Chapter 26 Sources of Magnetic Field 解根据对称性分析B=B2=dBsi dB B I db=do ldl 4 r
Chapter 26 Sources of Magnetic Field I x 2 0 d 4π d r I l B = r B d B B I l d p R o * 解 根据对称性分析 B = Bx = dBsin
Chapter 26 Sources of Magnetic Field ldl dB R r=R+x X P rb loI r cos adl 2 Rc2兀R dB- lo ld/ B dl 4兀 R db- lo l cos adl B 4兀 2(x2+R
Chapter 26 Sources of Magnetic Field x x R * p 2 0 cos d 4π d r I l Bx = = l r I l B 2 0 cos d 4π 2 2 2 cos r R x r R = + = = R l r IR B 2π 0 3 0 d 4π 2 3 2 2 2 0 （2 x R ） IR B + = 2 0 d 4π d r I l B = o B d r I l d
Chapter 26 Sources of Magnetic Field Discussions ()At center RIB x 2R x=0B 2R (2) The direction of B is along the axis direction(right-hand rule)as right figure and Fig 26-18 in P615
Chapter 26 Sources of Magnetic Field Discussions: (1) At center O (2) The direction of is along the axis direction (right-hand rule) as right figure and Fig.26-18 in P615. B I R o R I B 2 0 0 = x B0 R I B 2 0 x = 0 =
Chapter 26 Sources of Magnetic Field EXAMPLE 26-11 of P616: One quarter of a circular loop of wire carries a current and two straight sections whose extensions 3 intersect the center c of the arc R Find b at point c? Solution: For parts I and2,∵is∥F∴B1=B2=0 B3 Hoi(t/2)=Hol(direct into the screen 4R 8R .B=B1+ B2+ B (into the screen) 8R
Chapter 26 Sources of Magnetic Field ids //r B1 = B2 = 0 Solution: R i R i B o o 4 8 ( / 2) 3 = = (direct into the screen) R i B B B B o 8 1 2 3 = + + = (into the screen) EXAMPLE 26-11 of P616: One quarter of a circular loop of wire carries a current I, and two straight sections whose extensions intersect the center C of the arc. Find at point B C? 1 3 2 For parts 1 and 2
Chapter 26 Sources of Magnetic Field 3 AStretched-out solenoid (载流直螺线管轴线上的磁场) 如图所示,有一长为l,半径为R的载流密绕直螺 线管,螺线管的总匝数为N,通有电流.设把螺线管 放在真空中,求管内轴线上一点处的磁感强度
Chapter 26 Sources of Magnetic Field 3 A Stretched-out Solenoid (载流直螺线管轴线上的磁场) I I 如图所示，有一长为l , 半径为R的载流密绕直螺 线管，螺线管的总匝数为N，通有电流I. 设把螺线管 放在真空中，求管内轴线上一点处的磁感强度