1 É an = ēx (ī – Bct) + ēvý + ēzz ) 41€0 [172 (– Bct)2 + y2 + z2] NICO qyßc 3 1 B Mo ē, y + -ē, z 47 [y2 (– Bct)2 + y2 +

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1 E An Ex I Bct Evy Ezz 41 0 172 Bct 2 Y2 Z2 Nico Qyssc 3 1 B Mo E Y E Z 47 Y2 Bct 2 Y2 1 (106.8 KiB) Viewed 23 times

1 É an = ēx (ī – Bct) + ēvý + ēzz ) 41€0 [172 (– Bct)2 + y2 + z2] NICO qyßc 3 1 B Mo ē, y + -ē, z 47 [y2 (– Bct)2 + y2 + z2] for the electric and magnetic fields of a particle moving with velocity Bc along the +ī axis you found this result by transforming A' from the x frame where a ф A = 0 = = 4περ 2 Now st and again in the x-frame where E= q (x,y,z) p3 0) B = 0 4πεο And find Ē, B by the Lorentz Transformation E" uv = ע - A AFB α where
Ο Ο ΛΑ Υ βγ O 0 β) Ο Ο n 0 1 0 1 α 0 Ο Ο