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Sagot :

[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.

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