TODUCTO MAGETC EFFECT OF CUET & MAGETM The molecul theo of mgnetism ws given b Webe nd modified lte b Ewing. Oested, in 18 obseved tht mgnetic field is ssocited with n electic cuent. ince, cuent is due to motion of chges, it becomes mndto to conclude tht it is the moving chge which cetes the field. MAGET Two bodies even fte being neutl (showing no electic intection) m ttct / epel stongl if the hve specil popet. This popet is known s mgnetism. This foce is clled mgnetic foce. Those bodies e clled mgnets. One end of the mgnet (s A) is diected ppoimtel towds noth nd the othe end (s ) ppoimtel towds south. This obsevtion is mde evewhee on the eth. Due to this eson the end A, which points towds noth diection is clled OTH POLE nd the othe end which points towds south diection is clled OUTH POLE. The cn be mked s nd on the mgnet. Pole stength mgnetic dipole nd mgnetic dipole moment : A mgnet lws hs nd nd it is poles of two mgnets epel ech othe nd the nile poles of two mgnets ttct ech othe the fom ction ection pi. F F F F (i) Mgnetic Field () iot-svt s lw ( due to cuent element) id d 4 i d p (b) due to moving point chge qv 4 q v P (c) due to stight cuent cing wie sin 1 sin 4 1 P t i A (sin1 sin ) 4 1 A Cse (i) Fo infinite long stight cuent cing wie 9 1
Cse Fo semi infinite long stight cuent cing wie 1 9. 4 Emple 1 : Find the field stength t distnce fom n infinite stight wie tht cies cuent. olution : Figue shows the infinitesiml contibution to the field, d fom n bit cuent element. n ode to integte contibutions of ll the elements, we use the ngle, mesue fom the pependicul, s the vible. Fom the digm we see tht d l = dl sin = dl sin ( ) = dl cos... (i) ince l = tn, we hve dl = sec d.... Futhemoe, = sec On substituting these epession into eqution nd integting we find = o cos d 4 1 o = (sin 1 + sin ) 4 ote: Those ngles on opposite sides of the pependicul e ssigned opposite signs. This ensues tht the contibutions fom eithe side of the wie dd togethe the thn subtct. Fo n infinite wie, the limits of integtion e / to +/. Thus = o... (iv)... (iii) dl d dl dl l 1 Clculting the mgnetic field poduced b long stight wie we m use the ngle s the vible. MAGETC FELD DUE TO A CCULA LOOP (d) due to cuent cing cicul loop (i) At cente : Due to ech d element of the loop t C is inwds (in this cse) numbe of tuns in the loop C Diection of : The diection of the mgnetic field t the cente of cicul wie cn be obtined using the ight hnd thumb ule. f the finges e culed long the cuent, the stetched thumb will point towds the mgnetic field.
Anothe w to find the diection is to look into the loop long its is. f the cuent is in nticlockwise diection, the mgnetic field is towds the viewe. f the cuent is in clockwise diection, the field is w fom the viewe. emicicul nd qute of cicle 4 8 (e) (f) (iii) On the is of the loop P / = numbe of tuns (intege) A loop s mgnet : The ptten of the mgnetic field due to cuent cing cicul loop is compble with the mgnetic field poduced b b mgnet. The side M L 'L' (the side fom which the emeges out) of the loop cts s OTH POLE nd side M (the side in which the entes) cts s the OUTH POLE. is fo 4 t is simil to is due to mgnet m = 4 Mgnetic dipole moment of the loop = = A (M) M t the is of olenoid olenoid contins lge numbe of cicul loops wpped ound non-conducting clinde. (it m be hollow clinde o it m be solid clinde) ni (cos 1 cos ) whee n is numbe of tuns pe unit length P 1 M (i) Fo del solenoid (long solenoid) ni Compison between idel nd el solenoid 1
del olenoid el olenoid ni ni/ end cente of solenoid end (distnce fom cente) / / Emple : n the shown figue cuent i is flowing in stight conducto nd enteing long the dimete of the cicul loop of simil conducto though the point A. The cuent is leving the loop long nothe simil semi-infinite conducto pllel to the plne of the loop though the othe opposite end D of the dimete. i A E i P i G D i Z Y X Wht is the mgnetic field t the cente of the loop P? olution : The mgnetic field t the cente P due to the enteing cuent long dimete is zeo. The mgnetic field t P due to the semicicul loop AED is i 1. k 4... (1) The mgnetic field t P due to the semi-cicul segment AGD is i. k 4... () nd the field t P due to the semi infinite stight conducto is given b i. k 4... () The net field t P is o i o i o i 1. k. k. k 4 4 4 o i. k o i. 4 k o The net mgnetic field t P is long the pependicul to the plne of the loop downwd. (g) (h) Ampee s cicuitl lw The line integl. d on closed cuve of n shpe is equl to (pemebilit of fee spce) times the net cuent though the e bounded b the cuve... d (i) Line integl is independent of the shpe of pth nd position of wie with in it. The sttement.. d does not necessil men tht evewhee long the pth but onl tht no nett cuent is pssing though the pth (iii) ign of cuent: The cuent due to which is poduced in the sme sense s d i.e.. d positive will be tken positive nd the cuent which poduces in the sense opposite to d will be negtive. due to hollow cuent cing infinitel long clinde : ( is unifoml distibuted on the whole cicumfeence)
(i) fo ; ; in 1 (i) due to solid infinite cuent cing clinde Assume cuent is unifoml distibuted on the whole section e (i) J J = cuent densit J ( J ) 1 = Emple : uppose tht the cuent densit in wie of dius vies with ccoding to J=K, whee K is constnt nd is the distnce fom the is of the wie. Find the mgnetic field t point distnce fom the is when () < nd (b) > olution: Choose cicul pth cented on the is of the conducto nd ppl Ampee s lw. () To find the cuent pssing though the e enclosed b the pth integte d = JdA = (K ) (d) K ince d K d d o 4 K ok o 4 (b) f >, then net cuent though the Ampein loop is K K d K 4 4 4 4 d MAGETC FOCE O MOVG CHAGE () Mgnetic foce cting on moving chge pticle in mgnetic field When chge q move with velocit v, in mgnetic field, then the mgnetic foce epeienced b moving chge is given b fomul:
F q( v ) F qv sin, whee is ngle between v nd ote: (i) F is pependicul to both v nd ince F is pependicul to v, powe due to mgnetic foce on chge pticle is zeo ( P F. v ). (iii) ince the wok done b mgnetic foce is zeo in eve pt of the motion, the mgnetic foce cnnot incese o decese the speed (o kinetic eneg) of chged pticle. t cn onl chnge the diection of velocit. (iv) On sttion chged pticle, mgnetic foce is zeo. (v) f v, then lso mgnetic foce on chged pticle is zeo. t moves long stight line if onl mgnetic field is cting. (b) Motion of chged pticles unde the effect of mgnetic foce (i) f pticle is t est v F m pticle will emin t est v hee = o = 18 F v constnt (iii) (iv) m pticle will move in stight line with constnt velocit f v is pependicul to nd is unifom F m qv ince F m is lws pependicul to v pth of chged pticle will be cicle v F m mv p mk qvm q q q q whee, p momentum of pticle; K kinetic eneg of pticle; V = cceleting potentil. m q q T,, f q m m ote: (1) The plne of the cicle is pependicul to mgnetic field. f the mgnetic field is long z-diection, the cicul pth is in plne. The speed of the pticle does not chnge in mgnetic field. Hence, if v be the pticle, then velocit of pticle t n instnt of time will be v v i v j whee v v v () T, f nd e independent of v while the dius is diectl popotionl to v. Angle between v nd is othe thn, 9, o 18 (Helicl pth) f the velocit of the chge is not pependicul to the mgnetic field, we cn bek the velocit in two components v, pllel to the field nd v, pependicul to the field. The components v emins unchnged s the foce qv is pependicul to it. n the plne pependicul to the field, the pticle tces cicle of dius. The pth is heli. v = sin v V = cos mv q mv sin q A 1 A A A 4 A 5 Pitch
ote : m q T ; f q m mv cos pitch (p) = v T q Following points e wothnoting in cse of helicl pth (1) The plne of the cicle of the heli is pependicul to mgnetic field. () The is of the heli is pllel to mgnetic filled. () The pticle while moving in helicl pth in mgnetic field touches the line pssing though the stting point pllel to the mgnetic field fte eve pitch. O v (c) Chged pticle in unifom E & When chged pticle moves with velocit v in n electic field E nd mgnetic field, then. et foce epeienced b it is given b following eqution. F qe q( v ) Combined foce F is known s loentz foce (i) E v E v n bove sitution pticle psses undevited but its velocit will chnge due to electic field nd mgnetic foce on it is zeo. E nd unifom, pticle is elesed with velocit v t n ngle. v +q E, mv sin ; T q m q v sin +q z v v cos E, 1 qe m k = v cos t t i ( sint ) j ( cost )( ) v sin c z (d) Ccloid motion uppose tht points in the -diection, nd E in the z-diection. z E b c F q( E v ) q( Ez z z ) m m( z z)
q m E z, z Thei genel solution is ( t) C1 cost C sint ( E / ) t C z( t) C cost C1 sint C4 E E ( t) ( t sint ), z( t) (1 cost ) E ( t ) ( z ) E v The pticle moves s though it wee spot on the im of wheel, olling down the is t speed, v. The cuve geneted in this w is clled ccloid. otice tht the ovell motion is not in the diection of E, but pependicul to it. Emple 4 : A unifom mgnetic field of mt eists in the + X diection. A pticle of chge +ve nd mss 1.67 1-7 kg is pojected though the field in the + Y diection with speed of 4.8 16 6 m/s. () Find the foce on the chged pticle in mgnitude nd diection (b) Find the foce if the pticle wee negtivel chged. (c) Descibe the ntue of pth followed b the pticle in both the cses. olution : () F qv = e vj i = e V k = (1.6 1 19 ) (4.8 1 6 ) ( 1 ) sin 9 =. 1 14. The diection of the foce is in the (z) diection. (b) f the pticle wee negtivel chged, the mgnitude of the foce will be the sme but the diection will be long (+z) diection. n this cse, V V j (c) As V, the pth descibed is cicle, whee dius is given b = mv/q = (1.67 1 7 ) (4.8 1 6 )/(1.6 1 19 ) ( 1 ) = 1.67 m. MAGETC FOCE O A CUET CAYG WE uppose conducting wie, cing cuent i, is plced in mgnetic field. cuent element i d of the wie (figue). The mgnetic foce cting on this cuent element df id F i d i d is given b L id Conside smll f is unifom then, F i d F il Hee L d vecto length of the wie = vecto connecting the end points of the wie
P L Q P Q ote : () f cuent loop of n shpe is plced in unifom then F on it is ( L ) (b) Point of ppliction of mgnetic foce On stight cuent cing wie the mgnetic foce in unifom mgnetic field cn be ssumed to be cting t its mid point. This cn be used fo clcultion of toque., / CUET LOOP UFOM MAGETC FELD M M sin ; whee M A Wok done in otting loop in unifom field fom 1 to W M(cos 1 cos ) M Emple 5 : A unifoml chged disc whose totl chge hs mgnitude q nd whose dius is ottes with constnt ngul velocit of mgnitude. Wht is the mgnetic dipole moment? olution : The sufce chge densit is q/. Hence the chge within ing of dius nd width d is q q dq (d) (d) The cuent cied b this ing is its chge divided b the ottion peiod, dq q di [.d] / The mgnetic moment contibuted b this ing hs the mgnitude dm = di, whee is the e of the ing. dm = q di =..d M dm q. ( d) q 4 d