Hi there!
We can use Ampère's Law to derive an expression for the magnetic field strength produced by an infinitely long current-carrying wire:
[tex]\oint B \cdot dl = \mu_0 I_{encl}[/tex]
B = Magnetic Field Strength (T)
dl = Differential length along path
μ₀ = Permeability of Free Space (Tm/A)
I = Enclosed current (A)
The integral is a cross product, so the cosine of the angle between the magnetic field and the path of integration is used. However, for a straight current-carrying wire, the path is ALWAYS parallel to the magnetic field, so since cos(180) = 1, we can disregard the cross product.
Additionally, the path of integration is equivalent to:
[tex]l = 2\pi r[/tex]
[tex]B \cdot l = \mu_0 I_{encl}\\\\B * 2\pi r = \mu_0 I_{encl}\\\\B = \frac{\mu_0 I_{encl}}{2\pi r }[/tex]
Rearrange to solve for enclosed current:
[tex]I_{encl} = \frac{2\pi rB}{\mu_0} = \frac{2\pi (0.02)(0.0037)}{(4\pi *10^{-7})} = \boxed{370 A}[/tex]
