64 m kj izns'k jktf"kZ V.Mu eqDr fo'ofo|ky;] bykgkckn

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4 522 m kj izns’k jktf"kZ V.Mu eqDr fo’ofo|ky;] bykgkckn vf/kU;kl (Assignment) 2012-2013 Lukrd foKku dk;Z e …ch-,l-lh-‰ Bachelor of Science Programme (B.Sc.) fo"k; % HkkSfrd foKku fo"k; dksM % ;w-th-ih-,p-,l- Subject : Physics Subject Code : UGPHS dkslZ ’kh"kZd % dkslZ dksM % ;w-th-ih-,p-,l-&04 Subject Title : Electric & Magnetic Course Code : UGPHS-04 Phenomena Section - A [k.M & v vf/kdre vad % 30 Maximum Marks : 30 vf/kdre vad % 18 Maximum Marks : 18 uksV % nhkZ m kjh; iz’uA vius iz’uksa ds m kj 800 ls 1000 ’kCnksa esa fy[ksaA lHkh iz’u vfuok;Z gSaA Note : Long Answer Question. Answer should be given in 800 to 1000 words. Answer all questions. All questions are compulsory. r 1. State and discuss the Gauss law of electrostatics. Derive D = eo r r E + P using Gauss law in a general case where conductors containing free charges are placed in an infinite dietectric medium. Here D r in the displacement vector, P r is the polarization Vector, E r r B is the electric field and eo is the electric permittivity of the vacumm. How the dielectric constant is defined from this relation 6 fLFkr fo|qfrdh ds xkml fu;e dk mYys[k ,oa O;k[;k dhft;sA ,d lkekU; fLFkfr esa tgka vuUr ijkoSkqr ek/;e esa j[ks pkydksa ij Lora=k vkos’k gks] xkml fu;e dk mi;ksx djds O;qRiUu dhft, fd D r = c r r E + P ;gka D r fo|qr LFkkukUrj.k osDVj gS] E r fo|qr {ks=k dh rhozrk gS vkSj eo fuokZr dh fo|qr’khyrk gSA bl laca/k ls ij oS|qr fu;a=kkd dks dSls ifjHkkf"kr djrs gSa\ 2. Using Gauss law determine r E inside and outside the sphere and plot graph showing variation of ½ r E ½ with distance from centre, assuming the permittivity of spherical region to be that of vacuum. find the energy stored in the electric field of this charge distribution. 6 xkml ds fu;e dk mi;ksx djds r E dk eku xksys ds vanj rFkk ckgj Kkr dhft, rFkk ½ r E ½ es dsUnz ls nwjh ds lkFk gksus okys ifjorZu dks n’kkZrs gq, xzkQ [khafp;sA ;g eku yhft;s fd xksyh; {ks=k rFkk fuokZr ds ijkoS|qrkad eku cjkcj gSA bl vkos’k forj.k ds fo|qr {ks=k esa lafpr ¯tkZ Kkr dhft,A 3. Show by a simple model that (i) magmatization current is found is materials with inhomogeneous atomic currents only. (ii) find the total magnetization current density along any one axis and (iii) hence show r M is the magnetization and r Jm is this that Jm r = Curl M r , where magnetization current density. 6 ,d ljy ekWMy ds }kjk ;g n’kkZb;s fd (i) pqEcdhdj.k /kkjk dsoy vle:i ijek.kq& /kkjk j[kus okys inkFkksZa esa gh ik;h tkrh gSA (ii) igys fdlh ,d v{k dh fn’kk esa pqEcdh; /kkjk kuRo fudkfy;s fQj (iii) ;g n’kkZb;s fd Jm r = Curl kuRo gSA r M tgkaa M r pqEcdhdj.k rFkk Jm r pqEcdh /kkjk Section - B [k.M & c vf/kdre vad % 12 Maximum Marks : 12 uksV % ykq m kjh; iz’uA vius iz’uksa ds m kj 200 ls 300 ’kCnksa esa fy[ksaA lHkh iz’u vfuok;Z gSaA Note : Short Answer Question. Answer should be given in 200 to 300 words. Answer all questions. All questions are compulsory. 4. Find total energy absorbed by two coupled coils and prove that L 1 L 2 > M 2, Here L 1 and L 2 and self inductances of first and second coil and M is mutual inductance. 3
nks ;qxfer dq.Mfy;ksa }kjk iw.kZ vo’kksf"kr ¯tkZ dk eku fudkfy, rFkk fl) dhft;s fd L 1 L 2 >, M 2 , tgka L 1 vkSj L 2 izFke] ,oa f}rh; dq.Mfy;ksa dh Loizsjdrk;sa gSa rFkk M vU;ksU; izsjdRo gSaA 5. Using Biot-Savart law show that one can write r B = curl r A find an expression for r A . 3 ck;ks& lckVZ fu;e dj iz;ksx dj fn[kkb;s fd r B = curl r A , fy[kk tk ldrk gSA r A dk O;atd fudkfy;sA 6. Obtain an expression for electrostatic potential at a distant point due to an electric depole. 3 fdlh oS|qr f}/kzqo ls lqnwj fcUnq ij fLFkj oS|qr foHko ds fy, ,d O;atd izkIr dhft;sA 7. Obtain an expression for arial magnetic field of a solewod. 3 fdlh ifjukfydk esa v{kh; pqEcdh; {ks=k ds fy, ,d O;atd izkIr dhft,A ¾¾¾¾¾¾ 5
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Page 3: 3 5. Derive the wave equation for a
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Page 19 and 20: 19 - ;fn y = (sin 1 x) gks rks fn[k
Page 21 and 22: 21 é1 2 3ù A = ê ú ê 0 5 0 3
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74 8- lgh fodYi dks fVd djsaA ¼d½
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81 efJe
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83 efJe
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87 6. efJeÅeeLeea meneÙelee mesJe
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89 117 Gòej ØeosMe jepeef
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99 127 mRrj iznsk jktfkZ V.Mu eqDr
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153 OTHER INFORMATION : Volume 2 :
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181 efJe
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189 ╪ ≥ Γ ┼ ┘ 96 G≥ ej e
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64 m kj izns'k jktf"kZ V.Mu eqDr fo'ofo|ky;] bykgkckn

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