Answer:
The magnetic field at the center of the coil is [tex]B =53 \mu T[/tex]
Explanation:
From the question we are told that
The magnetic field of earth is [tex]B_e = 50 \mu T = 50 *0^{-6} \ T[/tex]
The diameter of the wire is [tex]d = 1.0\ m[/tex]
The radius is [tex]r = \frac{1}{2} = 0.5 \ m[/tex]
The number of turn is [tex]N = 200 \ turns[/tex]
The current it is carrying is [tex]I = 0.211 \ A[/tex]
The earth magnetic field strength at the center of the coil is mathematically represented as
[tex]B = N \frac{\mu_o * I }{2 * r}[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space
[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = 200* \frac{ 4\pi * 10^{-7} * 0.211 }{2 * 0.5}[/tex]
[tex]B = 5.304*10^{-5}[/tex]
[tex]B =53 \mu T[/tex]
