Question

Quiz Ch.29 Ampere's law you have a coaxial cable, made of an inner conclucting cylinder of radius a, and non uniform current density given by: J=ar' (the constant 2 is in units of A/m") The outer conducting cylindrical shell has inner radius b and outer radius c. It carries a total current I, with uniform current density J = TCC-by) for this configuration looks a "view from the top" as follows Find: Cuse Ampere's law) a) B (the magnetic field at distance rca. b) B at acreb c) B at barac d) B at r>c Coutside the array)

Answer 1

For inner conducting cylinder cwrent density J = 272 For onder cuorend conducting cylinder density - I Tile?627. Remember cworent in the inner conducting cylinder and outer condudding Shell flows in opposite direction. (as to find B at mea we consider a circular ac, ampherian loop, with nadius na from symmetry we can say B = B (m) Bis directed along the tangent to the look, rin screw relation, Scanned with CamScanner

gives Amphere ciscuintal law SEdT - HOS 3.05 ci = SB(2) de = Mefits 2 B(2) fola Mo S 22?ds - Bin), 2717 = 240 ?sds B(21 27th = apar? Tim? * B1m = Homs nea 0 For B at acaat we consider Хоор c, velween Į B. df =q, Irodal na and rab, as s Bladet Mo ( current enclosed by inner cylinder 1/snafar ฯ/ระมหนา fitwinner/ ਹਰ ਪਲ rafins al B(ਮ) 2x – 1. ਨੂੰ Scanned with CamScanner

O consider 6 and look c3 betw C. Ec»)/die HelI-I') current enclosed by cylinder of radius enclosed by outer Shell of rea inner radius band outer Sܝܕܩܡ B. 2T17 B.211m = MALI- JS) = 44, (2- Hier) () = Monti- = Mos i 1.06. The one that loance) o consider ca ad me I Bidł MolII2 I enclose enclosed a by Shell be - by inner i linter of inner стале : B ) = 0 Sosy ouder radiua cs Scanned with CamScanner