复旦大学：《电动力学》课程电子讲义（PDF演示版）第四章 静磁场 04-02 静磁场的矢势及其微分方程和边值关系

经典电动力学导论 Let there be light A(r)=A(z)er B=V×A 0A(z) dx dy dz b(ze az A(z)00 对回路1应用安培环路定律∮B·d=u0Inc得:B(z)=B0(B与z无关) 对回路2应用安培环路定律∮B·d=10Ienc得:2Bl=μ0Ienc=0al 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1oÙµ·^| § 4.2 A~(r~) = A(z) eˆx, B~ = ∇ × A~ =         eˆx eˆy eˆz ∂ ∂x ∂ ∂y ∂ ∂z A(z) 0 0         = ∂A(z) ∂z eˆy = B(z) eˆy 飴 1 A^S‚´½Æ I B~ · d~l = µ0 Ienc µB(z) = B0 (B0 † z Ã') 飴 2 A^S‚´½Æ I B~ · d~l = µ0 Ienc µ2Bl = µ0Ienc = µ0αl E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light A(r)=A(z)er B=V×A e0 0A(z) az A(z)00 对回路1应用安培环路定律∮B·d=u0Inc得:B(z)=B0(B与z无关) 对回路2应用安培环路定律∮B·d=10Ienc得:2Bl=μ0Ienc=0al + poa ey for z<o 从而:B= Hoc e for z>0 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1oÙµ·^| § 4.2 A~(r~) = A(z) eˆx, B~ = ∇ × A~ =         eˆx eˆy eˆz ∂ ∂x ∂ ∂y ∂ ∂z A(z) 0 0         = ∂A(z) ∂z eˆy = B(z) eˆy 飴 1 A^S‚´½Æ I B~ · d~l = µ0 Ienc µB(z) = B0 (B0 † z Ã') 飴 2 A^S‚´½Æ I B~ · d~l = µ0 Ienc µ2Bl = µ0Ienc = µ0αl l µB~ =    + 1 2 µ0α eˆy for z < 0 − 1 2 µ0α eˆy for z > 0 E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light A(r)=A(z)er B=V×A e0 0A(z) b(ze A(z)00 对回路1应用安培环路定律∮B·d=u0Inc得:B(z)=B0(B与z无关) 对回路2应用安培环路定律∮B·d=10Ienc得:2Bl=μ0Ienc=0al + poa ey for z<o Oa已 for z<0 从而:B= 比较:E 11 Hoc e for z>0 gge for z>0 ∈ 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1oÙµ·^| § 4.2 A~(r~) = A(z) eˆx, B~ = ∇ × A~ =         eˆx eˆy eˆz ∂ ∂x ∂ ∂y ∂ ∂z A(z) 0 0         = ∂A(z) ∂z eˆy = B(z) eˆy 飴 1 A^S‚´½Æ I B~ · d~l = µ0 Ienc µB(z) = B0 (B0 † z Ã') 飴 2 A^S‚´½Æ I B~ · d~l = µ0 Ienc µ2Bl = µ0Ienc = µ0αl l µB~ =    + 1 2 µ0α eˆy for z < 0 − 1 2 µ0α eˆy for z > 0 'µE~ =    + 1 2 1 0 σq eˆz for z < 0 − 1 2 1 0 σq eˆz for z > 0 E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light A(r)=A(z)er B=V×A 0A(z) dx dy dz b(ze A(z)00 对回路1应用安培环路定律∮B·d=u0Inc得:B(z)=B0(B与z无关) 对回路2应用安培环路定律∮B·d=10Ienc得:2Bl=μ0Ienc=0al + poa ey for z<o Oa已 for z<0 从而:B= 比较:E 11 Hoc e for z>0 gge for z>0 ∈ 例2:两平行无穷大平板,均匀带电士σa,以速度 =υen运动,求(1)空间磁场;(2)作用于上平板 单位面积的磁力;(3)当速度多大时,磁力等于电力。 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1oÙµ·^| § 4.2 A~(r~) = A(z) eˆx, B~ = ∇ × A~ =         eˆx eˆy eˆz ∂ ∂x ∂ ∂y ∂ ∂z A(z) 0 0         = ∂A(z) ∂z eˆy = B(z) eˆy 飴 1 A^S‚´½Æ I B~ · d~l = µ0 Ienc µB(z) = B0 (B0 † z Ã') 飴 2 A^S‚´½Æ I B~ · d~l = µ0 Ienc µ2Bl = µ0Ienc = µ0αl l µB~ =    + 1 2 µ0α eˆy for z < 0 − 1 2 µ0α eˆy for z > 0 'µE~ =    + 1 2 1 0 σq eˆz for z < 0 − 1 2 1 0 σq eˆz for z > 0 ~ 2µü²1ጲ†§þ!‘>±σq§±„Ý v~ = v eˆx $ħ¦(1) m^|¶(2) Š^uþ²† ü ¡È^å¶(3) „ÝõŒž§^åu>å" E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light A(r)=A(z)er B=V×A 0A(z) dx dy dz b(ze A(z)00 对回路1应用安培环路定律∮B·d=u0Inc得:B(z)=B0(B与z无关) 对回路2应用安培环路定律∮B·d=10Ienc得:2Bl=μ0Ienc=0al + poa ey for z<o Oa已 for z<0 从而:B= 比较:E 11 Hoc e for z>0 gge for z>0 ∈ 例2:两平行无穷大平板,均匀带电士σa,以速度 +og =υen运动,求(1)空间磁场;(2)作用于上平板 单位面积的磁力;(3)当速度多大时,磁力等于电力。 q 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1oÙµ·^| § 4.2 A~(r~) = A(z) eˆx, B~ = ∇ × A~ =         eˆx eˆy eˆz ∂ ∂x ∂ ∂y ∂ ∂z A(z) 0 0         = ∂A(z) ∂z eˆy = B(z) eˆy 飴 1 A^S‚´½Æ I B~ · d~l = µ0 Ienc µB(z) = B0 (B0 † z Ã') 飴 2 A^S‚´½Æ I B~ · d~l = µ0 Ienc µ2Bl = µ0Ienc = µ0αl l µB~ =    + 1 2 µ0α eˆy for z < 0 − 1 2 µ0α eˆy for z > 0 'µE~ =    + 1 2 1 0 σq eˆz for z < 0 − 1 2 1 0 σq eˆz for z > 0 ~ 2µü²1ጲ†§þ!‘>±σq§±„Ý v~ = v eˆx $ħ¦(1) m^|¶(2) Š^uþ²† ü ¡È^å¶(3) „ÝõŒž§^åu>å" E￾ŒÆ ÔnX  Mï
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