A) [tex]2.0\cdot 10^{-6} T[/tex]
The magnetic field strength around a current-carrying wire is given by the formula
[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]
where
[tex]\mu_0 = 1.257 \cdot 10^{-6} H/m[/tex] is the vacuum permeability
I is the current in the wire
r is the radial distance from the wire
In this problem, we have
I = 200 A is the current in the wire
and we want to calculate the magnetic field strength at a distance of
r = 20 m
from the wire, so:
[tex]B=\frac{(1.257\cdot 10^{-6})(200)}{2\pi (20)}=2.0\cdot 10^{-6} T[/tex]
B) 4 %
The magnetic field strength at the surface of the Earth is
[tex]B_e = 5\cdot 10^{-5} T[/tex]
While the strength of the magnetic field generated by the wire at ground level is
[tex]B=2.0\cdot 10^{-6} T[/tex]
Therefore, in percentage, it is:
[tex]\frac{B}{B_e}=\frac{2.0\cdot 10^{-6}}{5\cdot 10^{-5}}\cdot 100 = 0.04 \cdot 100 = 4 \%[/tex]
