6. THE TATC MAGNETC FELD 6- LOENTZ FOCE EQUATON Lorent force eqution F = Fe + Fm = q ( E + v B ) Exmple 6- An electron hs n initil velocity vo = vo y in uniform mgnetic flux density B = Bo. () how tht the electron moves in circulr pth. (b) Find the rdius r o of the circulr pth in terms of electron chrge e, electron mss m nd the initil speed v o. olution () The mgnetic force on the electron is Fm = e v B. Becuse F m is perpendiculr to v, it only chnges the direction of v, nd corresponds to the centripetl force on the electron. (b) The mgnetic force on the electron Fm = evobo ( e < 0) The force on the electron, mss ccelertion dv dv d m = m dt d dt where =ω t nd ω is ngulr velocity. ewrite it using v = vo nd d / d = d( vo ) d( t) m ω = mvo ( ) ω d dt 6- Proprietry of Prof. Lee, Yeon Ho
From two equtions eb ω= o m Using vo = roω mvo ro = ebo where e < 0. 6- BOT-AVAT LAW Biot-vrt lw dl dh = 4π : = r r For stedy current flowing in wire dl H = C 4π (6-7) 6- Proprietry of Prof. Lee, Yeon Ho
d l in terms of J nd Js dl = Jdv = J ds s For current density J J dv H = V 4π For surfce chrge density J s s ds H = J 4π (6-9) (6-0) All the unit vectors in the unprimed coordintes The integrtion in the primed coordintes. Exmple 6- H of stedy current in stright wire Determine H t point P; ( x, y, ) due to n infinitely long stright conducting wire of negligible thickness crrying stedy current nd lying long -xis. olution The line current hs cylindricl symmetry, nd Biot-vrt lw only gives dh in direction. o we expect H = ( ) H The trnsltion vector, : = r r' = The vector differentil length, : dl = d 6-3 Proprietry of Prof. Lee, Yeon Ho
From the definition d ( ) d H = = 4 π 4 π or =+ =+ = 3 = 3/ H = π ( + ) (6-) (6-) H lso hs cylindricl symmetry, being independent of nd. For finite line current d H = = 4π 4π + 3/ ( + ) = = eferring to Fig. 6-4 H = θ + θ 4π [ cos cos ] (6-3) This formul cn be pplied even for θ or θ lrger thn 90 o. 6-4 Proprietry of Prof. Lee, Yeon Ho
Exmple 6-3 Find H, t point p;(, b,0) in the first qudrnt, due to n infinite right-ngled wire crrying stedy current. olution For y < 0 : θ =α, θ = 0 nd = H = [ cos α + ]( ) 4π For x < 0 : θ = 0, θ =β nd = b H = [ + cosβ]( ) 4πb n Fig. 6-5 b cos α= sin( α 90 o ) = + b cosβ= sin ( β 90 o ) =. + b The mgnetic field intensity t p is b H = H + H = + 4 + π 4 b + b π + b = + b ( + b) 4πb ( ) ( ) 6-5 Proprietry of Prof. Lee, Yeon Ho
Exmple 6-4 Determine H t point p;0,0, ( b ) due to circulr wire of rdius centered t the origin in xy -plne, nd crrying stedy current. olution n cylindricl coordintes : = b dl = d. From Eq. (6-7) dl = π d ( b ) H = = C 0 3 4π = 4π = πb = π = d + d = 0 3 0 3 4π = 4π The first integrl cncels to ero due to ( = ) = ( =+π ) H = + b 3/. (6-4) (6-5) 6-6 Proprietry of Prof. Lee, Yeon Ho
6-3 AMPEE CCUTAL LAW The closed line integrl of H is equl to the totl current enclosed by the pth of integrtion. Hi d l = : ight-hnd rule for C nd C The totl enclosed current The continuity eqution is Ji d s = 0 (6-7) The totl current enclosed by C is equl to the totl current cross the surfce bounded by C. Ampere s circuitl lw C Hidl = Jids : ight-hnd rule for C nd Applying tokes s theorem Hids = Jids (6-9) The point form of Ampere s circuitl lw H = J (6-0) To determine H by Ampere s circuitl lw, the closed pth C should be chosen such tht the mgnetic field intensity H is t lest constnt on the pth C. 6-7 Proprietry of Prof. Lee, Yeon Ho
Exmple 6-5 Find H, by Ampere s circuitl lw, due to n infinitely long stright wire lying long - xis nd crrying stedy current. olution The cylindricl symmetry H independent of nd Biot-vrt lw. dh due to d l only in direction We expect H = ( ) : H depends only on H On C of rdius π Hi l ( ) C i π = 0 = 0 d = H d = H d = H = = π H The sme result with tht by Biot-vrt lw 6-8 Proprietry of Prof. Lee, Yeon Ho
Exmple 6-6 An infinitely long coxil cble crries uniform stedy current. The rdii of the cylindricl surfces re, b nd c, respectively. Find H in the region > 0 by Ampere s circuitl lw. olution The uniform current Current elements t nd Cylindricl symmetry Only H -component We cn expect H = ( ) : H is constnt on the circle C n the region 0 H The current enclosed by C, π <, Hidl = H i( d ) = πh C π π The result H = H = : ( 0 < ) π The enclosed current : ( b) c : (b c ) c b 0 : ( c ) The result = π H ( b = 0 ) (6-3b) 6-9 Proprietry of Prof. Lee, Yeon Ho
c H = π c b (b c ) (6-3c) H = 0 ( c ) (6-3d) Exmple 6-7 Find H of toroidl core hving N turns of wire crrying stedy current. The toroid hs the inner rdius of nd the outer rdius of b. olution ottion symmetry bout -xis Constnt H on C cross = plne H nd H, smll Lrge enclosed current N H, lrge An idel toroidl coil H = H inside H = 0 outside. n the region << b π Hidl = H id = H d= πh C C 0 The totl enclosed current is N N H = π N H = H = : ( < < b) π H = 0 (No current enclosed) : < nd > b 6-0 Proprietry of Prof. Lee, Yeon Ho
6-4 MAGNETC FLUX DENTY n free spce =μ o 7 B H : 4 0 The totl mgnetic flux Φ= Bi d s. μ = π, permebility of free spce o An infinite line current An rbitrry distribution of H, concentric circles H, lwys closed. Liner property Guss s lw for the mgnetic field Bi d s = 0 Point form of Guss s lw i B = 0 Mxwell s equtions for the sttic electric nd mgnetic fields i D = v E = 0 i B = 0 H = J n integrl forms C C Dids = dv V Eidl = 0 Bids = 0 Hidl = Jids v 6- Proprietry of Prof. Lee, Yeon Ho