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(C) The magnetic field $B$ due to an infinitely long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For the wire along the

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$\frac{\mu_0 I}{12 \pi}$

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(C) The magnetic field $B$ due to an infinitely long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For the wire along the $x$-axis,the perpendicular distance to point $P(2, 3)$ is $d_x = 3 \ m$. Using the right-hand thumb rule,the direction of the magnetic field at $P$ is outward (denoted by $\odot$).$B_x = \frac{\mu_0 I}{2 \pi (3)} = \frac{\mu_0 I}{6 \pi} \odot$For the wire along the $y$-axis,the perpendicular distance to point $P(2, 3)$ is $d_y = 2 \ m$. Using the right-hand thumb rule,the direction of the magnetic field at $P$ is inward (denoted by $\otimes$).$B_y = \frac{\mu_0 I}{2 \pi (2)} = \frac{\mu_0 I}{4 \pi} \otimes$The net magnetic field $B_{	ext{net}}$ is the difference between the two fields because they are in opposite directions:$B_{	ext{net}} = B_y - B_x = \frac{\mu_0 I}{4 \pi} - \frac{\mu_0 I}{6 \pi}$$B_{	ext{net}} = \frac{3 \mu_0 I - 2 \mu_0 I}{12 \pi} = \frac{\mu_0 I}{12 \pi}$

Two mutually perpendicular insulated conducting wires carrying equal currents $I$ intersect at the origin. The resultant magnetic induction at point $P(2 \ m, 3 \ m)$ will be:

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