Question

2. A modified coaxial cable consists of a solid cylinder (radius 'a') with a uniform current density and a concentric cylindrical conducting thin shell (radius 'b'). The outer and inner current have an equal magnitude, but are opposite in direction. Io (along outside) (along the axis) (off-axis view) In terms of radial distance 'r', and the relevant parameters in the diagram above, A) Derive an expression for the magnetic field inside the solid cylinder (r <a) B) Derive an expression for the magnetic field between the cylinders (b >r>a), C) Derive an expression for the magnetic field outside the cylindrical shell (r > b) D) If a= 5.00 cm, b = 8.00 cm and I = 3.00 A, determine where the strongest magnetic field would be, and its magnitude.

Answer 1

"Radius of solid cylinder is a. Radius of conducting shell is b. Cuttent density of solid cylinder is J = ""current per unit are"" J = I_0/pi a^2 A) first we draw a Ampere's loop of radius r such that r < a from Ampere's law contourintegral B. dl = mu_0 I_increases where I_increased be the current in closed by the ampere's loop I_increased = J times crosssectional area of Ampere's loop I_increased = I_0/pi a^2 times pi r62 Now B times 2pi r = mu_0 times I_0 r^2/a^2 B = mu_0 I_0 r/2pia^2 r < a where mu_0 be the permeability of free state. (B) Now we draw a Amphere's loop of radius r such that b > r > a current in closed by the Ampere's loop is I_increased = I_0 from Ampere's law B times 2pi r = mu_0 I_0 B = mu_0 I_0/2pi r a < r < b (C) current in closed by the Ampere's loop is I_increased = I_0 - I_0 = 0 because current in opposite direction then from Ampere's law "