Answer:
1.25 kA
Explanation:
The magnitude of the magnetic field produced by a current-carrying wire is given by:
[tex]B=\frac{\mu_0 I}{2 \pi r}[/tex]
where
[tex]\mu_0 = 1.26 \cdot 10^{-6} H/m[/tex] is the vacuum permeability
[tex]I[/tex] is the current in the wire
r is the distance from the wire at which the field is calculated
In this problem, we have:
[tex]B = 5.0\cdot 10^{-4} T[/tex] is the intensity of the magnetic field
[tex]r = 0.50 m[/tex] is the distance from the wire
Re-arranging the equation, we can find the intensity of the current in the wire:
[tex]I=\frac{2 \pi r B}{\mu_0}=\frac{2 \pi (0.50 m)(5.0\cdot 10^{-4}T)}{1.26\cdot 10^{-6} H/m}=1246 A=1.25 kA[/tex]
