A coaxial cable consists of two cylinders of radii R1 and R2. The conductor with

Question

A coaxial cable consists of two cylinders of radii R1 and R2. The conductor with radius r < R1 has a non-uniform current density given by J(r) = kr2 , where k is a known constant. The current in this region flows out of the figure. The outer cylindrical shell for R1 < r < R2 has a total current I distributed uniformly throughout its cross section. The current in this region flows into the figure. Let current flowing out of the figure be positive current flow. Find an expression for B(r) for

(a) r < R1 ;

(b) R1 < r < R2.

Simplify the final expressions for B(r) as much as possible.

The question on the link below for more information  

http://capone.mtsu.edu/phys2120/HW_Problems/CH22-II-HWP.pdf

Explanation / Answer

a)

Here , let assume a loop of radius r with same centre as the cable

and r < R1

current inside the loop Iin

Iin = integrate(J(r)*dA) from 0 to r

Iin = integrate(kr^2*2*pi*r*dr) from 0 to r

Iin = integrate(2*pi*kr^3 dr) from 0 to r

Iin = pi*k*r^4/2

Using amphere circuital law ,

B . 2*pi*r = Iin * u0

B . 2*pi*r = u0* pi*k*r^4/2

B = u0*k*r^3/4

the magnetic field for r < R1 is

B(r) = u0*k*r^3/4

b)

For R1 < r < R2

Here , Iin = pi*k*R1^4/2 - I*(r^2 - R1^2)/(R2^2 - R1^2)

Using amphere's circuital law

B. 2*pi*r = u0*(pi*k*R1^4/2 - I*(r^2 - R1^2)/(R2^2 - R1^2))

B = u0*(pi*k*R1^4/2 - I*(r^2 - R1^2)/(R2^2 - R1^2))/2*pi*r

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