Answer:
0.25 m
Explanation:
The intensity of the magnetic field around a current-carrying wire is given by:
[tex]B=\frac{\mu_0 I}{2 \pi r}[/tex]
where
where
[tex]\mu_0 = 4\pi \cdot 10^{-7}Tm/A[/tex] is the permeabilty of free space
I is the current
r is the distance from the wire
In this problem, we know:
[tex]B=4 \mu T=4 \cdot 10^{-6} T[/tex] is the magnetic field
[tex]I=5.0 A[/tex] is the current in the wire
Re-arranging the equation, we can find the distance of the field from the wire:
[tex]r=\frac{\mu_0 I}{2 \pi B}=\frac{(4\pi \cdot 10^{-7})(5.0 A)}{2\pi(4\cdot 10^{-6} T)}=0.25 m[/tex]
