A coaxial cable carries a current I = 0.02A through a wire of radius Rw=.2mmin one direction and an identical current through the outer sheath of radius Rs=3.4mm in the opposite direction. Determine the magnetic field everywhere.

a) B = 10⁻¹ r, b) B = 4 10⁻⁹ / r, c) B=0

Explanation:

For this exercise let's use Ampere's law

∫ B. ds = μ₀ I

Where I is the current locked in the path. Let's take a closed path as a circle

ds = 2π dr

B 2π r = μ₀ I

B = μ₀ I / 2μ₀ r

Let's analyze several cases

a) r
Since the radius of the circumference is less than that of the wire, the current is less, let's use the concept of current density

j = I / A

For this case

j = I / π Rw² = I'/π r²

I' = I r² / Rw²

The magnetic field is

B = (μ₀ / 2π) r²/Rw² 1 / r

B = (μ₀ / 2π) r / Rw²

calculate

B = 4π 10⁻⁷ / 2π r / 0.002²

B = 10⁻¹ r

b) in field between Rw
In this case the current enclosed in the total current

I = 0.02 A

B = μ₀ / 2π I / r

B = 4π 10⁻⁷ / 2π 0.02 / r

B = 4 10⁻⁹ / r

c) the field outside the coaxial Rs
In this case the waxed current is zero, so

B = 0