Electromagnetic Waves Notes

Electromagnetic Waves BASIC EQUATIONS OF ELECTRICITYAND MAGNETISM The whole concept of electricity and magnetism can be explained by the four basic equations we have deal so far. Q (1)  E ds = (Gauss law for electrostatic) (2)  B ds = 0 (Gauss law for magnetism) (3)  B dl = µ0i (Ampere’s law for Magnetism) (4)  E dl = 0 (Ampere’s law for electrostatics) The above stated equation are true for non-time varying fields 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 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  E dl = – d dt 35 CONCEPT OF DISPLACEMENT CURRENT (i) Let us consider a circuit containing the capacitor as shown in fig. 21.1 (iii) (iv) i i B (ii) B B 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   (v)   In this law it is assumed that conduction 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. Electromagnetic Waves 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. Thus, there is a continuous flow of current in a capacitive circuit also, 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. 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   (vi) It is called the Ampere-Maxwell's circuital law. The displacement current is defined as ID = where f E is the electric flux linked Ex. 1 In a plane electromagnetic wave, the electric field oscillates sinusoidal at a frequency 21010 Hz and amplitude 48 V/m. The amplitude of oscillating magnetic field will be: 1 (A) Wb/m2 16 (B) 16  10–8 Wb/m2 (C) 12  10–7 Wb/m2 1 (D) Wb/m2 12 Sol. (B) Oscillating magnetic field B= = 16  10–8 Wb/m2 between the plates of the capacitor at any instant. Therefore Ampere-Maxwell circuital law may be expressed as    Ex. 2 A parallel plate capacitor of plate separation 2 mm is connected 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 36 Sol. (ii) (D) ID = or ID = = 1.062  10–4A Ex. 3 Sol. 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 (A) Displacement current and conduction current are equal.  ID 2.2A (iii) Electromagnetic Waves 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 monopoles do not exist. Faradays' 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,   But q Line integral of electric  field. (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. MAXWELL'S EQUATIONS AND ORENTZ 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 1 closed surface is equal to times the net charge enclosed by Mathematically,    This equation is called Maxwell's first equation.     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. Ex. 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 Sol. (B) Intensity of electromagnetic wave given is by I Em = = = 62.6 V/m 37 Ex. 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 Sol. (C) The maximum value of magnetic field is given by (iii) Electromagnetic Waves In 1988, Hertz demonstrated the production of electromagnetic apparatus is shown schematically in fig. Bm = 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 0 times the total current (i.e., sum of conduction and displacement currents threading the surface bounded by that closed path) Mathematically,    (v) This equation is called Maxwell's fourth equation. 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. PRODUCTION OF ELECTROMAGNETIC WAVES (i) According to Maxwell, an accelerated charge sets up a magnetic field in its neighborhood. 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. (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 arrangement 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. 38 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 arrangements; in 1899 Marconi first established wireless communication across the English Channel i.e. across a distance of about 50 km. 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 1 c= m/s (iii) (iv) (v) (vi) Electromagnetic Waves 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.  The instantaneous magnitude of E and  B in an electromagnetic wave are related by the expression E B Electromagnetic waves carry energy. The rate of flow of energy crossing a unit area  is described by the Pointing vector S    where Electromagnetic waves carry momentum and hence can exert pressure (P) on surfaces, which is called radiation vector  S , incident on a perfectly absorbing surface P c and if incident on a 2S c 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 w is the angular frequency of the wave and k is wave number which are given by 2 w = 2wf and k = perfectly reflecting surface P = (vii) (viii) The intensity of a sinusoidal plane electro-magnetic wave is defined as the average value of Pointing vector taken over one cycle. Sav = 39 Electromagnetic Waves (ix) (x) (xi) 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. Electromagnetic waves obey the principle of superposition. The electric vector of an electromagnetic field is responsible for all optical effects. For this reason electric vector is called a light vector. Ex. 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 Sol. (B) Average energy density is given by uE = 1 4 = 2.2  10–12 J/m3 Sol. (D) ID = ID =  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 2 1 with E.F. = 0E 2 We know that energy density associated with M.F. = Thus total energy of EM wave is given by U = UE + UB or 2 1 B2 0E + 2 2µ0 putting the values of E.F. and M.F. 1 U= 0E02 2 sin2(t–x/c)+ = Ex. 7 To establish an instantaneous displacement current of 2A in the space between two parallel plates of 1F 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 B2 2µ0 2µ0 If we take the average value over a long period of time =  Uav 1 E02 4 1 2 1 2 B 4µ0 0 The above equation can also be written as Uav 2 B20 = or Uav = 2µ0 40 Electromagnetic Waves Intensity: Energy crossing per unit area per unit time perpendicular to the direction of propagation is called intensity of wave. Energy contained in the volume : I = 2 1 0 E0 c 2 E0 = 2I –4 or ×10 N/C Ex.10 In a EM wave the amplitude of electric field is 10 V/m. The frequency of wave is 4 U = Uav × vol = intensity = 1 2 I  5 Ч 10 Hz. The wave is propagating along Z-axis. Then the average energy density of Magnetic field is- 2 1 0E (ACT) 2 –10 (A) 2.21 × 10 U = CE02 or –8 (B) 2.21 × 10 J/m 2 –8 (C) 2 × 10 J/m 1 2 B 2µ0 0 –10 (D) 2 × 10 Sol. Ex.8 The intensity of light from a source is 2 500/W/m . Find the amplitude of electric field in this wave2 (A) 3 × 10 N/C N/C Sol. 2I E0 = 2 (B) 2 3 2 3 (C) × 10 N/C 2 2 1 (B) I = 0CE0 2 1 (D) 2 3 ×10 N/C –4 (C) I = P A  UB = E 2µ 4µ0 4 –10 10 J/m 3 or (A) 3 × 10 Sol. E0 1 µ Hence B0 = UB = 2.21 × A point source of 2 watts is radiating uniformly in all direction in vacuum. Find the amplitude of electric field at a distance 2m from it3 × 10 3 E0 E = C B0 = 0 B0 C or 2 (C) 3 3 J/m B20 (A) UB = 4µ0 Also 3 × 10 E0 = 2 3 × 10 N/C Ex.9 J/m (B) 3 × 10 –4 or 2 –2 –2 (D) 3 × 10 = 2 1 W/m 8 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 41 Electromagnetic Waves (ii) Various parts of electromagnetic pectrum S.No. Radiation Discover How Wave Frequency Energy produced length range range 3  1022 Hz 107eV Properties Application range 1. &Rays Henry Becquerrel Due to 10–14 m decay of to and radioactive –10 Madam nuclei to 10 m. 18 3  10 Hz 4 – 10 eV (a) High (a) Gives penetrating information on nuclear structure power (b) Uncharged Curie (c) Low ionising power 2. X-Rays Roentgen Due to collisions of high 6  10–10 –9 m to10 m 5  1019 Hz 2.4  105 to eV to 3  1017 Hz 1.2  103 energy eV electrons with heavy targets (b) Medical treatment etc. (a) Low (a) Medical penetrating power diagnosis and treatment (b) other properties (b) Study of crystal structure similar to (c) Indus-trial radiography &rays except wavelength 3. Ultraviolet Ritter Rays By ionised 6  10–10m gases, sun to are lamp spark etc. –7 3.8  10 3  1017 Hz 2103 eV to to 3eV 19 (a) All properties (b) Photoelectric effect 5  10 Hz m (a) To detect adulteration, (b) Sterilization of water due to its destructive act-ion on bacteriaa 4. Visible light Newton Outer orbit 3.8  10–7 8  1014 Hz 3.2 eV to (a) Sensitive (a) To see electron transitions in atoms, m to to 1.6eV to human eye objects –7 7.8  10 14 4  10 Hz m gas discharge tube, (b) To study molecular structure incandescent (a)Violet 3.910–7 7.69  1014 m to 4.55 Hz to –7 10 m 6.591014 Hz 42 Electromagnetic Waves –7 (b) Blue 4.5510 m to 4.92 –7 (c) Green 4.9210–7 6.10  1014 m to 5.77 Hz to 10 m 5.201014 Hz 5.7710–7 5.20  1014 m to 5.97 Hz to 10 m 5.031014 Hz 5.9710–7 5.03  1014 m to 6.22 Hz to –7 (f) Red Hz to 6.101014 Hz –7 (e)Orange 6.59  10 10 m –7 (d) Yellow 14 10 m 4.821014 Hz 6.2210–7 4.82  1014 m to Hz to –7 7.8010 3.841014 Hz m 5. Infra-Red waves William Herschell (a) Rearrangement of outer orbital's 7.810–7 –3 m to 10 m 41014 Hz to 11 310 Hz 1.6 eV to –3 1.6 eV electrons in atoms and molecules (a) Thermal effect (a) Used in industry, me (b) All properties similar dicine and astro nomya of those of (b) Used for fog or have light except  photography (b)Change of molecular vibrational and rotational energies. (c) By bodies at high temperature 6. Microwav es Hertz special electronic devices such as klystron tube 10–7 m to 31011 Hz to 10–3eV to (a) Pheno (a) Radar and 0.3m 9 –5 mena of reflection, telecom munication refraction and diffraction (b) Analysis of 10 Hz 10 eV fine details of molecular structure 43 Electromagnetic Waves 7. (A) Radio waves Marconi Oscillating circuits 0.3m to few kms. 9 –3 10 Hz to few Hz 10 ev to 0 of Radio spectrum Super High Ultra High (b) (UHF) Very High (c) (VHF) (a) Radio communication properties (b) Radar, more than particle like properties Radio and sate llite communi cation (Microw aves) Radar and television broad cast Subparts (a)SHF (a) Exhibit waves like short distance 0.1m to 0.1m 31010 Hz to communication, Television 3109 Hz communication 0.1m to 1m 3109 Hz to 1m to 10m 3108 Hz to 3108 Hz 3107 Hz (B) High 10m to 3107 Hz to Medium frequency 100m 3106 Hz distance Medium frequency 100m to 1000m 6 310 Hz to Low Frequency 1000m to 10000m 3105 Hz to Very low frequency 10000m to 3104 Hz to 30000m 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% along with a little portion of argon, carbon dioxide, water vapor, hydrocarbons, sulphur compounds and dust particles. (iii) The density of atmospheric air goes on decreasing gradually as we go up. 3105 Hz long range 3104 Hz 3104 Hz (iv) (a) (b) communication Marine and navigation use, communication. Long distance communication. The earth's atmosphere has no sharp boundary. However, it has been divided into various regions as given below: Troposphere: It extends upto a height of 12 km from earth's surface. The temperature in this region decreases from 298K to 220K and conductivity increases. All climatic changes occur in this region. Stratosphere: It extends from 12 km to 50 km after troposphere. At the upper part of this region, approximately 20km 44 (c) (d) (v) 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. Mesosphere: It extends from 50km to 80 km after stratosphere. In this region the temperature decreases from 280K to 180K 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. 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) (a) (b) Electromagnetic Waves Propagation of Radio waves: Low frequency waves-the AM band: Radio waves 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. 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 possibleonly 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 = d2 2 Re where d is the radius of the area to be covered for TV telecast and Re is the radius of earth. 45 Electromagnetic Waves Ex. 11 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/km 2? (A) 39.5  105 (B) 19.5  106 (C) 29.5  107 (D) 9  104 Sol. (A) Radius of the area covered by T.V. telecast d= 2h R e Total population covered ¾ d2  population density = 2hRe  population density 1000 = 2  3.14  100  6.4  106  6 10 5 = 39.503  10 Ex. 12 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) -ray region Sol. (C)  = This wavelength belongs to X-ray region. 46