Question

A hollow cylindrical (inner radius = 1.0 mm, outer radius = 3.0 mm) conductor carries a current of 80 A parallel to its axis. This current is uniformly distributed over a cross section of the conductor. Determine the magnitude of the magnetic field at a point that is 2.0 mm from the axis of the conductor.

Answer 1

Answer:

The magnetic field is   $B = 3 mT$

Explanation:

From the question we are told that

     The inner radius is  $r_i = 1.00 mm =1*10^{-3} \ m$

      The outer radius is  $r_2 = 3.00 \ mm = 3.0 *10^{-3} \ m$

       The distance from the axis of the conductor is $d =2.0 \ mm = 2.0 *10^{-3} \ m$

      The current carried by the conductor is  $I = 80 A$

According to Ampere's circuital law , the magnetic field at a point that is  $r_3$  from the axis of the conductor

                  $B = \frac{\mu_oI}{2 \pi d } [\frac{d - r_1}{r_2 -r_1} ]$

Where $\mu_o$ is the permeability of free space with a value of $\mu_o = 4 \pi *10^{-7} N/A^2$

substituting values

                  $B = \frac{(4 \pi *10^{-7})(80)}{2 * 3.142 * 2 *10^{=3} } [\frac{(2^2 - 1 ^2 )*10^{-3}}{(3^2 - 1^2) *10^{-3}} ]$

                 $B = 3 mT$