Maxwell's fourth equation. (v) Lorentz: The vector sum of electric force and magnetic force on any charged particle is called the Lorentz force. The above five equations give a complete description of all electromagnetic interactions. 3. PRODUCTION OF ELECTROMAGNETIC WAVES (i) According to Maxwell, an accelerated charge sets up a magnetic field in its neighbourhood. The magnetic field, in turn, produces an electric field in that region. Both these fields vary with time and act as sources for each other. (ii) As oscillating charge is accelerated continuously, it will radiate electromagnetic waves continuously. (iii) In 1988, Hertz demonstrated the production of electromagnetic apparatus is shown schematically in fig. (iv) An induction coil is connected to two spherical electrodes with a narrow gap between them. It acts as a transmitter. The coil provides short voltage surges to the spheres making one positive and the other negative. A spark is generated between the spheres when the voltage between them reaches the breakdown voltage for air. As the air in the gap is ionised, it conducts more rapidly and the discharge between the spheres becomes oscillatory. (v) The above experiment arrangment is equivalent to an LC circuit, where the inductance is that of the loop and the capacitance is due to the spherical electrodes. (vi) Electromagnetic waves are radiated at very high frequency( 100 MHz) as a result of oscillation of free charges in the loop. (vii) Hertz was able to detect these waves using a single loop of wire with its own spark gap (the receiver). (viii) Sparks were induced across the gap of the receiving electrodes when the frequency of the receiver was adjusted to match that of the transmitter. 4. HISTORY OF ELECTROMAGNETIC WAVES (i) In year 1865, Maxwell predicted the electromagnetic waves theoretically. According to him, an accelerated charge sets up a magnetic field in its neighborhood. (ii) In 1887, Hertz produced and detected electromagnetic waves experimentally at wavelength of about 6m. (iii) Seven year later, J.C. Bose became successful in producing electromagnetic waves of wavelength in the range 5 mm to 25 mm. (iv) In 1896, Marconi discovered that if one of the spark gap terminals is connected to an antenna and the other terminal is earthed, the electromagnetic waves radiated could go upto several kilometers. (v) The antenna and the earth wires form the two plates of a capacitor which radiates radio frequency waves. These waves could be received at a large distance by making use of an antenna earth system as detector. (vi) Using these arrangments; in 1899 Marconi first established wireless communication across the English channel i.e.across a distance of about 50 km. 5. PROPERTIES OF ELECTROMAGNETIC WAVES (i) The electric and magnetic fields satisfy the following wave equations, which can be obtained from Maxwell's third and fourth equations. (ii) Electromagnetic waves travel through vacuum with the speed of light c, where c = m/s (iii) The electric and magnetic fields of an electromagnetic wave are perpendicular to each other and also perpendicular to the direction of wave propagation. Hence, these are transverse waves. (iv) The instantaneous magnitude of and in an electromagnetic wave are related by the expression (v) Electromagnetic waves carry energy. The rate of flow of energy crossing a unit area is described by the Poynting vector where (vi) Electromagnetic waves carry momentum and hence can exert pressure (P) on surfaces, which is called radiation vector , incident on a perfectly absorbing surface and if incident on a perfectly reflecting surface P = (vii) The electric and magnetic fields of a sinusoidal plane electromagnetic wave propagating in the positive x-direction can also be written as E = Em sin (kx – ) B = Bm sin (kx – ) where is the angular frequency of the wave and k is wave number which are given by = 2 f and k = (viii) The intensity of a sinusoidal plane electro-magnetic wave is defined as the average value of Poynting vector taken over one cycle. Sav = (ix) The fundamental sources of electromagnetic waves are accelerating electric charges. For example radio waves emitted by an antenna arise from the continuous oscillations (and hence acceleration) of charges within the antenna structure. (x) Electromagnetic waves obey the principle of superposition. (xi) The electric vector of an electromagnetic field is responsible for all optical effects. For this reason electric vector is called a light vector. Illustration 6: In an electromagnetic wave, the amplitude of electric field is 1V/m. The frequency of wave is 5 × 10–14 Hz. The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be: (A) 1.1 × 10–11 (B) 2.2 × 10–12 (C) 3.3 × 10–13 (D) 4.4 × 10–14 Solustion: (B) Average energy density is given by uE = = = 2.2 × 10–12 J/m3 Illustration 7: To establish an instantaneous displacement current of 2A in the space between two parallel plates of 1μF capacitor, the potential difference across the capacitor plates will have to be changed at the rate of: (A) 4 × 104 V/s (B) 4 × 106 V/s (C) 2 × 104 V/s (D) 2 × 106 V/s Solution: (D) ID = ;k ID = ∴ 6. ELECTROMAGNETIC SPECTRUM (i) The orderly distribution of electromagnetic waves (on the basis of wavelength or frequency) in the form of distinct groups having widely differing properties is called the electromagnetic spectrum. The following table gives a complete and detailed picture of electromagnetic spectrum (ii) Various parts of electromagnetic spectrum S. No. Radiation Discover How produced Wavelength range Frequency range Energy range Properties Application 1. &Rays Henry Becquerrel and Madam Curie Due to decay of radioactive nuclei 10–14 m to 10–10m. 3 × 1022 Hz to 3 × 1018 Hz 107eV – 104 eV (a) High penetrating power (b) Uncha-rged (c) Low ionising power (a) Gives information on nuclear structure (b) Medical treatment etc. 2. X-Rays Roentgen Due to collisions of high energy electrons with heavy targets 6 × 10–10 m to 10–9 m 5 × 1019 Hz to 3 × 1017 Hz 2.4 × 105 eV to 1.2 × 103 eV (a) Low penetrating power (b) other properties similar to &rays except wavelength (a) Medical diagnosis and treatment (b) Study of crystal structure (c) Indus-trial radio-graphy 3. Ultraviolet Rays Ritter By ionised gases, sun are lamp spark etc. 6 × 10–10 m to 3.8 × 10–7 m 3 × 1017 Hz to 5 × 1019 Hz 2×103 eV to 3eV (a) All pro-perties (b) Photo-electric effect (a) To detect adul-teration, (b) Sterili-zation of water due to its des-tructive act-ion on bac-teriaa 4. Visible light (a)Violet (b) Blue (c) Green (d) Yellow (e)Orange (f) Red Newton Outer orbit electron transitions in atoms, gas disch- arge tube, incandes- cent 3.8 × 10–7 m to 7.8 × 10–7 m 3.9×10–7 m to 4.55 ×10–7 m 4.55×10–7 m to 4.92 ×10–7 m 4.92×10–7 m to 5.77 ×10–7 m 5.77×10–7 m to 5.97 ×10–7 m 5.97×10–7 m to 6.22 ×10–7 m 6.22×10–7 m to 7.80×10–7 m 8 × 1014 Hz to 4 × 1014 Hz 7.69 × 1014 Hz to 6.59×1014 Hz 6.59 × 1014 Hz to 6.10×1014 Hz 6.10 × 1014 Hz to 5.20×1014 Hz 5.20 × 1014 Hz to 5.03×1014 Hz 5.03 × 1014 Hz to 4.82×1014 Hz 4.82 × 1014 Hz to 3.84×1014 Hz 3.2 eV to 1.6eV (a) Sens-itive to hu-man eye (a) To see objects (b) To study molecular structurea 5. Infra-Red waves William Herschell (a) Rearr- angement of outer orbitals electrons in atoms and mole-cules (b)Change of molecu-lar vibrati-onal and rotational energies. (c) By bodies at high temp- erature 7.8×10–7 m to 10–3 m 4×1014 Hz to 3×1011 Hz 1.6 eV to 1.6–3eV (a) Thermal effect (b) All prop erties simi lar of those of light exc ept λ (a) Used in industry, me dicine and astro nomya (b) Used for fog or have photographya 6. Microwaves Hertz special electronic devices such as klystron tube 10–7 m to 0.3m 3×1011 Hz to 109 Hz 10–3eV to 10–5eV (a) Pheno mena of reflection, refraction and diffra ction (a) Radar and telecom -munication (b) Analysis of fine details of molecular structurea 7. (A) Radio waves Subparts of Radio spectrum Super High (a) SHF Ultra High (b) (UHF) Very High (c) (VHF) Marconi Oscillating circuits 0.3m to few kms. 0.1m to 0.1m 0.1m to 1m 1m to 10m 109 Hz to few Hz 3×1010 Hz to 3×109 Hz 3×109 Hz to 3×108 Hz 3×108 Hz to 3×107 Hz 10–3ev to 0 (a) Exhibit waves like properies more than particle like properties (a) Radio communication (b) Radar, Radio and sate llite communi cation (Microw aves) Radar and television broad cast sho rt distance co mmunication, Television communication (B) High frequency Medium frequency Low Frequency Very low frequency 10m to 100m 100m to 1000m 1000m to 10000m 10000m to 30000m 3×107 Hz to 3×106 Hz 3×106 Hz to 3×105 Hz 3×105 Hz to 3×104 Hz 3×104 Hz to 3×104 Hz Medium distance communication Marine and navigation use, long range communication. Long distance communication. 7. EARTH'S ATMOSPHERE AND ELECTROMAGNETIC WAVES (i) The gaseous envelop surrounding the earth is called earth's atmosphere. (ii) It mainly consists of nitrogen 78% and oxygen 21% alongwith a little portion of argon, carbon-di-oxide, water vapour, hydrocarbons, sulphur compounds and dust particles. (iii) The density of atmospheric air goes on decreasing gradually as we go up. (iv) The earth's atmosphere has no sharp boundary. However, it has been divided into various regions as given below: (a) Troposphere% It extends upto a height of 12 km from earths surface. The temperature in this region decreases from 298K to 220K and conductivity increases. All climatic changes occur in this region. (b) Stratosphre% It extends from 12 km to 50 km after troposhpere. At the upper part of this region, approximately 20km thick, most of ozone of atmosphere is concentrated. This layer is called as ozone layer. This layer absorbs very large portion of ultraviolet radiations coming from sun, therefore its temperature increases from 220K to 280K. (c) Mesosphere% It extends from 50km to 80 km after stratosphere. In this region the temperature decreases from 280K to 180K (d) Ionosphere% It extends from 80 km to 400 km after mesosphere. The temperature of this region rises from 180K to 700K. In this region ultraviolet radiation coming from sun cause ionisation, therefore this part mostly consists of free electrons and positive ions. The concentration of free electrons is found to be very large in a region beyond 110 km from earth's surface which extends vertically for a few kilometers and is called Kennelly Heaviside layer. Beyond this layer the concentration of free electrons decreases considerably until a height of about 250 km. Beyond it there is another layer of electrons, called Appleton layer. (v) Greenhouse effect% The atmosphere is transparent to visible radiations, but most ifrared (heat) radiations are not allowed to pass through. The energy from the sun heats the earth which then starts emitting radiations like any other hot body. However, since the earth is much colder than sun, its radiations are mainly in the infra red region. These radiations are unable to cross the lower atmosphere and are reflected back. Low lying clouds also reflect back the infra red radiations. As such, the earth's surface warm at night. This phenomenon is called the Green house effect. (vi) Propagation of Radio waves% (a) Low frequency waves-the AM band% Radiowaves having wavelengths of 10m or more (frequency less than 30 Mhz) are said to constitute the AM band. The lower atmosphere is transparent to these waves, but the ionosphere reflects them back. A signal transmitted from a certain point can be received at another point in two possible ways-directly along the surface of the earth (called sky wave) and after reflection from ionosphere (called sky wave). Waves having frequencies upto about 1500kHz (Wavelength above 200m) are mainly transmitted through ground because low frequency sky waves lose their energy very quickly than the sky waves. Therefore, higher frequencies are mainly transmitted through sky. These two regions of the AM band are called medium wave and short wave bands respectively. (b) High frequency waves-Television transmission% Above a frequency of about 40MHz the ionosphere does not reflect the wave toward the earth. The television signals have frequencies in the range 100-200 MHz. Therefore TV transmission via the sky is not possible-only direct reception via the ground is possible. Therefore, in order to have larger coverage, the transmission has to be done through very tall antennas. The height of transmitting antenna for TV telecast is given by h = where d is the radius of the area to be covered for TV telecast and Re is the radius of earth. Illustration 8: A T.V. tower has a height of 100m. How much population is covered by T.V. broadcast, if the average population density around the tower is 1000/km2? (A) 39.5 × 105 (B) 19.5 × 106 (C) 29.5 × 107 (D) 9 × 104 Solution: (A) Radius of the area covered by T.V. telecast d = Total population covered ¾ πd2 × population density = 2πhRe × population density = 2 × 3.14 × 100 × 6.4 × 106 × = 39.503 × 105 Illustration 9: An electromagnetic radiation has an energy 14.4 Kev. To which region of electromagnetic spectrum does it belong? (A) Infra red region (B) Visible region (C) X-ray region (D) γ-ray region Solustion: (C) λ = This wavelength belongs to X-ray region. Hence the correct answer will be(C) 8. ASSIGNMENT 1. If E and B are the electric and magnetic field vectors of electromagnetic waves then the direction of propagation of electromagnetic wave is along the direction of: (A) (B) (C) (D) None of these Sol. (C) 2. The charge on a parallel plate capacitor is varying as q = q0 sin 2πnt The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is: (A) (B) (C) 2πnq0 cosπnt (D) Sol. (C) ID = q0sin2πnt = 2πnq0 cos2πq0 cos2πnt 3. The value of magnetic field between plates of capacitor, at distance of 1m from centre where electric field varies by 1010 V/m/s will be: (A) 5.56T (B) 5.56μT (C) 5.56mT (D) 55.6nT Sol. (D) B = 4. The electromagnetic waves do not transport: (A) Energy (B) charge (C) momentum (D) information Sol. (B) 5. A capacitor is connected in an electric circuit. When key is pressed, the current in the circuit is: (A) Zero (B) Maximum (C) any transient value (D) depends on capacitor used Sol. (B) 6. Displacement current is continuous: (A) when electric field is changing in the circuit (B) when magnetic field is changing in the circuit (C) in both types of fields (D) through wires and resistance only Sol. (A) 7. Instantaneous displacement current 1A in the space betwen the parallel plates of 1μF capacitor can be established by chaning the potential difference at the rate of: (A) 0.1 V/s (B) 1 V/s (C) 106 V/s (D) 10–6 V/s Sol. (C) ID = or = 106 V/s 8. The magnetic field between the plates of a capacitor when r > R is given by: (A) (B) (C) (D) 'kwU; Sol. (C) According to Ampere's law, when r > R B = 9. The magnetic field between the plates of a capacitor is given by B = : (A) r R (B) r R (C) r < R (D) r = R Sol. (C) According to Ampere's law, when r > R B = 10. The conduction current is the same as displacement current when the source is: (A) A.C. only (B) D.C. only (C) Both A.C. and D.C. (D) neither for A.C. nor for D.C. Sol. (B) 11. The wave function (in S.I. units) for an electromagnetic wave is given as: (x, t) = 103 sinπ (3 × 106x – 9 – 1014t) The speed of the wave as: (A) 9 × 1014 m/s (B) 3 × 108 m/s (C) 3 × 1016 m/s (D) 3 × 107 m/s Sol. (B) c = = 3 × 108 m/s 12. In the above problem, wavelength of the wave is: (A) 666 nm (B) 666 (C) 666μm (D) 6.66 nm Sol. (A) (x – 3 × 108 t) Comparing it with = asin (x – vt) ∴ = 3 × 106π or λ = = 666 nm 13. The Maxwell's four equations are written as: (i) (ii) (iii) (iv) The equations which have sources of and (A) (i), (ii), (iii) (B) (i), (ii) (C) (i) and (iii)a (D) (i) and (iv) Sol. (D) 14. Out of the above four equations which do not contain source field are: (A) (i) and (ii) (B) (ii) only (C) all of four (D) (iii) only Sol. (B) 15. Out of four Maxwell's equations above, which one shows non-existence of monopoles? (A) (i) and (iv) (B) (ii) only (C) (iii) only (D) only Sol. (B) 16. Which of the above Maxwell's equations shows that electric field lines do not form closed loops? (A) (i) only (B) (ii) only (C) (iii) only (D) (iv) only Sol. (A) 17. In an electromagnetic wave the average energy density is associated with: (A) electric field only (B) magnetic field only (C) equally with electric and magnetic fields (D) average energy density is zero Sol. (C) 18. In an electromagnetic wave the average energy density associated with magnetic field will be: (A) (B) (C) (D) Sol. (B) 19. In the above problem, the energy density associated with the electric field will be: (A) (B) (C) (D) Sol.% (D) 20. If there were no atmospher, the average temperature on earth surface would be: (A) Lower (B) Higher (C) same (D) 0oC Sol. (A) The green house effect would not have been possible without atmosphere. Hence temperature would be lower. 21. In which part of earth's atmosphere is the ozone layer present? (A) Troposphere (B) Stratosphere (C) Ionosphere (D) Mesosphere Sol. (B) 22. Kenneley's Heaviside layer lies between: (A) 50Km to 80 Km (B) 80Km to 400 Km (C) beyond 110 Km (D) beyond 250 Km Sol. (C) 23. The ozone layer in earth's atmosphere is crucial for human survival because it: (A) has ions (B) reflects radio signals (C) reflects ultraviolet ray (D) reflects infra red rays Sol. (C) 24. The frequency from 3 × 109 Hz to 3 ×1010 Hz: (A) High frequency band (B) Super high frequency band (C) Ultra high frequency band (D) High frequency band Sol. (B) 25. The frequency from 3 to MHz is known as: (A) Audio band (B) Medium frequency band (C) Very high frequency band (D) High frequency band Sol. (B) 26. The AM range of radiowaves have frequency: (A) less than 30 MHz (B) More than 30 MHz (C) less than 20000 Hz (D) More than 20000 Hz Sol. (A) 27. The displacement current flows in the dielectric of a capacitor (A) becomes zero (B) has assumed a constant value (C) is increasing with time (D) is decreasing with time Sol. (C) 28. Select wrong statement from the following Electromagnetic waves: (A) are transverse (B) travel with same speed in all media (C) travel with the speed of light (D) are produced by acceleration charge Sol. (B) 29. The waves related to tele-communication are: (A) infra red (B) visible light (C) microwaves (D) ultraviolet rays Sol. (C) 30. Electromagnetic waves do not transport: (A) energy (B) charge (C) momentum (D) information Sol. (A) 31. The nature of electromagnetic wave is: (A) longitudinal (B) longitudinal stationary (C) transverse (D) transverse stationary Sol. (C) 32. Greenhouse effect keeps the earth surface: (A) cold at night (B) dusty and cold (C) warm at night (D) moist Sol. (C) 33. A parallel plate capacitor consists of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source. The charging is being charged and is equal to 0.15 A. The rate of change of potential difference between the plates will be: (A) 8.173 × 107 V/s (B) 7.817 × 108 V/s (C) 1.873 × 109 V/s (D) 3.781 × 1010 V/s Sol. (C) = = 1.873 × 109 V/s 34. In the above problem, the displacement current is: (A) 15A (B) 1.5A (C) 0.15A (D) 0.015A Sol. (C) ID = IC = 0.15A 35. The wave emitted by any atom or molecule must have some finite total length which is known as the coherence length. For sodium light, this length is 2.4cm. The number of oscillations in this length will be: (A) 4.068 × 105 (B) 4.068 × 106 (C) 4.068 × 107 (D) 4.068 × 108 Sol. (B) No. of oscillations in coherence length = 36. In the above problem, the coherence time will be: (A) 8 × 10–8s (B) 8 × 10–9s (C) 8 × 10–10s (D) 8 × 10–11s Sol. (D) The coherence time t = ls- 37. A parallel plate capacitor made to circular plates each of radius R = 6cm has capacitance C = 100pF. The capacitance is connected to a 230V A.C. supply with an angular frequency of 300 rad/s. The r.m.s. value of conduction current will be: (A) 5.7μa (B) 6.3μA (C) 9.6μA (D) 6.9μA Sol. (D) IRMS = = ωCERMS = 300 × 10–10 × 230 = 6.9μA 38. In the above problem, the displacement current will be: (A) 6.9μA (B) 9.6μA (C) 6.3μA (D) 5.7μA Sol. (A) ID = IC = 6.9μA 39. In Q. 37, the value of B at a point 3 cm from the axis between the plates will be: (A) 1.63 × 10–8T (B) 1.63 × 10–9T (C) 1.63 × 10–10T (D) 1.63 × 10–11T Sol. (D) B0 = = = 1.63 × 10–11T 40. A plane electromagnetic wave of frequency 40 MHz travels in free space in the X-direction. At some point and at some instant, the electric field has its maximum value of 750 N/C in Y-direction. The wavelength of the wave is: (A) 3.5 m (B) 5.5 m (C) 7.5 m (D) 9.5 m Sol. (C) 41. In the above problem, the period of the wave will be: (A) 2.5 μs (B) 0.25 μs (C) 0.025 μs (D) None of these Sol. (C) T = = = 0.025μs 42. In Q. 40, the magnitude and direction of magnetic field will be: (A) 2.5 μT in X-direction (B) 2.5 μT in Y-direction (C) 2.5 μT Z-direction (D) none of these Sol. (C) Bm = = 2.5μT Z-directiona 43. In Q. 40, the angular frequency of e.m.f. wave will be:(in rad/s) (A) 8π × 107 (B) 4π × 106 (C) 4π × 105 (D) 8π × 104 Sol. (A) ω = 2πf = 2 × π × 4 ×10 = 8π × 107 rad/s. 44. In Q. 40, the propagation constant of the wave will be: (A) 8.38m–1 (B) 0.838m–1 (C) 4.19m–1 (D) 0.419m–1 Sol. (B) K = m–1 45. The sun deliverse 103 W/m2 of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions 8m × 20m, will be: (A) 6.4 × 103W (B) 3.4 × 104W (C) 1.6 × 105W (D) None of these Sol. (C) power P = 5A = 103 × 8 × 20 = 1.6 × 105W 46. In the above problem, the radiation force on the (A) 3.33 × 10–5N (B) 5.33 × 10–4 N (C) 7.33 × 10–3 N (D) None of thesea Sol. (B) F = PA = 47. In Q. 45, the solar energy incident on the roof in 1 hour will be: (A) 5.76 × 108J (B) 5.76 × 107J (C) 5.76 × 106J (D) 5.76 × 105J Sol. (A) E = power  time = 1.6 × 105 × 3600 = 5.76 × 108J 48. The sun radiates electromagnetic energy at the rate of 3.9 × 1026W. Its radius is 6.96 × 108m. The intensity of sun light at the solar surface will be: (A) 1.4 × 104 (B) 2.8 × 105 (C) 4.2 × 106 (D) 5.6 × 107 Sol. (D) Isurface = = = 5.6 ×107 W/m2 49. In the above problem, if the distance from the sun to the earth is 1.5 × 1011 m, then the intensity of sunlight on earth's surface will be-(in W/m2) (A) 1.38 × 103 (B) 2.76 × 104 (C) 5.52 × 105 (D) buesa ls dksbZ ugha Sol. (A) Iearth = = 1.38 × 103 W/m2 50. A laser beam can be focussed on an area equal to the square of its wavelength. A He-Ne laser radiates energy at the rate of 1nW and its wavelength is 632.8 nm. The intensity of focussed beam will be: (A) 1.5 × 1013 W/m2 (B) 2.5 × 109 W/m2 (C) 3.5 × 1017 W/m2 (D) None of these Sol. (B) Area through which the energy of beam passes = (6.328 × 10–7)2 = 4 × 10–13 m2 ∴ I = = 2.5 × 109 W/m2 51. A flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave: Ex = 36sin (1.20 × 107z – 3.6 × 1015t) V/m The average intensity of the beam will be: (A) 0.86 W/m2 (B) 1.72 W/m2 (C) 3.44 W/m2 (D) 6.88 W/m2 Sol. (B) Iav = = = 1.72 W/m2 52. An electric field of 300 V/m is confined to a circular area 10 cm in diameter. If the field is increasing at the rate of 20 V/m-s, the magnitude of magnetic field at a poit 15cm from the centre of the circle will be: (A) 1.85 × 10–15 T (B) 1.85 × 10–16 T (C) 1.85 × 10–17 T (D) 1.85 × 10–18 T Sol. (D) B = = = 1.85 × 10–18T 53. A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 10m from the lamp will be: (A) 1.34 V/m (B) 2.68 V/m (C) 5.36 V/m (D) 9.37 V/m Sol. (A) Sav = ∴ = = 1.34 V/m 54. A plane electromagnetic wave of wave intensity 6W/m2 strikes a small mirror of area 40 cm2, held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be: (A) 6.4 × 10–7 kg–m/s (B) 4.8 × 10–8 kg–m/s (C) 3.2 × 10–9 kg–m/s (D) 1.6 × 10–10 kg–m/s Sol. (D) In one second P = = 1.6 × 10–10 Kg-m/s 55. In the above problem, the radiation force on the mirror will be: (A) 6.4 × 10–7 N (B) 4.8 × 10–8 N (C) 3.2 × 10–9 N (D) 1.6 × 10–10 N Sol. (D) ∴ Momentum per sec is force ∴ F = 1.6 × 10–10 Newton 56. In the above problem, the wavelength of the wave will be: (A) 1.5m (B) 66.6m (C) 1.5cm (D) 66.6cm Sol. (C) Wavelength of electromagnetic wave = 1.5 cm Hence correct answer will be (C) 57. In Q. 5, the energy density at a distance 3.5m from the source will be_ (in joule/m3) (A) 1.73 × 10–5 (B) 1.73 × 10–6 (C) 1.73 × 10–7 (D) 1.73 × 10–8 Sol. (D) Energy density at 3.5m is given by = = 1.73 × 10–8 Hence the correct answer will be(D) 58. A 100 pF capacitor is connected to a 230V, 50 Hz A.C. source. The r.m.s. value of conduction current will be: (A) 7.2 × 10–6A (B) 3.6 × 10–5A (C) 1.8 × 10–4A (D) 0.9 × 10–3A Sol. (A) The r.m.s. value of conduction current or I = 2 × 3.14 × 50 × 100 × 10–12 × 230 = 7.2 × 10–6 A Hence the correct answer will be(A) 59. What should be the height of transmitting antenna if the T.V. telecast is to cover a radius of 128 km? (A) 1560m (B) 1280m (C) 1050m (D) 79m Sol. (B) Height of transmitting antenna h = Hence the correct answer will be(B) 60. The area to be covered for T.V. telecast is doubled, then the height of transmitting antena (T.V. tower) will have to be: (A) doubled (B) halved (C) quardupled (D) kept unchanged Sol. (A)The area of transmission surrounding the T.V. tower A = πd2 = π(2hRe) A h Hence the correct answer will be(A)