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Chapter 29 Magnetic Fields Force on a Charge Moving in a Magnetic Field, Formula The properties can be summarized in a vector equation: FB qv B FB is the magnetic force q is the charge v is the velocity of the moving charge B is the magnetic field Direction: Right-Hand Rule #1 The fingers point in the direction of v B comes out of your palm Curl your fingers in the direction of B The thumb points in the direction of v B which is the direction of FB q0 q0 Direction: Right-Hand Rule #2 Alternative to Rule #1 Thumb is in the direction of v Fingers are in the direction of B Palm is in the direction of FB On a positive particle You can think of this as your hand pushing the particle More Differences Between Electric and Magnetic Fields Work The electric force does work in displacing a charged particle The magnetic force associated with a steady magnetic field does no work when a particle is displaced This is because the force is perpendicular to the displacement WFB FB d r (qv B) vdt 0 FB v Work in Fields, cont. The kinetic energy of a charged particle moving through a magnetic field cannot be altered by the magnetic field alone When a charged particle moves with a given velocity through a magnetic field, the field can alter the direction of the velocity, but not the speed or the kinetic energy 1 2 1 2 K mv f mvi 0 v f vi 2 2 Wnet K Wnet WFB 0, K 0 자기장 내에서의 전자 운동 예제 29.1 v 8 106 m/s , q 1.6 1019 C, B 0.025T z FB qv B qvB sin 60 (k̂ ) q - v y 60° FB x B : xy-plane (1.6 10 19 )(8 106 )(0.025) sin 60 (k̂ ) 2.8 10 14 (k̂ )( N) Force on a Charged Particle Equating the magnetic and centripetal forces: 2 mv v2 FB qvB Fnet ma, Fnet FB qv B, a (r̂ ) r r Solving for r: mv r qB FB v , v const r is proportional to the linear momentum of the particle and inversely proportional to the magnetic field More About Motion of Charged Particle The angular speed of the particle is v qB ω r m The angular speed, w, is also referred to as the cyclotron frequency The period of the motion is 2πr 2π 2πm T v ω qB v, r 에 무관 Motion of a Particle, General If a charged particle moves in a magnetic field at some arbitrary angle with respect to the field, its path is a helix Same equations apply, with v v y2 v z2 Use the active figure to vary the initial velocity and observe the resulting motion PLAY ACTIVE FIGURE 균일한 자기장에 수직으로 운동하는 양성자 예제 29.2 m 1.67 1027 kg, q 1.6 1019 C, B 0.35T, r 14cm v r v? qvB mv2 /r v qBr/m (1.6 10-19 )(0.35)(0.14) / (1.67 10- 27 ) 4.7 106 (m / s) 예제 29.3 전자빔의 휘어짐 m 9.111027 kg, q 1.6 1019 C, r 7.5cm,V 350V Bin (A) B=? mv2 mv qvB B r qr (9.1110-31 )(1.11107 ) B (1.6 10-19 )(0.075) r v 4 8.4 10 (T) (A) ω=? v 1.1110 ω 1.5 108 (rad / s) r 0.075 7 V v0 1 2 qV mv 2 v 2qV/m 1.11107 m / s Particle in a Nonuniform Magnetic Field The motion is complex For example, the particles can oscillate back and forth between two positions This configuration is known as a magnetic bottle Velocity Selector, cont. When the force due to the electric field is equal but opposite to the force due to the magnetic field, the particle moves in a straight line This occurs for velocities of value v=E/B FB FE FB FE FB FE Bin FB FE v0 E/B 스크린 x1 t1 x1 /v0 전자 - q 1.6 1019 C t 2 x2 /v0 x2 v y a y t1 a y x1 +++++++++++++++ v0 ay - -q v0 qEx1 mv0 --------------- a y qE/m q x1 , x2 ,v0 , E, B m y2 v y t 2 y2 y1 1 2 qEx12 y1 a y t1 2 2mv02 qEx1 x2 mv02 Cyclotron, 2 D1 and D2 are called dees because of their shape A high frequency alternating potential is applied to the dees A uniform magnetic field is perpendicular to them qvB mv2 /r v qBr/m 1 2πr 2πm T f v qB r 1 2 mv 2 q2 B2 2 r 2m K Force on a Wire, equation The magnetic force is exerted on each moving charge in the wire F qvd B The total force is the product of the force on one charge and the number of charges F qvd B nAL nqvd AL B nqvd AL B IL B nqvd A I Force on a Wire, Arbitrary Shape Consider a small segment of the wire, ds The force exerted on this segment is dFB I ds B The total force is b FB I ds B a 예제 29.4 반원형 도선에 작용하는 자기력 m 1.67 1027 kg, q 1.6 1019 C, B 0.35T, r 14cm θ dθ r B ds θ I 반원에서 자기력 dFB IdL B dFB IdLB sin θ I (ds sin θ ) B FBcurve IB (ds sin θ ) IB ds sin θ IB (2r ) 직선에서 자기력 FB IL B FBstr ILB IB (2r ) curve str FB FB FB 0 예제 29.4 반원형 도선에 작용하는 힘 반지름 R인 반원의 폐회로를 구성하고 있는 도선에 전류 I가 흐른다. 회로는 xy 평면에 놓여 있고, 균일한 자기장이 +y 방향을 따라 작용한다. 도선의 직선 부분과 곡선 부분에 작용하는 자기력을 구하라. 풀이 직선 부분에 작용하는 힘 곡선 부분에 작용하는 힘 F1 ILB 2IRB (+z방향) b FB I ds B a 0 0 F2 I ( Rd ) B sin IRB sin d 2IRB Torque on a Current Loop, 3 The forces are equal and in opposite directions, but not along the same line of action The forces produce a torque around point O b b τ max=F2 +F4 2 2 b b =( IaB ) +( IaB ) 2 2 =IabB τ τ max sin θ Torque on a Current Loop, Equation The maximum torque is found by: b b b b τ max F2 F4 (I aB ) (I aB ) 2 2 2 2 I abB The area enclosed by the loop is ab, so τmax = IAB This maximum value occurs only when the field is parallel to the plane of the loop Torque on a Current Loop, General Assume the magnetic field makes an angle of < 90o with a line perpendicular to the plane of the loop The net torque about point O will be τ = IAB sin Use the active figure to vary the initial settings and observe the resulting motion PLAY ACTIVE FIGURE Torque on a Current Loop, Summary The torque has a maximum value when the field is perpendicular to the normal to the plane of the loop The torque is zero when the field is parallel to the normal to the plane of the loop IA B where A is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop τ τ max sin θ IA B Magnetic Dipole Moment The product IA is defined as the magnetic dipole moment, , of the loop μ IA Often called the magnetic moment SI units: A · m2 Torque in terms of magnetic moment: Analogous to B for electric dipole p E τ τ max sin θ IA B B 예제 29.5 코일의 자기 쌍극자 모멘트 A 5.4cm 8.5cm 45.9cm2 , N 25turns , I 15mA , B 0.35T B⊥ A (A) 코일의 자기 쌍극자 모멘트의 크기 μ=NIA=(25)(0.015 )(0.00459) = 1.72 × 10-3 Am 2 (B) 코일에 작용하는 토크의 크기 τ = μ×B =Bsin90 = (1.72 × 10-3 )(0.35) = 6.02 × 10-4 (Nm) B