Study Material with Classroom Practice solutions. To Electromagnetic Theory CONTENTS. 01 Static Fields Maxwell Equations & EM Waves 06 11

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2 Pg No. Stud Mtil with lssoom Pctic solutions To lctomgntic Tho ONTNTS hpt No. Nm of th hpt Pg No. Sttic Filds 5 Mwll qutions & M Wvs 6 Tnsmission ins Wvguids lmnts of ntnns 7

3 hpt. ns: V cos cos î ĵ sin â â Fom divgnc thom.d V.nˆ ds v kˆ sin Sttic Filds â.ddv... cos sin = cos sin dv = ddd Putting ths vlu in qution w hv V.nˆ ds. ns: (c) lss oom Pctic Solutions d d d dd d Fo th givn t I =.d, I is vlutd ov th pth shown in th Fig., s follows I =.d, =, +.d, fom to, = fom to.d, =, fom.d, / d d d / / to, = fom to / / d t = / = / ( ). ns: (d) F= + sin = F + F + F = + sincos = + sincos () sin. F = F F. F F Z =,. F =. F =. F / / Fig.. ns: (c) D â D 6 â Z D D ρ s â n n ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

4 â D ρ s â n â Z ρs m 5. ns: (d) V = + V 6 D.D = â 6 â â ( 6 ) ( = (t, ) = = 6. ns: (d) Givn V(,, ) = (,,) in f spc gd (V) V â ) : : lctomgntic Tho,,,, o in i, j, k nottion, i j k 7. ns: (b) Fo vlid,. = = = =. ns: (d) shown in th Fig. Th two infinitl long wis ointd s V (,, ) V V V/ m V / m,, = Diction of th lctic fild is givn b th unit vcto in th diction of. Th infinitl long wi in th - pln cing cunt long th diction poducs th mgntic fild t th oigin in th diction of =. Th infinitl long wi in th - pln cing cunt long th diction poducs th mgntic fild t th oigin in th diction of =. wh, nd unit vctos long th, nd s spctivl. nd componnts of mgntic fild non-o t th oigin. ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

5 : 5 : Postl oching Solutions 9. ns: (). divgnc lss vcto m b cul of som oth vcto.dl l s. ds s. ds is qul to mgntic flu though sufc.. ns: (c) In gnl, fo n infinit sht of cunt dnsit k /m H = k n H = =. ns: (b) = (Sinc = hs noml componnt ) t.. W know mgntic flu dnsit is continuous n n n. Sufc chg, k = H H t t H H t t t t t (. t n..6.. H.6.. H [.6.. ) ] / m. ns: (b) Tngntil componnts of lctic filds continuous t t sin sin () D D n n S D n n Dn ngg. Publictions () Fom () ( ) = s s =. ns: () =... n. = P = Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt P = Noml componnt of lctic flu dnsitis continuous coss chg f intfc D D n n cos cos () 6 tn tn tn 5

6 hpt Mwll qutions & M Wvs mpl.:. sint â V / m t = sintâ V / m t = t = â = â = i.., lin polition nd lso vticl polition with spct to ˆ is. H 5cost â / m This is lin polition. sint â sint â phs diffnc btwn â componnt nd â componnt is So tht it is lin polition Not: fo phs diffnc &, ispctiv of thi mplituds it must b in lin polition.. 55cost â 55sint â Phs diffnc btwn â componnt is mplituds sm. So it is cicul polition t = nd lt = t = 55â â â 55â It is W diction i.. HP ngg. Publictions â componnt nd sin t â 5cos t Phs diffnc = mplituds = not sm So it is llipticl polition. To dcid diction of ottion follow blow pocdu. t =, nd t = t = = â Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt â 5â â = â 5â = â â It is nti clock wis diction i.., ight Hnd llipticl Polition. j tz â jâ cos cost jsint jcost jsint t â sint â â â â Mgnituds of mplituds sm, phs diffnc is ; So it is cicul polition. Now w pocd to dcid diction of ottion. H cos t â sin t â t = & lt = t = = = â â â â â â

7 : 7 : Postl oching Solutions = â â i.., w gt clock wis ottion i.., ft Hnd icul Polition 7. not vlid M wv psnttion. 5cos t â t = & = t t 5â â = 5â 9. = â i.., lin polition j t Im â jâ cost jsint â = Im jcost jsint â = sin(t) â + cos(t) â t = & = t = â â â â = â â â â So it is ight Hnd llipticl Polition. sint â sint 5 â lt = & = t t = o â sin 5 â = â â â sin 5 = â â t â o t = â sin5 â = â â o t â sin5 â = â â Not: = 6.76 is th mimum vlus diction obtind b d t = & t = d t â â t = â â So it is HP. sint â sint 5 â Vlid M wv but polition cn not dfind. This is vlid M wv psnttion but it is not stisf non of th polition pincipl lss oom Pctic Solutions. ns: (c) Givn ful = (t t)mwb Mgnitud of inductd mf ' = t = () = 6mWb t sc ' d dt This fo on tun; but fo tuns ' N =.6 volts 6mWb tsc ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

8 : : lctomgntic Tho. ns: (d) Givn, = cos ( 6 t ) V/m H = cos ( 6 t ) /m = ; = W know tht, H H = Mdium, Mdium, i Glss slb n = = = Fig. = / m H = cos ( 6 t ) /m = 6 = = =.d/m. ns: (b) This qustion lts to noml incidnc of UPW on th i (mdium ) to glss (mdium ) intfc s shown in Fig. n =.5 = = If n nd n th fctiv indics nd v nd v th vlocitis n v μ n v μ fo Fo n =, n =.5.5 μ μ μ flction cofficint, i P % P 5 i f spc =, =, = Incidnt. wv flctd. wv i. ns: (d) Noml incidnc is shown in Fig. Fig. losslss ( = ) non-mgntic( = ) dilctic( > ) Intfc 5 Givn: m = 5 min in mdium. m VSW, S 5 min S 5 S 5 flction cofficint, i, 5 5 η μ ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

9 : 9 : Postl oching Solutions 7 9 π 6 π ( π) Ω Intinsic impdnc of th dilctic mdium, = 5 5. ns: () Givn: j5 (â jâ ) in f spc. jβ ( ) ω β 5 c ω 5 c 5 d/s f =.9 GH. GH t = 9 =, = j lds b 9 t = t = cos (t) thn = cos (t + 9) ft Hnd scw is to b tund in th diction long th cicl s tim incss so tht th scw movs in th diction of popgtion,. Th wv is lft cicull polid. 6. ns: (b) H =.cos (t-) â Wv is pogssing long + X diction (+X) H H t = t = t = 7.cos( t ) -jβ ^ s =.η H s =. * P vg = s Hs =. η ^ =. (π) ^ w/m = pln ds =dd ˆ W vg = 7. ns: () P PQ p Pp Q P vg.ds wtts S =. () = dd -jβ ^ (.) (π) π(5) =.59 Wtts = 59. mw 6 mw P Q = : P P. ns: (b) = = f.5 f f f ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

10 : : lctomgntic Tho.5 =. 75 m Simill m 9. ns: (b) 5 = π 5.5 = 9.7 Sinc >> hnc s wt is good conducto Wh ttnution is 9%, tnsmission is %, thn - =. Wh is ttnution constnt ωμσ = =.75 = ln(.).75 =. =.7m. ns: (b). 59. ns: (c) is minimum H is mimum i.., c is th option Tn Tn [pfct conducto Tn ] H J H Tn S n Tn HTn JS n [pfct conducto HTn ]. ns: (d) -. 6 H=.5 cos ( t-)ˆ /m (+X) =η=- H H Wv fqunc = 6 dins/s Phs constnt = d/m d/m = =.m. Th wv is tvling long +X diction, Givn wv is polid long Y. It hs Y-componnt of lctic fild. ns: () Th noml incidnc of pln wv tvling in positiv diction is shown t th intfc = in Fig. i md () md () Incidnt wv flctd wv f spc ( =, =, = ) = < Givn: wh i i i = Fig. Tnsmittd wv t losslss mdium ( =, =, = 9 ) > cos ( t ) V / m = d/s,, v Fo f spc, v = v = m/s = d/m H i i ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

11 : : Postl oching Solutions i cos ( t ) H i Hi Hi H, H i 9 Wh H Hi H cos ( t ) π cos ( t ) /m π Not tht H is flctd wv which tvls in ngtiv diction, which cosponds to + tm with = in th pssion fo H.. ns: (b) wst s ngl θ =tn ε ε - 5. ns: (d) Givn tht t = Wh t is lctic fild of tnsmittd wv is lctic fild of flctd wv t If i is lctic fild of incidnt wv. η ut - = i η +η nd i nd lso i so = ε = ε tn = o t this ngl th is no flctd wv whn wv is plll polid. n sin i = n sin t sin i sin sin t = sin i o sin t = t = 6 t i ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

12 hpt Tnsmission ins mpl.7: (i) 5 lin, th impdnc t th junction is invs impdnc of shot cicuit i. fo 5 mt lin th impdnc t th junction is sm s lod impdnc bcus fo this lin chctistic impdnc is qul to lod impdnc so th nt impdnc t th junction is plll combintion of infinit nd i. onl Now th Z in is onl bcus fo this lin chctistic impdnc is qul to lod impdnc (ii) Fo th lin th impdnc t th junction is invs of lod impdnc i. Fo th lin th impdnc t th junction is sm s lod impdnc i. so th nt impdnc t th junction is plll combintion of nd i. onl So now th Z in is bcus fo th ngg. Publictions 5 th Z in is sm s junction impdnc bcus th impdnc is ptd fo v n wh n is n intg (iii) s bsd on th bov nlsis th Z in is mpl.: Y = U = S = [U Y] [U + Y] = = = mpl.5: Z = U = S = [Z U] [Z + U] = = =. ns: (b) Z jztn β Z in = Z Z jz tn β Phs vlocit ω υ p β f υ p β f π β υ. ns: () lss oom Pctic Solutions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt = p π β π. (Givn l=m) tnβ Z Z in ( j)ω m Z jx Z = m

13 : : Postl oching Solutions V s = j j -. t t j = -j j ω π υ P β β Givn f = 5 MH υ p = m/s. ns: (b) V j = -j i = (o) j j. j j j. j. =. j. j j j. 7 j. = l j Z Not: In th options. is givn. ut coct nsw is. S / Z = 5 Fig. P Z Q j Z i / Z i / j. ns: (c) Fom th voltg SW pttn, V min =, V m =, VSW = S = Z = = 5 t th sistiv lod b Fo sistiv lods S fo fo = S = 5 = fo > = /S = 5/ =.5 fo > s voltg minimum is occuing t th lod point, =.5 5. ns: () flction cofficint: ns: (d) Th intconnction of T s is shown in Fig. 5 Zi 5 5 Zi. 5 Z = 5.5 = 5 Z = 5 Z = 5 ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

14 : : lctomgntic Tho Z flction cofficint t PQ Z t th input S, 5 flction cofficint, = 7 s = Z Z j iii) Opn lin mns Z =, Givn tht. tnπ Z jz tn Z in = Z Z jz tn Z in = Z iv) Fo mtchd lin of n lngth Z = Z Z jz tn Z in = Z = Z Z jz tn 7. ns: (d) Z in = Z Z Z ngg. Publictions jz jz tn tn i) Fo shotd lin, Z = = / = Z in = Z in = j Z jz Z Z ii) Fo shotd lin mns Z = Givn tht = Z Z in = Z Z in = Z. ns: (c) Th lin is mtchd s Z = Z = 5 nd hnc flctd wv is bsnt. Fo th tvling wv, givn: Phs diffnc fo lngth of mm = / d Fqunc of cittion = GH Phs vlocit, v p = 9 d/sc = Phs-shift p unit lngth = d / m v p.6 9. ns: (b)..99 S.99. Fo cipocl; S = S It is stisfid. Fo losslss lin S S.. 9 m / s =.9 It is loss lin Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

15 hpt Wvguids lss oom Pctic Solutions. ns: (b) vnscnt mods mns no wv popgtion. Dominnt mod mns, th guid hs lowst cut-off fqunc. TM nd TM not possibl, th minimum vlus of m, n fo TM t lst, spctivl. ns: () Th mod which hs lowst cutoff fqunc is clld dominnt mod T. t GH ll mods vnscnt. t 7GH dgnt mods possibl T nd TM dgnt. c f c T 5 GH. t 6 GH dominnt mod will popgt. t GH high od mods possibl. ns: () Givn: In ctngul WG of coss-sction : ( b) ω μ π π H sin sin (ω t β ) ŷ h Th wv is tvling in th -diction hving componnt onl s function of. s th is no componnt of in th diction of popgtion, th wv is Tnsvs lctic (T). omping th sin tm in with th m π gnl pssion: sin m = s th is no function of in, n = Th mod of popgtion in th WG is T ngg. Publictions. ns: (d) Givn =.755, b =.5, f = GH, c = m/s ut off fqunc Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt c m n f c = b Fo T, mod c f c = =.5 GH f > f c (T mod) so it popgts Fo T mod c f (T ) = = [f c (T )] = 6. GH f > f c [T ] so it popgts Fo T mod c f (T) = b c = 6.77GH b f > f c (T ] so it popgt Fo T mod c f c[t] = = 7.7 GH b f > f c (T ) so it popgt So, ll mods possibl to popgt. 5. ns: () Givn = 6cm, b = cm f = GH ut off fqunc c m n f c = + b

16 : 6 : lctomgntic Tho T : f c = T : f c = c c b =.5 GH =.75 GH c T : f c = + b c TM : f c = + b =.5 GH =.5 GH 6. ns: () m m n n b b Fo TM wv popgting long -diction nd H = TM c m n TM f c b Substitut c = cm/sc m =, = 6 cm n =, b = cm w gt f c = 5. GH fc TM f f GH nd =. & f c = 5. GH Substitut ll th bov vlus nd w gt TM = ns: (c) o W vg = η.b; T T c =, 9 f c = =.9 =.5cm / c.7cm So w gt T Putting ll th vlus W vg =.kw. ns: () 9. ns: () g c f c.9gh c.cm 9.9 Substitut g = cm, c =. cm W gt, = 5.7 cm f 5.7 =. GH. ns: () c.9gh 6.5cm b. ns: () cm f c f.9. =.57 d/cm = 5.7 d/m ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

17 5 hpt lmnts of ntnns lss oom Pctic Solutions W d =. ns: (c) ntnn civs W of pow: P = W MS vlu of incidnt fild = mv/m Pow dnsit, P d ( ) = W / m η 77 P ffctiv ptu, Pd m ( ) 77. ns: (b) osslss ntnn dictiv gin = 6 d = Input pow to th ntnn = mw fo losslss w gt % fficinc Wd G o Win Do W d = W in W d = mw. ns: (c) P d = W d = ngg. Publictions sin â = sin = sin d W / m sin dd U = P d = D m =. ns: (d) Wh P sin sin Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt U W m d = = = Dm =.5 W W 5. ns: (b) d sin m = P d.ds d d. â = â d =, l = G D =, G p =? = G p = d d 6. ns: (c) G D =.75 = D g = d= P T = 7.5 kw D g W/m.75 dition intnsit ditd Pow

18 : : lctomgntic Tho D g = U = U W d U = P d 7.5 P d : Pow dnsit w hv to find P d t = m U P d = 7.5 = W/m ( ) 7. ns: (d) W d = kw m = mv/m = km = 9% P d ( ) =.99 5 U m = ( ).99 5 = 769. Um D W d 769. D = G.9 D G = = 9.7. ns:. Givn: ntnn lngth, l = cm Fqunc, f = GH Distnc, = Wv lngth, f d 9 = cm, hnc th givn ntnn is Htin dipol. In th f fild, th tngntil lctic fild jidsin is givn b, 9. ns: (c) Givn: j77..v / cm ngth of dipol, =. s it is v smll, compd with wvlngth, hnc it cn b ppoimtd to Htin dipol d d = (.) d =. ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt

19 . ns: (d) n sin F sin tk limit n sin n. n t sin t. n ngg. Publictions n. ns: (b) In bod sid th WFN is givn b WFN (d) Wh, = lngth of th = (n ) d Givn: n = 9 Spcing, d WFN (9 ) WFN = 57.9 o. ns: (d) Th dictivit of n-lmnt nd fi is givn b D Wh, = (n )d : 9 : Postl oching Solutions nd ( n, v lg ) nd D D Dictivit, (in d) =. ns: 7.7 U m Dictivit, D Pd Givn: U(, ) = sin sin ;, U m = Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt P d sin sin sin = D sin sin dd dd D = 6 Dictivit, (in d) = log6 = ns: 79 Fo Htin dipol th dictivit, D is givn b D =.5 D

20 : : lctomgntic Tho.5 =.9 Wvlngth, m =.9 9 =.7 m ptu of ntnn is givn b P P Wh, P = pow civd t th ntnn lod tminls. P = pow dnsit of incidnt wv P P.7 6 P =.79 W/m (o) 79 nw/m Givn: No. of lmnts, n = Spcing, d Diction of min bm (o) pincipl lob, m = 6 o phs function, is givn b = dcos + To fom mjo lob. = = dcos m cos6 Th phs shft btwn th lmnts quid is 5. ns: (c) odsid diction o m = 6 o nd m / is of ngg. Publictions Hdbd Dlhi hopl Pun hubnsw ngluu ucknow Ptn hnni Vijwd Vig Tiupti uktpll olkt