Answer:
The electric current in the wire is 0.8 A
Explanation:
We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:
[tex]B=\frac{2\pi*a }{u*I}[/tex]
B= Magnetic field due to a straight and long wire that carries current
u= Free space permeability
I= Electrical current passing through the wire
a = Perpendicular distance from the wire to the point where the magnetic field is located
Magnetic Field Calculation
We cleared (I) of the formula (1):
[tex]I=\frac{2\pi*a*B }{u}[/tex] Formula(2)
[tex]B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }[/tex]
a =8cm=0.08m
[tex]u=4*\pi *10^{-7} \frac{Weber}{A*m}[/tex]
We replace the known information in the formula (2)
[tex]I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }[/tex]
I=0.8 A
Answer: The electric current in the wire is 0.8 A
