物理与光电工程学院：《大学物理》课程PPT教学课件（讲稿，英文第三版）Chapter 26 Sources of Magnetic Field

Chapter 26 Sources of Magnetic Field Chapter 26 Sources of Magnetic Field 1. Biot-savart's law 2. Magnetic Flux and Gauss'Law 3. Ampere's Law 4. Magnetic Materials

Chapter 26 Sources of Magnetic Field Chapter 26 Sources of Magnetic Field 1. Biot-Savart’s Law 2. Magnetic Flux and Gauss’ Law 3. Ampere’s Law 4. Magnetic Materials

Chapter 26 Sources of Magnetic Field New words and expressions Biot-Savart's law 毕奥一萨伐尔定律 solenoid 螺线管 toroid 螺绕环 Amperes Loop Law 安培环路定理 paramagnetism 顺磁质 diamagnetism 抗磁质 ferromagnetism 铁磁质 magnetization 磁化强度 isotropic medium 各向同性介质 permeability of magnetic materials 磁导率 hysteresis 磁滞 magnetization Curve 磁化曲线 hysteresis Loop 磁滞回线

Chapter 26 Sources of Magnetic Field New words and Expressions Biot-Savart’s Law 毕奥—萨伐尔定律 solenoid 螺线管 toroid 螺绕环 Ampere’s Loop Law 安培环路定理 paramagnetism 顺磁质 diamagnetism 抗磁质 ferromagnetism 铁磁质 magnetization 磁化强度 isotropic medium 各向同性介质 permeability of magnetic materials 磁导率 hysteresis 磁滞 magnetization Curve 磁化曲线 hysteresis Loop 磁滞回线

Chapter 26 Sources of Magnetic Field 26-6&26-5. Biot-Savart's Law Applications (P613-616) 1.Biot- Savart'sLaw毕奥一萨伐尔定律: (电流元在空间产生的磁场 The current element dl Id dB generates a magnetic field db given by: Idu sin e dB dB 4兀 2 ldl×F rId dB

Chapter 26 Sources of Magnetic Field 26-6&26-5. Biot-Savart’s Law & Applications (P613-616) 1. Biot-Savart’s Law毕奥—萨伐尔定律: (电流元在空间产生的磁场 The current element Idl generates a magnetic field dB given by: I P * I l  d B  d  r  I l  d r  B  d 2 0 d sin 4π d r I l B   = 3 0 d 4π d r I l r B     = 

Chapter 26 Sources of Magnetic Field 真空磁导率0=4×10NA2 Magnetic permeability constant in a vacuum The principle of superposition of magnetic field, can be applied for a complete circuit 任意载流导线在点P处的磁感强度 磁感强度叠加原理 B=「d_bldl×F

Chapter 26 Sources of Magnetic Field 真空磁导率 7 2 0 4π 10 N A − −  =   Magnetic permeability constant in a vacuum The principle of superposition of magnetic field, can be applied for a complete circuit: 3 0 d 4π d r I l r B B      = =    任意载流导线在点 P 处的磁感强度 磁感强度叠加原理

Chapter 26 Sources of Magnetic Field 2. The Application of Biot-Savart Law 1) A Long Straight Wire(载流长直导线的磁场P614): For the field near a long straight wire D carrying a current I, Determine the dz个 magnetic field dB Solution: O dB方向均沿 x轴的负方向

Chapter 26 Sources of Magnetic Field 1) A Long Straight Wire (载流长直导线的磁场P614): For the field near a long straight wire carrying a current I, Determine the magnetic field. 2. The Application of Biot-Savart Law: y x z I P C D o 0 r * B  d r  z dz Solution: 方向均沿 x 轴的负方向 B  d

Chapter 26 Sources of Magnetic Field The magnitude of the differential magnetic field produced at p by the current-length element ldl located a distance ro from P is given by: db- lo ldz sin 0 De 02 4兀 2 ldzsin e b=dB= CD dB dz=rd618∞ =-n coter=n/ in0 O C B sin ede

Chapter 26 Sources of Magnetic Field y x z I P C D o 0 r * B  d 2 0 d sin 4π d r I z B   =   = = CD r I z B B 2 0 d sin 4π d   z = −r0 cot,r = r0 /sin   2 0 dz = r d /sin 1 r   2   = 2 1 sin d 4π 0 0      r I B z dz The magnitude of the differential magnetic field produced at P by the current-length element Idl located a distance r0 from P is given by:

Chapter 26 Sources of Magnetic Field B=AπTo sin ade COS cOS Discussions. (1 The magnetic field Dde firom all current element of straight wire are in the same direction and B into the screen x p y B的方向沿x轴的负方向

Chapter 26 Sources of Magnetic Field （ 1 2 ） 0 0 cos cos 4π    = − r I B 的方向沿 x 轴的负方向.   = 2 1 sin d 4π 0 0      r I B 1  2  P C D y x z o I B  + Discussions: (1) The magnetic field from all current element of straight wire are in the same direction and into the screen

Chapter 26 Sources of Magnetic Field (2) For straight and infinite wire(无限长载流直导 线),的1=0,62=Then B=/0(cos 0, - 0) 4丌 70 0,→>0B=2兀r0 ,→>兀 B The magnitude of magnetic field of infinite straight wire at a point with distance R is in inverse B proportion to the distance

Chapter 26 Sources of Magnetic Field π 0 2 1 → →   0 0 2π r I B  = （ 1 2 ） 0 0 cos cos 4π    = − r I B (2) For straight and infinite wire (无限长载流直导 线), 1=0,2=,then B  I B  I —— The magnitude of magnetic field of infinite straight wire at a point with distance R is in inverse proportion to the distance

Chapter 26 Sources of Magnetic Field ◆无限长载流长直导线的磁场 B 01 tB B 2元 ◆电流与磁感强度成右螺旋关系 半无限长载流长直导线的磁场 O,→ 2 B P 6,→>兀 兀F r.P

Chapter 26 Sources of Magnetic Field I B r I B 2π 0 = 电流与磁感强度成右螺旋关系 半无限长载流长直导线的磁场 r I BP 4π 0 = 无限长载流长直导线的磁场 r * P I o π 2 π 2 1 → →   I X B

Chapter 26 Sources of Magnetic Field 2)Current Loop p615 Determine b for point on the axis of a circular loop of wire of radius r carrying a current 1. 真空中,半径为R的载流导线,通有电流I,称圆 电流.求其轴线上一点P的磁感强度的方向和大小 X

Chapter 26 Sources of Magnetic Field 2) Current Loop p615 I x 真空中 , 半径为R 的载流导线 , 通有电流I , 称圆 电流. 求其轴线上一点 p 的磁感强度的方向和大小. p R o * Determine for point on the axis of a circular loop of wire of radius R carrying a current I. B 