Answer:
The magnetic field is [tex]B = 3 mT[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r_i = 1.00 mm =1*10^{-3} \ m[/tex]
The outer radius is [tex]r_2 = 3.00 \ mm = 3.0 *10^{-3} \ m[/tex]
The distance from the axis of the conductor is [tex]d =2.0 \ mm = 2.0 *10^{-3} \ m[/tex]
The current carried by the conductor is [tex]I = 80 A[/tex]
According to Ampere's circuital law , the magnetic field at a point that is [tex]r_3[/tex] from the axis of the conductor
[tex]B = \frac{\mu_oI}{2 \pi d } [\frac{d - r_1}{r_2 -r_1} ][/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a value of [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]
substituting values
[tex]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}} ][/tex]
[tex]B = 3 mT[/tex]
