Question

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.

Answer 1

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