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(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$.For the wire along the $x$-axis carrying current $I_1 = 8

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$2 	imes 10^{-7} \,T$

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(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$.For the wire along the $x$-axis carrying current $I_1 = 8 \,A$, the point $P(2, 4)$ is at a perpendicular distance $r_1 = 4$ units. Using the right-hand rule, the magnetic field $B_1$ at $P$ is directed into the plane $(\otimes)$.$B_1 = \frac{\mu_0 I_1}{2\pi r_1} = \frac{(4\pi 	imes 10^{-7}) 	imes 8}{2\pi 	imes 4} = 4 	imes 10^{-7} \,T$ (into the plane).For the wire along the $y$-axis carrying current $I_2 = 6 \,A$, the point $P(2, 4)$ is at a perpendicular distance $r_2 = 2$ units. Using the right-hand rule, the magnetic field $B_2$ at $P$ is directed out of the plane $(\odot)$.$B_2 = \frac{\mu_0 I_2}{2\pi r_2} = \frac{(4\pi 	imes 10^{-7}) 	imes 6}{2\pi 	imes 2} = 6 	imes 10^{-7} \,T$ (out of the plane).The net magnetic field $B_{net} = |B_2 - B_1| = |6 	imes 10^{-7} - 4 	imes 10^{-7}| = 2 	imes 10^{-7} \,T$.

Two long wires carrying currents of $8 \,A$ and $6 \,A$ are placed along the $x$-axis and $y$-axis respectively. Find the magnitude of the magnetic field at the point $P(2, 4)$. (Take $\mu_{0} = 4\pi \times 10^{-7} \,T \cdot m/A$)

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