$B = dfrac{mu_0 times I}{2 pi times r}$

where $mu_0$ is the permeability of free space and its value is $4 pi times 10^{-7}:T.m/A$. $r$ is the radial distance of the point from the wire.

This relation can be rewritten for the radial distance $r$ as

$r = dfrac{mu_0 times I}{2 pi times B}$

If the wire carries a current of $7.2:A$ then the radial distance of the point, at which the field produced by the wire equals the magnitude of the Earth’s magnetic field ($B = 5 times 10^{-5}:T$), is given by

$r = dfrac{ ^2 cancel{4 pi} times 10^{-7} times 7.2}{cancel{2 pi} times 5times 10^{-5}} = dfrac{2 times 7.2}{5} times 10^{-2} = 2.9 times 10^{-2}:m = 2.9:cm$

$B = dfrac{mu_0 times I}{2 pi times r}$

where $mu_0$ is the permeability of free space and its value is $4 pi times 10^{-7}:T.m.A^{-1}$. $r$ is the radial distance of the point from the wire.

This relation can be rewritten for the current $I$ as

$I = dfrac{2 pi times r times B}{mu_0}$ ; Substituting the value of $mu_0$,

$I = dfrac{cancel{2 pi} times r times B}{^2 cancel{4 pi} times 10^{-7}} = dfrac{r times B}{2} times 10^{7}$

Given data are

$B = 7.1 times 10^{-6}:T$ and $r = 12.5:cm = 0.125:m$.

$I = dfrac{0.125 times 7.1}{2} times 10^{-6} times 10^7 =4.44:A$