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(C) The magnetic field due to a long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.Using the right-hand thumb rule:$1$. Fo

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(C) The magnetic field due to a long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.Using the right-hand thumb rule:$1$. For the vertical wire carrying current in the $+y$ direction,the magnetic field at point $P$ (located at $(d, d)$) is directed into the plane,i.e.,$-\hat{k}$ direction.$2$. For the horizontal wire carrying current in the $+x$ direction,the magnetic field at point $P$ is directed out of the plane,i.e.,$+\hat{k}$ direction.Thus,the net magnetic field is $\overrightarrow{B}_{net} = \overrightarrow{B}_1 + \overrightarrow{B}_2 = \frac{\mu_0 I}{2 \pi d}(-\hat{k}) + \frac{\mu_0 I}{2 \pi d}(\hat{k}) = 0$.

Two very long,straight and insulated wires are kept at a $90^o$ angle from each other in the $xy$-plane as shown in the figure. These wires carry a current of equal magnitude $I$,whose directions are shown in the figure. The net magnetic field at point $P$ will be:

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