[tex]\large\mathcal{ANSWER:}\\[/tex]
[tex]\qquad \qquad \qquad \sf \bold{ B = \dfrac{\mu_{\circ}}{2 \pi}\bigotimes}[/tex]
[tex]\large\mathcal{SOLUTION:} \\ [/tex]
We have been given two straight current carrying conductors which are at same perpendicular distance from a point P.
When current flows through a conductor, a magnetic field is produced.
The intensity for a magnetic field prodcued by current carrying conductor at a certain point is given by:
[tex] \qquad \qquad \hookrightarrow \: \: \: \sf B = \dfrac{\mu_{\circ}I}{2\pi r}[/tex]
'I' is the current in Amperes and 'r' is the distance in metres.
For the condcutor with 5 A current, the magnetic field intensity at P will be:
[tex]\begin{gathered} \qquad \qquad\sf{B_{1} = \dfrac{\mu_{\circ}\times 5}{2\pi \times 2.5} \bigotimes} \\ \\ \\ \qquad \qquad\sf{B_{1} = \dfrac{2 \mu_{\circ}}{2\pi} \bigotimes}\\ \\ \\ \qquad \qquad \qquad\sf{B_{1} = \dfrac{\mu_{\circ}}{\pi} \bigotimes} \\ \end{gathered} [/tex]
For the condcutor with 2.5 A current, the magnetic field intensity at P will be:
[tex]\begin{gathered} \qquad \qquad\sf{B_{2} = \dfrac{\mu_{\circ}\times 2.5}{2\pi \times 2.5} \bigodot}\\ \\ \\ \qquad \qquad\sf{B_{2} = \dfrac{ \mu_{\circ}}{2\pi} \bigodot}\end{gathered} [/tex]
Net magnetic field at P will be:
[tex]\begin{gathered}\qquad \qquad \sf B_{1}-B_{2}=\dfrac{\mu_{\circ}}{\pi} - \dfrac{\mu_{\circ}}{2\pi} \bigotimes \\ \\ \\\qquad \qquad \sf B = \dfrac{2\mu_{\circ}-\mu_{\circ}}{2\pi} \bigotimes \\ \\ \\ \qquad \qquad\sf \implies \bold{B = \dfrac{\mu_{\circ}}{2\pi} \bigotimes}\end{gathered} [/tex]
The direction for the magnetic field is inside the plane because [tex]\sf B_{1}[/tex] was greater in magnitude.
