Question

Quiz Ch.29 Ampere's law. you have a coaxial cable, made of an inner conclucting cylinder of radius a, and non miform current density given by: J=ar² (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 I, total wrrent with uniform current density situ J J= I ( -) for this configuration looks a "view from the top" as follows . Find: (use Ampere's law) a) B (the magnetic field at distance rca. b) B at acrab c) B at berac d) B at ryc Coutside the array)

Answer 1

29 the inner cylinder is cursent density of J = 22 first we calculate total current of the inner cylinder and ring of viadius ut We cut an elemental such that rcq. width dur. ving is of this elemental cross-sectional autua 2 41 elemental strip is this mow curuiunt in dI = J.JA

di = 21² x 21tu du di = 4Ture du (9) loop of uradius of Ampere's We draw an such that uca - Ampure's loop loop is by the Ampere's inclosed total currunt 4" 5" opdan Im T - 44 4TT Iim I'm. = Tut now from Ampure's law -

f Bode = Mo I'm = Mo Tut 3 B. 2tY 2B = Mou3 79. 8 B Mov? = Your 3 look of radius, Now we such that draw an Ampere's acul Lb. s loob. - Ampute's Ampure's loop is the incelesed by total current 19 I di 4 3 dur - dan Im = = Im - dr. In - 4*

from Ampore's law B. 25 = Mo Im. = Mortat B. 25/9 B- Moa B Hot acurab = qCulb loop of - (c) Now we draw An vadius ut such that Ampere's buikc. loop - Ampure's total culturent inclosed by the I Ampere's loop is f (-b?) I TAS-) Im = Tat – I (²62) (c²_62)

now from Ampurs's law. B. RITU = Mo Im. B. RITUT = 40 | Tot - T1²-62) (e?-62) B = Ho | Tat - I (0²-6²) , bculcc ZTU an Ampere's loop of radius of such d) Now that we draw sic thy Ly Ampure's loop. total current inclosed by Ampure's loop is I'm = Trat - I from Ampure's law B. RTV - Molim = Moltrat - I) B = Mo 1 Tat-I) oc 2 Tur