At what distance from a long straight wire carrying a current of $$12\,A$$  will the magnetic field be equal to $$3 \times {10^{ - 5}}Wb/{m^2}$$   ?

Solution :
Total magnetic field due to current carrying straight wire at any point $$P$$ is given by
$$B = \frac{{{\mu _0}i}}{{4\pi r}}\left( {\sin {\phi _1} + \sin {\phi _2}} \right)$$
When the conductor is of infinite length and point $$P$$ lies near the centre of conductor, then $${\phi _1} = {\phi _2} = {90^ \circ }$$
$$\eqalign{ & \therefore B = \frac{{{\mu _0}i}}{{4\pi r}}\left( {\sin {{90}^ \circ } + \sin {{90}^ \circ }} \right) \cr & = \frac{{{\mu _0}i}}{{4\pi r}} \cdot \frac{{2i}}{r} \Rightarrow r = \frac{{{\mu _0}i}}{{2\pi B}} \cr} $$
Here, current $$\left( i \right) = 12\,A,$$
magnetic field $$\left( B \right) = 3 \times {10^{ - 5}}\,Wb/{m^2}$$
$$\therefore $$ Perpendicular distance from wire to the point,
$$\eqalign{ & r = \frac{{4\pi \times {{10}^{ - 7}} \times 12}}{{2 \times \pi \times \left( {3 \times {{10}^{ - 5}}} \right)}} \cr & = 8 \times {10^{ - 2}}m \cr} $$