Answer:
The magnetic field is [tex]B = 2.319 *10^{-3} \ T[/tex]
Explanation:
From the question we are told that
The radius of the wire is [tex]r = 0.8 \ cm = 0.008 \ m[/tex]
The current is [tex]I = 106 \ A[/tex]
The position considered is d = 0.07 cm = 0.0007 m
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * I}{2\pi * \frac{r^2}{d} }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex] 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = \frac{ 4\pi * 10^{-7} * 106 }{2 * 3.142 * \frac{0.008^2}{0.0007} }[/tex]
=> [tex]B = 2.319 *10^{-3} \ T[/tex]
