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Pradeep 4 Fundamental ehysies: OO orgs, 8/18 Say, 8.14. MICROWAVE OVEN —, , It is a device used in the kitchen for ee eae ea eetee aun owen of, , microwave oven is to create microwave radiation of suitable frequ ‘ . ei ce, , oked, can be kept. This radiation may match the resonant frequency o rotation Of Watey, oct en 3 GHz, In this situation, energy from the waves is vie (0 the, Kinetic energy of molecules. The microwave of this frequency is of energy equival len heating up Water,, When microwaves of this frequency fall on food item containing water like fruit, vegeta - Meet, cereals, ete., placed in oven, the water molecules absorb these radiations. Their energy aha ese molecules, share their energy with neighbouring food molecules. Due to it, the entire food gels heated., , Tn a microwave oven one should use porcelain vessel and not metal container, for Cooking the food, items. It is required to avoid the danger of getting an electric shock from accumulated electric charges ang to, avoid the melting of metal from heating., , In a microwave oven, the porcelain container containing food for cooking remains uneffected and coo}, whereas only the food items gets heated. It is so because the large molecules of Porcelain container vibrate, and rotate with much smaller frequency than that of microwave and thus cannot absorb microwaves,, , In this situation energy from the waves is transferred efficiently to the kinetic energy of molecules,, , , , QUESTIONS, , , , Ca Suppose the eyes of an alien being are sensitive to microwaves. Do you expect the alien, being to have larger or smaller eyes than ours ?, , Ans. The wavelength of microwaves is much larger than that of visible light. The microwaves can enter, the eyes and produce the Sensation of sight if the size of the eyes and the ratina are comparable to, , be It is a common belief that the exposure of radiations for long time is dangerous to health. Are, the meals cooked in microwave oven not dangerous to health ?, , Ans. As is known, the exposure of high energy radiation like ‘YTfays, X-rays and ultraviolet rays for long time is, dangerous to health as they spoil the blood cells of the body. Microwaves are low energy waves with, energy much less than that of ‘Tays and X-rays. The microwaves do not have the energy necessary to, , ionise molecules in the way, that y-rays, X-rays can do. Therefore, the meals cooked in microwave oven, are not dangerous to health., , DISPLACEMENT CURRENT,, , , , T° AMPERE-MAXWELL's Law (i $ Bal =uc41,), Formulae used. (iii) Magnetic field induction B at a point outside a, Straight wire at a perpendicular distance r from a, (i) Ipeq do, =<, Age. Straight wire carrying current / is given by, dt dt My 27, B= 026, 7" ag {t)-fo4w on, dt\d d a where € = 8.85 x [Q-12 CN! m?,, * Jp =cl ope’ | Gg A ; Hg = 4 mx 107 Wh Av! ne! or (TA7! m), a B= qin C= == Units used. Vin volt, A in m?, Cin Sarad, dor rin, a m, Tin ampere and Bin tesla or Wo m?,, , , , Scanned with CamScanner

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ELECTROMAGNETIC WAVES, , 8/19, , , , Example f] There is a parallel plate, capacitor of capacitance 2-0 LF. The voltage, between the plates of parallel plate capacitor is, changing at the rate of 6-0 V s-!, What is the, displacement current in the capacitor ?, , Solution, Here, C = 2.0 tF = 2 x 10° F,, dV, , —s= -1, dt =6'V's, , Displacement current,, , IE d(V, Ll, =e, A =e, AX | —, peo a 8 ata, , e,A, oi av. av, dat dt, = (2x 10°) x 6=12x 10A, =12pA, , Examplef] A parallel plate capacitor, C = 0:2 pF is connected across an a.c. source of, angular frequency 400 rad s-!, The value of, conduction current is 2 mA. Find the rms value of, the voltage from the source and the displacement, current in the region between the two plates., , Solution. Here, C = 0-2 uF = 0-2 x 10°F, , =2x107F,, = 400 rad/s, ppg =2mA=2x 107A, Vins = Arms * Xe, , , , 1, =I x—=(2x103)x—_—___|_, rms “oC 400 (2x1077), =25V, Displacement current = conduction current, =2mA, , Example] (a) Fig. 8.10 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. Use Ampere’s law (modified to include, displacement current as given in the text) and the, symmetry in the problem to calculate magnetic, field between the plates at a point (i) on the axis, (ii) 6.5 cm from the axis (iii) 15 cm from the axis., , FIGURE 8.10, _ FIGURE 8.10, , \, , , , (b) At what distance from the axis is the, magnetic field due to displacement current, greatest ? Obtain the maximum value of the field., , Solution. Here, R = 0.12 m,/=0.15 A, , +. Area of the plate, A = 1. R2 = 1 x (0.12)? m?,, , (a) Consider a loop of radius r between the two, circular plates, placed coaxially with them., , Then area of the loop, A’=" 77, , =, By symmetry magnetic field induction B is, equal in magnitude and is tangentially to the circle at, , every point., In this case, only displacement current Ip will, cross the loop. Therefore, using Ampere’s Maxwell, , law, we have, , 2 GO, , B.dl = ply, , 2 1r B= Up x (current passing, through the area A’), , sud nr, =o DaR? for r<R, , =Holp for r>R, 2, , thus, paHotD” Holo” watr<R), , R?2nr 2nR?, , I, , and B=/0°D wi) (If r > R), , 2ar, (i) On the axis, r=0.—, Using (i), we get, B=0, , (ii) For a point 6.5 cm from the axis,, r=6.5cm=6.5 x 10 m., Using (i), we have,, Be 4nx1077 x0.15x6.5x 10, 2nx(12x107?)?, =135 x 107T, (iii) For a point 15 cm from the axis,, r=15cm=0.15 m., , Scanned with CamScanner

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8/20, Using (ii), we have,, , _ 4x10? x0.15, ~ 2nx0.15, , ()) From equations (7) and (ii) we note that B is, maximum if r= R = 12 em = 0.12 m, , Ho ln _ 4nx107? x0.15, mx AR 2nx0.12, =2.5x 10-7T, , Exampte [] (a) Use the Biot-Savart law to, determine the magnetic field due to conduction, current outside the plates (refer to Fig. 8.10) at, Points 6.5 cm, 12 cm and 15 cm from the wire. Do, the answers match with those in Ex. 3? Explain., , (b) If the conduction wire has a radius of 1.0, mm, what is the maximum value of magnetic field, due to the conduction current ? [When you, compare the answer to Ex. 3 (b) and 4 (b), you, will appreciate why it is not easy to notice magnetic, field due to the displacement current]., , (c) Suppose the thin wire in Fig. 8.10. is, replaced by rods each of radius 12 cm (i.e. we now, have two long cylindrical rods separated by a small, gap). Will magnetic field configurations for r>R, be identical for the regions between the plates and, outside the plates ?, , Solution. (<) Using Biot-Savart law, magnetic, field induction at a point outside a straight current, carrying wire is given by, Ho!, , 2nur, , Now, since J = Tp, this formula is the same as in, Ex. 3 for r > R, Therefore, for r = 12 cm and, 15 cm, answers are the same as in Ex. 3. For, , r= 6.5 cm (r < R), the two formulae differ. Here, B=46 x 107’ T, greater than B in Ex. 3,, , (6) Bis maximum at the surface of the wire,, , Bato! _4nx10-7 x0,15, 2mr exo, (Here, r= 1.0 mm = 10-3 m], This is much greater than the maximum value, of B in Ex. 3. Due to this Teason, it is not easy to, notice magnetic field due to displacement current., , (c) Yes, since J = Tp » magnetic field, configuration are identical for r > R., , B =2x107T, , B, , , , , , B, , =3.0x 106T, , , , ies (XIN), , , , Pradecp Fundamental Phy., as =, te, , Example A parallel plate Capacity, 7, area 50 cm? and plate separation 3.9 tm, i red initially to 80 LC. Due to a radiogyiys, ‘ ee earby, the medium between the Plates, , ache condueling and the plate loses the cha :, initially at the rate of 15 x 10 a @ Whats, the magnitude and direction of displacemen, current ? (ii) What is the magnetic field between, NCERT Solved Exampi,, , , , the plates ? 23 gaia », Solution. Conduction current within the Plates, , is from the positive plate to negative plate Of paralle|, , plate capacitor., (i) Displacement current,, , doy d dE, , =€, —= =€, — (FA) =e, A, , Tp =€o dt So an } 0 dt, d(_q | dg, , = —|— ]=e, a—_ 4, «“5(aca] One, A at, , de -8, a Ip == 15x10 A, dq, , Since charge is decreasing with time, so oF, , dE er, and hence a 0. It shows that the direction of Ip, , is opposite to that of electric field and hence opposite, to the conduction current. But the magnitude of, displacement current is same as that of conduction, current. The net current between the plates is zero., So total current, J’ = J + In=0, , (ti) Using Ampere’s Law, $ B- d= Hy!’ =0, , Hence magnetic field within the plates is Zero., aa DE, , Formulae used. () Energy of a photon,, E=hy =he/h, Ey= Boc, , VELocity, AMPLITUDE AND ENERGY, NSITY OF ELECTROMAGNETIC WAVES, , (ii), , Where Ey = Amplitude (or maximum value) of, electric field,, , By = Amplitude (or maximum value) of, Magnetic field, , + — i 1 E, , 2 0, , —__, , Scanned with CamScanner

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ELECTROMAGNETIC WAVES 8/21, yd 2 2 8 \2, and up lg? _ (By V2)? es bo, of ee es (3x10 : = 1-86, 2No Hy Ay ryt, (2-2x108)? x1-0, , where up, Ug = average energy density of electric, and magnetic field, , E, B =r.m.s. value of electric field and, magnetic field respectively, , 2 3, Ey, By = amplitude of E and B respectively., Total average energy density = up + Up =2 Ue, Bg, 2Yg, , Units used. 4 is taken in m, cin ms“, vin Hz, Ein, NC71, Bin tesla, up and ug in Jnr., , ], =2up=5€o EB =, , , , Example a A plane electromagnetic wave, of frequency 25MHz travels in free space along, the x-direction. At a particular point in space and, , OK A, time the electric vector is, E = 6.3 V/mj., , ~, Calculate B at this point., , [CBSE 2006 (C)] NCERT Solved Example, , E 63V/im, Solution. B = —, , =—— =21x10°T, c 3x108 m/s *, , iF, As E isalong y-direction and wave is travelling, >, along x-direction, therefore, B is along z-direction,, > A, ie, B =21x 10 k tesla., , Example fj Electromagnetic waves travel, in a medium at a speed of 2-2 x 108 ms“!. The, relative permeability of the medium is 1-0. Find, the relative permittivity of the medium., , Solution. Here, v = 2:2 x 108 ms“, p, = 1-0;, , , , , , , , , , e,=?, 1 1 1 1, vp=——= = x, Vue Yo Heo, vHoSo vr, = _¢, He,, , Example [i] A plane electromagnetic is, propagating in the x-direction has a wavelength, of 5-0 mm. The electric field is in the y-direction, and its maximum magnitude is 30 Vmr!, Write, the suitable equations for the electric and magnetic, fields as a function of x and ¢., , Solution. Here, 4 = 5-0 mm = 5 x 103 m;, , Ey = 30 Vm, , 2, Angular frequency w= 2tv = ene, , _ 2x 3142x (3x108), 5x103, Max. magnitude of magnetic field,, , = 3-77 x 10!! rads“, , Ey 30, B= =, c 3x10, Equation for electric field along y-axis can be, written as, E=E,= Eg sin @ (t-x/c), = 30 sin 3-77 x 1041 (¢ — x/c) Vm™, , Equation for magnetic field along z-axis can be, written as, , , , = |d -7, gz = 10x10 5, , B= B,= Bg sin w (t —x/c), = (1-0 x 1077) sin 3-77 x 104 (¢ - x/C), , Tesla, ‘Examplef) Light with an energy flux of, 18 watt/cm? falls on a non-reflecting surface at, normal incidence. If the surface has an area of, 20 cm?, find the average force exerted on the, surface during a 30 minute time span, when no, , incident light is reflected. How will your result be, modified if the surface is a perfect reflector ?, , NCERT Solved Example, , Solution. Total energy falling on the surface,, U= 18 x 20 x 30 x 60 J =6-48 x 10°J, , Total momentum delivered to the surface is, , Scanned with CamScanner

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8/22, , The average force exerted on the surface is, , 9 :, Fa 2. MOXIO™ 12x ON, t 30x60, If the surface is a perfect reflector, the change, of momentum will be =p-(-p), =2p=2x 216 x 103kg ms!, Now average force,, , _ 2x2:16x103, "30x60, Example [{] Calculate the peak values of, electric and magnetic fields produced by the, radiation coming from a 100 watt bulb at a, distance of 3 m. Assume that the efficiency of the, bulb is 2-5% and it isa Point source ?, , NCERT Solved Example, ee ree,, , =2-4 x 10°N, , Solution. Useful Intensity,, , y= Power _ 100 (2-5/100) = 25 Wen?, 4n(3)2 36m, , Half of this intensity () belongs to electric field, and half of that to magnetic field. Therefore,, , area, , I 1 :, 2740 A0¢, , 27, , FO Veo e, , , , 2x (2:5/36 m), , 1, —— |x(3x108, (oe hc * 3, , = 4:08 V7!, , , , , Ey 4.08, By= 2a _,, 8, OS Zoh = 136 x 1087, , ‘Example il A plane electromagnetic wave, in the visible region is moving along z-direction,, The frequency of the wave is 6 x 1014 Hz, and the, electric field at any point is varying sinusoidally, with time with an amplitude of 2 V m-!, Calculate, (i) average energy density of the electric field and, (ii) average energy density of the magnetic field., , Dradeoe's Fundamental Physics (x1) q, 14, Solution. Here, v = 6 x 10" Hz, E, =2 Vat, , (:) Average energy density of the electric Fiely, , 1, _ 760 RR = Z * (B85 x 10°12) x 99, , = 8-85 x 107? J m-3, (ii) Average energy density of magnetic field, , 2 2, _ 0 1 ole” 1, “Fu, 4 A, 4 yc?, , 22, 4" Gaxl0 yx x10)?, = 885 x 10-12 J m3, Example [Y] A laser beam has intensity, 3-0 x 104 W m™~. Find the amplitudes of electric, and magnetic fields in the beam., Solution. Here, J = 3-0 x 10!4 W m-2,, Ey=?, By=?, Intensity of the plane electromagentic wave is, , 1, = oe 2, T=u,,c= € Ege, , 2, 2 ew (20 2x3x10!4, “ J. = SE, Oo Yeqe (8-85 x 107!2) x (3x 108), , = 4-75 x 108 V m-1, , Ey _ 4-75108, , Ba Be = 46, oe “3x08 = 158T, , Example [R] 4 beam of light travelling, , along x-axis is described by the magnetic field,, B.=5x 109 T sin w (¢— x/c), Calculate the maximum electric and, magnetic forces on a charge , i.e. alpha particle, , moving along y-axis with a speed of 3 x 107 m/s,, charge on electron = 1-6 x10%C, , Solution. Here, Maximum magnetic field,, Bo=Sx10°T;, charge on alpha Particle, g=+2¢, =2X 16 x 10-19 = 3.9 x 10-19,, ¥=3x 107 ms"!, , , , , , Scanned with CamScanner