8 20.2 magnetic forces on currents and torque on a current loop in a magnetic field 1. Magnetic forces on a current-carrying wire Electric currents are charges in motion The force exerted on the wire is a manifestation of the B B basic magnetic interaction on the moving charged particle in it (a) 8 20.2 magnetic forces on currents and torque on a current loop in a magnetic field Choose a wire segment of length d and cross sectional area a current I=ngA<v> The average magnetic force on a charge carrier q(p)×B dl The total differential magnetic force on the segment of the dFm=n4dy(可)xB
11 §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 1. Magnetic forces on a current-carrying wire Electric currents are charges in motion. The force exerted on the wire is a manifestation of the basic magnetic interaction on the moving charged particle in it. Choose a wire segment of length dl and cross sectional area A current I = nqA < v > The average magnetic force on a charge carrier q v B r r × The total differential magnetic force on the segment of the wire F nA lq v B r r r d m = d × dl I Fm r v r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field
8 20.2 magnetic forces on currents and torque on a current loop in a magnetic field The total force on the whole wire from df=nadig(v)x B =nq4v)dl×B=ldl×B We have Fm=JdFm=「xB In uniform magnetic field Fm=(/di)xB ILx B 8 20.2 magnetic forces on currents and torque on a current loop in a magnetic field If the wire is not straight line rom Fm=(di)xB b B We will get 7。 F=LX B f=BILsin e
12 The total force on the whole wire nqA v l B I l B F nA lq v B r r r r r r r = × = × = × d d from d m d ∫ ∫ = = × wire Fm dFm Idl B r r r r We have In uniform magnetic field IL B F I l B r r r r r = × = × ∫( d ) wire m Fm r I θ §20.2 magnetic forces on currents and torque on a current loop in a magnetic field a b I l r d L r I θ B r F IL B r r r m = × Fm = BILsinθ If the wire is not straight line ∫I l = IL r r d F I l B r r r = × ∫( d ) wire m from We will get §20.2 magnetic forces on currents and torque on a current loop in a magnetic field
8 20.2 magnetic forces on currents and torque on a current loop in a magnetic field 2. Torque on a current loop in a magnetic field 2=L,I(cos 8)i-sin Ok y L4 =LIcos 8i +sin k F2 F B -l2 cos 6i+sin6kl×Bk lL2Bcos 8j B= BK F2=4×B m2Icos6i+sin Bk]x Bk Fi IL2 Bcos 8j F,+F4=0 on a current loop in a magnetic field nd torque 8 20.2 magnetic forces on currents l4=l2 F l28 B= BK lIF b=Bk 4j B= BK F1=m1xBF3=l3×B =lL jx Bk =1(-Lj)xBk F+F3=0 I, Bi IL, BI
13 2. Torque on a current loop in a magnetic field ]ˆ sin ˆ [cos ]ˆ sin ˆ [( cos ) 4 2 2 2 l l i k l l i k θ θ θ θ = + = − − r r Il B j Il i k Bk F Il B ˆ cos ˆ ]ˆ sin ˆ [cos 2 2 2 2 θ θ θ = = − + × = × r r r Il B j Il i k Bk F F Il B ˆ cos ˆ ]ˆ sin ˆ [cos 2 2 4 2 4 θ θ θ = − = + × = − = × r r r r F2 + F4 = 0 r r I I 2l r 4l r B Bk ˆ = r θ x y z F2 r F4 r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field Il Bi Il j Bk F Il B ˆ ˆ ˆ 1 1 1 1 = = × = × r r r Il Bi I l j Bk F Il B ˆ ˆ )ˆ ( 1 1 3 3 = − = − × = × r r r 0 F1 + F3 = r r I I l j ˆ 1 l j ˆ − 1 B Bk ˆ = r θ x y z F1 r F3 r l j ˆ 1 l j ˆ − 1 B Bk ˆ = r θ x z 2l 4 2 l = l ⊗ B Bk ˆ = r F1 r F3 r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field
8 20.2 magnetic forces on currents and torque on a current loop in a magnetic field r=(/2)cos 6i +sin Bk] r3=(12/2)-c0s6-sink l4=l2 F2-1 B= Bk B= Bk 20.2 magnetic forces on currents and torque on a current loop in a magnetic field FI Icos a+sinl×1Bi =(I/21l2Bsin的 F F-c0s6-sink×(-l1B)i 2 =(/2)1l2Bsin历 T=T+T3=(,l,Bsin 0)j=(AB sin 8)j
14 ]ˆ sin ˆ ( / 2)[ cos ]ˆ sin ˆ ( / 2)[cos 3 2 1 2 r l i k r l i k θ θ θ θ = − − = + r r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field l j ˆ B Bk 1 ˆ = r θ x z 2l 4 2 l = l ⊗ B Bk ˆ = r F1 r F3 r 1r r3 r r O l j ˆ − 1 I l l B j i k Il B i l r F I l l B j i k Il Bi l r F ˆ ( / 2) sin ˆ ] ( ) ˆ sin ˆ [ cos 2 ˆ ( / 2) sin ˆ ]ˆ sin ˆ [cos 2 1 2 1 2 3 3 3 1 2 1 2 1 1 1 θ θ θ τ θ θ θ τ = = − − × − = × = = + × = × r r r r r r Il l B j IAB j ˆ ( sin ) ˆ ( sin ) τ = τ 1 +τ 3 = 1 2 θ = θ r r r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field
8 20.2 magnetic forces on currents and torque on a current loop in a magnetic field let A -asin 0)i+(acoso A A is the area vector that perpendicular to the surface of the loop. l4=l2 A is the magnetic dipole F L moment of the current loop B= BK then T=MAxB=×B A B= BK T=(ABsin0) The result is valid for planar current loops of any shape. 820.2 magnetic forces on currents and torque on a current loop in a magnetic field Example 1: find the magnitude of magnetic dipole moment produced by an electron rotating with an angular frequency a along a circle of radius r Solution: Equivalent current produced by the moving electron is =gv= ya 2 The equivalent magnetic dipole moment is 2x i=g@2 ∥=L=9 2
15 IA A A i A k r r r = = − + µ θ θ ˆ ( cos ) ˆ let ( sin ) is the magnetic dipole moment of the current loop. µ r A r is the area vector that perpendicular to the surface of the loop. A r l j ˆ 1 l j ˆ − 1 B Bk ˆ = r θ x z 2l 4 2 l = l ⊗ B Bk ˆ = r F1 r F3 r A r then IA B B r r r r r τ = × = µ × IAB j ˆ τ = ( sinθ ) r The result is valid for planar current loops of any shape. §20.2 magnetic forces on currents and torque on a current loop in a magnetic field Example 1: find the magnitude of magnetic dipole moment produced by an electron rotating with an angular frequency ω along a circle of radius R. − e µ r Solution: Equivalent current produced by the moving electron is π ω ν 2 q I = q = The equivalent magnetic dipole moment is 2 2 2 2 q R R q IA ω π π ω µ = = = §20.2 magnetic forces on currents and torque on a current loop in a magnetic field