Two infinite length wires carries currents $8A$ and $6A$ respectively and placed along $X$ and $Y$-axis. Magnetic field at a point $P\,(0,\,0,\,d)\,m$ will be
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(d) Magnetic field at $P$ due to wire $1$, ${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\,(8)}}{d}$
and due to wire $2$, ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\,(6)}}{d}$
$==>$ ${B_{net}} = \sqrt {B_1^2 + B_2^2} = \sqrt {{{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{16}}{d}} \right)}^2} + {{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{12}}{d}} \right)}^2}} $
$ = \frac{{{\mu _0}}}{{4\pi }} \times \frac{2}{d} \times 10 = \frac{{5{\mu _0}}}{{\pi d}}$
and due to wire $2$, ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\,(6)}}{d}$
$==>$ ${B_{net}} = \sqrt {B_1^2 + B_2^2} = \sqrt {{{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{16}}{d}} \right)}^2} + {{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{12}}{d}} \right)}^2}} $
$ = \frac{{{\mu _0}}}{{4\pi }} \times \frac{2}{d} \times 10 = \frac{{5{\mu _0}}}{{\pi d}}$
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