ELECTROMAGNETISM. at a point whose position vector with respect to a current element i d l is r. According to this law :

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1 ELECTROMAGNETISM ot-svt Lw: Ths w s used to fnd the gnetc fed d t pont whose poston vecto wth espect to cuent eeent d s. Accodng to ths w : µ d ˆ d = 4π d d The tot fed = d θ P whee ˆ s unt vecto n the decton of vecto. Mgnetc fed due to cuent cyng conducto: µ = {sn α + sn β} 4π The decton of gnetc fed cn be obtned fo the ght hnd thub ue. µ qv ˆ Mgnetc fed of ovng chge =. 4π A α β P ot-svt s w s the gnetc nogue of couob s w n Eectosttcs. The chge eeent dq ppeng n Couob s w s sc but the cuent eeent d ppeng n ot-svt s w s vectod. Mgnetc fed ong the s of ccu oop: P µ n = ( + ) 3/ µ nπ When >>, = 4π 3 A cuent cyng oop cts s gnetc dpoe whose gnetc oent s M = na, whee A s e of the co. µ Mgnetc fed ong the s of soenod = n {sn α + sn β} whee β n s α nube of tuns pe unt ength. If soenod s vey ong α = β = 9 = µ n. Foce on ovng chge n unfo gnetc fed : The foce on chge q ovng wth veocty v n gnetc fed s gven by F = qv..e. F = qvsn θ whee θ s nge between v nd. Ntue of pth foowed by chged ptce n gnetc fed : If θ = o 18, F =, the pth s stght ne. v If θ = 9, F = v, the pth s cce of dus =, nd fequency fequency). f q = (cycoton π

2 If θ es between nd 9, the chged ptce descbes hec pth of dus v sn θ/q; nd ptch = vcos θ π. Afte te t, the devton w be θ = t, n tes of ength of the gnetc fed. 1 θ = sn. Loentz Foce: The foce epeenced by ovng chge wth veocty v n eectc fed E nd gnetc fed, F = q{e + v }. When E s pe to nd ptce veocty s pependcu to both of these feds, the pth of ptce s he wth ncesng ptch. When eectc fed E s pependcu to nd the ptce s eesed t est fo ogn, the pth of ptce s cycod. Foce on cuent cyng conducto n gnetc fed F = ( ). Foce between two pe wes cyng cuent s gven by F µ 1 =. π 1 Toque on cuent oop: The toque epeenced by cuent cyng oop n gnetc fed. τ = M Whee M s gnetc dpoe oent = na. Potent enegy of gnetc dpoe cuent cyng oop n gnetc fed s U = M. d Foce on gnetc dpoe cuent cyng oop non-unfo gnetc fed F = (M ). d 1. The Mgnetc fu (φ) though gven sufce A s gven by φ = N A = NA cos θ. dφ. Fdy s ws of eectognetc nducton sys e =. dt Whee e = nduced ef nd φ = gnetc fu. Accodng to Lenz w, negtve sgn shows tht ef nduced w oppose the chnge n gnetc fu (cusng the ef). 3. Moton ef: The ef nduced due to oton of conducto n gnetc fed s gven by e = ( v), whee s the gnetc fed n the egon. s the ength of conducto nd v s the veocty of conducto. The fou gven bove s vey usefu to fnd the nduced ef.

3 The decton of nduced cuent cn be gven by Feng Rght Hnd Rue. If the thub nd fst two fnges of the ght hnd e sped out so tht they pont n thee dectons t ght nges to one nothe, the fst fnge gvng the decton of gnetc fed, the thub ndctng the decton of the oton of the conducto, then second fnge ndctes the decton of nduced ef o cuent. 4. Moton ef nduced n ottng conducto: A conductng od of ength ottes wth constnt ngu speed ω bout one end P n unfo gnetc fed. Consde segent of od of ength d t dstnce fo P. ω Induced ef n ths segent de = v d = ω d. d Q Sung the ef nduced coss segents we get, tot ef coss the od. e = d = ω d ω e = Fo Feng Rght Hnd Rue we cn see tht Q s t hghe potent nd P s t owe potent. 5. Sef nducton: Wheneve the eectc cuent psses though co chnges, the gnetc fu nked wth t so chnges. As esut, n ef s nduced n the co. e cefu n checkng the poty of ths nduced ef (o ced bck ef). ) The gnetc fu poduced n co s decty popoton to the cuent fowng n t,.e., φ α I o φ = LI. The constnt of popotonty L s defned s the coeffcent of sef nducton. dφ d b) The nduced e..f. geneted n the co s gven by e = = L dt dt c) The nductnce of ong soenod o tood s gve by L = s ength nd A s e of coss- secton. µ N A. Whee N s tot nube of tuns, Ccut s de ON o cuent s ncesng Ccut s de OFF o cuent s decesng d e = L dt d e = L dt 6. Cobnton of nductos. ) Cos n sees LS = L1 + L L1L ) Cos n pe LP = L + L 1 7. Coeffcent of utu nducton (M): ) The coeffcent of utu nducton between two cos s equ to tht gnetc fu nked wth the secondy co whch s poduced s esut of unt cuent fow n the py co. φ M 1 = when I 1 =1 p, then M1 = φ I1 b) The coeffcent of utu nducton s nuecy equ to tht nduced e..f. n the secondy co, whch s poduced s esut of unt te of chnge of cuent n the py co, e M = di 1 when 1 =1 p/s, then M 1 =e di1 dt dt

4 µ N N A c) Mutu nductnce of two soenods o cos s M = 1. N 1, N e nube of tuns n py co, secondy co, s ength nd A s e of coss secton of the cos. d) If L 1 nd L e sef nductnces of two cos, the utu nductnce between the cos M= L 1L. 7. Oscton n L-C ccut: When chged cpcto C hvng n nt chge q s dschged though n nducto L then C q Ld q d q d q 1 = o L = o + q =. L C dt C dt dt LC Ths equton s s to Spe Honc Moton. So, chge osctes n the ccut wth ntu 1 π fequency ω =, q = q cos ωt, = qωcos ω t + LC. Apee s Lw: S to the Guss s w of eectosttcs, ths w povdes us shotcut ethods of fndng gnetc fed n cses of syety. Accodng to ths w, the ne nteg of gnetc fed ove the cosed pth ( d ) s equ to µ tes the net cuent cossng the e encosed by the pth. d = µ I cosed opp encosed d = µ (I + I I ) 1 3 I 5 d I 1 I 3 I I 4 Pevous questons: 1. A thn febe we of ength L s connected to two djcent fed ponts nd ces cuent I n the cockwse decton, s shown n the fgue. When the syste s put n unfo gnetc fed of Hnt: stength gong nto the pne of the ppe, the we tkes the shpe of cce. The tenson n the we s (IIT-1) ) IL b) *c) IL π T cos θ/ T Id d) IL π IL 4π d θ/ A θ/ θ/ θ θ/ O T sn θ/ T T cos θ/

5 T Id IR IL = = = θ π d L θ =, R = R π. A stedy cuent I goes though we oop PQR hvng shpe of ght nge tnge wth PQ = 3, PR = 4 nd QR = 5. If the gntude of the gnetc fed t P due to ths µ oop s k I Q, fnd the vue of k. 48 π Anyss: µ I µ I = [cos53 + cos37 ] = 7 1 4π 48π 5 K = /5 37 P 4 R 3. A gnetc fed = ˆ j ests n the egon < < nd = ˆ j, n the egon < < 3, whee s postve constnt. A postve pont chge ovng wth veocty v = vˆ, whee v s postve constnt, entes the gnetc fed t =. The tjectoy of the chge n ths egon cn be ke, ) z b) z c) z d) z 3 3 Anyss: fo < < < < 3 pth w be concve upwd pth w be concve downwd ; So () coect. 4 A ptce of ss nd chge q ovng wth veocty v entes Regon II no to the boundy s shown n the fgue. Regon II hs unfo gnetc fed pependcu to the pne of the ppe. The ength of the Regon II s. Choose the coect choce(s). *) The ptce entes Regon III ony f ts veocty v > Regon I Regon II Regon III b) The ptce entes Regon III ony f ts veocty v < v *c) Pth ength of the ptce n Regon II s u when veocty v = *d) Te spent n Regon II s se fo ny veocty v s ong s the ptce etuns to Regon I.