Consider a pipe that carries current where the current density is non-uniform. T

Question

Consider a pipe that carries current where the current density is non-uniform. This means that the current density at the center of the pipe is zero. The current density increases steadily outward and it is equal to I0 at the boundary of the pipe. (a) What is the electric field outside the pipe at a distance r from the center of the pipe? (b) What is the electric field inside the pipe at a distance r from the center of the pipe? (c) Using the above, calculate the electric field at 18 cm from the center of the pipe, given the following data. --The radius of the pipe is 6.5 cm --Value of the current density at the boundary of the pipe is 3.1 amp /cm3 . Hint: Use Ampere’s law

Explanation / Answer

here current density j = io*r/R

by amperes law, integral B.dr = uo integral j*2pi r dr

                       B.2pi r = uo*2pi*io/R* integral r^2 dr

                      B.r = uo*ior^3/(3R)

a) for r > R, B = uo*io*R^2/(3r)    where R is radius

b) for r<R, B = uo*io*r^2/(3R)  

c) B = uo*io*R^2/(3r) = 4 pi*10^-7*3.1e4*6.5e-2^2/(3*18e-2)

                          = 3.048*10^- 4 T

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