(③)无限长载流平板 解 Idx d/ dB. b dB dB d 461d 2π 2πbysec0 B,=B,=∫dB=∫dBcos0 。 dx dx ysec2ade 0 arctan 21 0=4 b arctanI b (3) 无限长载流平板 P B  d Bx  解 d b I x I d d = r I B  = 2 d d 0   BP = Bx = Bx d  = dBcos   = 1 0 0 d θ P b I B   y b b I 2 arctan 0  =  d sec d 2 x = y r    2 sec d 0 by I x  =   = 2 0 2 0 sec d 2 2 b x by I   x y O dx y b 2 1 = arctan B  d 1