F is on moving chrged prticle. Chpter 29 Mgnetic Fields Ech mgnet hs two poles, north pole nd south pole, regrdless the size nd shpe of the mgnet. Like poles repel ech other, unlike poles ttrct ech other. Compss: north pole points to the north of the Erth. Mgnetic poles re lwys found in pir. ( Monople?) A piece of iron cn e mgnetized in strong mgnetic field. F = 0, if v. (sin " = 0) Unit: tesl, it is lso clled weer per squre meter (W/m2) T= W N N = 2 = m C m / s A m A common non-s unit: guss (G), 1 T = 104 G 29.1 The mgnetic field Mgnetic field is vector, (like E ). The direction of the force cn e determined y the Right-Hnd Rule. Direction of is the direction in which the north pole of compss needle points t tht loction. Mgnetic field lines re from N pole to S pole outside mgnet, nd continue inside the mgnet. Mgnetic forces on positive chrge nd negtive chrge re in opposite direction. The mgnitude of is defined in terms of the mgnetic force: F = qv Difference etween E nd : (1) The electric force is lwys in the direction of the E ; The mgnetic force is to. F is to oth nd v (cross product) F = qvsin" F q v (2) The electric force cts on chrged prticle regrdless its velocity; The mgnetic force only cts on moving chrged prticle.
(3) For mgnetic force, F ds = (F v)dt = 0. Therefore, stedy cn only chnge the direction of the velocity vector, ut it cnnot chnge the speed or kinetic energy of the chrged prticle. Exmple: A proton moves with speed of 8 # 10 6 m/s long the x xis. The mgnetic field is 2.5 T s shown. () Clculte initil mgnetic force. F = qvsin = (1.6 "10 #19 C)(8 "10 6 m / s)(2.5t)(sin60 ) = 2.8 "10 #12 N F is in the +z direction. () Clculte the ccelertion of the proton. = F m p = 2.8 10"12 N 1.67 10 "27 kg = 1.7 1015 m / s 2 is in the +z direction. x v +e 60 z F y 29.2 Motion of Chrged Prticle n Mgnetic Field Mgnetic force is to the velocity of the prticle. The work done on the prticle y the field is 0. A chrged prticle moves in circle whose plne is to. F cts like centrl force, which chnges the direction of v, ut not the mgnitude. From the Newton s 2 nd Lw: F = qv = mv2 r r = mv q Angulr frequence: = v r = q m The period of its motion: T = 2r v = 2 " = 2m q Exmple: A proton is moving in circulr orit of rdius 14 cm. = 0.35 T, to the velocity of proton. Find the oritl speed of the proton. v = qr m = (1.6 10"19 C)(0.35T)(14 10 "2 m) 1.67 10 "27 = 4.7 10 6 m / s kg
29.3 Applictions nvolving Chrged Prticles Moving in Mgnetic Field n the presence of oth electric field E nd mgnetic filed, the totl electro-mgnetic force (clled the Lorentz force) cting on chrge is F = qe + qv " Velocity Selector: * A chrged prticle hving velocity vector tht hs component prllel to uniform mgnetic field moves in helicl pth. v (velocity component prllel to ) remins constnt. v(velocity component perpendiculr to ) produce circulr mv motion with rdius r = ". q Vn Allen Rdition elts rdition elts The Vn Allen consist of chrged prticles surrounding the Erth in doughnut-shped regions. The prticles re trpped y the Erth s nonuniform mgnetic field. The prticles spirl from pole to pole. My result in urors v= E/ Only chrged prticles with the desired velocity (v = E/) cn pss through velocity selector. Mss Sectrometer: m/q = r0/v m/q = r0/e Mss/chrge rtio cn e determined.
29.4 Mgnetic Force on Current-Crrying Conductor Cse Study 1: A curved wire crrying current. The totl force on the wire, " F = $ # % ds' & ( d s L' F = L'" L is the length perpendiculr to field. Force on chrge moving t speed v d, F i = (q i vd ) Cse Study 2: An closed loop crrying current. Force on segment of wire, crrying current, df = (qv d ) nads = (nqv d A) ds df = ds A F = ( ) " # 0 ds The totl mgnetic force on ny closed current loop in uniform mgnetic field is zero. d s Totl force on wire F = ds " where ds is in the direction of.
Exmple: () Find the net force on the loop F net must = 0 This is the mximum torque: mx = () Force on the stright portion. F s = L, The direction is out of pper. (c) Force on the curved portion. L n generl, = F 1 2 sin" + F 2 = sin" = Asin" 2 sin" F1 /2 F c + F s = 0 F c = L, the direction is into the pper. 29.5 Torque on Current Loop in Uniform Mgnetic Field Force on the sides of length : F 4 = F 2 = Force on the sides of length : F 1 =F 3 = 0 = Asin" Where, A = is the re of the loop. hs the mximum (A) when " = 90. =0 when " = 0. F1 /2 2 sin " F2 F2 Torque out the center: " = F 4 2 + F 2 2 Define: (1) A, vector # to the plne of the loop, A = re of the loop. The direction of A is determined y the right-hnd rule. = () 2 + () 2 =
(2) Mgnetic moment of the loop: µ = A (unit: A m 2 ) Then, the torque: = µ " Exmple: A rectngulr coil, 5.4 $ 8.5 cm, hs 25 turns of wire. Current is 15 ma. () Find the mgnitude of it mgnetic moment. µ = na = (25)(15 10 "3 A)(5.4 8.5 10 "4 m 2 ) = 1.72 10 "3 Am 2 () Suppose = 0.35 T, nd it is to the plte of the loop. Wht is the torque? " = µ = (1.72 #10 $3 Am 2 )(0.35T ) = 6.02 #10 $4 Am 2 T = 6.02 #10 $4 N % m (c) Wht is the torque when " = 60? = µsin" = (6.02 #10 $4 N %m)(sin60 ) = 5.21#10 $4 N % m