Answer:
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 <Rw
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 <r <Rs
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 <r
In this case the waxed current is zero, so
B = 0
