7.Magnetic Field from a Sheet of Surface Current dx 2= odx dx dBx=dy dB, dB=dB+dB2 ￥dB2 r=(2+y2)112 dK=Jody' d/2 /2 n= (x2+y2)1/2 dy d→ (a) (b) K:=Kol K2=-Ko 架 imJo△=K。 4+0 +d B2.- 与Kg d-a d y -。K司 B,= Hoko(y-d) 子，k B=B,+B2 (c) (d) A uniform surface current of infinite extent generates a uniform magnetic field oppositely directed on each side of the sheet.The magnetic field is perpendicular to the surface current but parellel to the plane of the sheet.(b)The magnetic field due to a slab of volume current is found by superimposing the fields due to incremental surface currents.(c)Two parallel but oppositely directed surface current sheets have fields that add in the region between the sheets but cancel outside the sheet.(d)The force on a current sheet is due to the average field on each side of the sheet as found by modeling the sheet as a uniform volume current distributed over an infinitesimal thickness A. From a line current I I H二2t i。=-sin0ix+cos4iy Thus from 2 symmetrically located line currents dHx= dI z(sin) 2πx2+y2 6.641,Electromagnetic Fields,Forces,and Motion Lecture 3 Prof.Markus Zahn Page 11 of 126.641, Electromagnetic Fields, Forces, and Motion Lecture 3 Prof. Markus Zahn Page 11 of 12 7. Magnetic Field from a Sheet of Surface Current From a line current I I H 2 r φ = π _ _ _ x y i sin i cos i φ =− φ + φ Thus from 2 symmetrically located line currents ( ) x ( ) 1 2 2 2 dI dH sin 2x y = − φ π +