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(B) The magnetic field $B$ due to an infinitely long wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For both wires, the distance fro

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$\frac{\mu_{0} I}{2 \pi} \hat{j}$

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(B) The magnetic field $B$ due to an infinitely long wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For both wires, the distance from the origin $O(0,0)$ is $r = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}$.The magnetic field vector is perpendicular to the position vector $\vec{r}$ and the current direction $\hat{k}$.For wire $L$ at $(-1, 1)$, the position vector is $\vec{r}_L = -\hat{i} + \hat{j}$. The magnetic field $\vec{B}_L$ is proportional to $\vec{I} 	imes \vec{r}_L = \hat{k} 	imes (-\hat{i} + \hat{j}) = -\hat{j} - \hat{i}$.For wire $M$ at $(-1, -1)$, the position vector is $\vec{r}_M = -\hat{i} - \hat{j}$. The magnetic field $\vec{B}_M$ is proportional to $\vec{I} 	imes \vec{r}_M = \hat{k} 	imes (-\hat{i} - \hat{j}) = -\hat{j} + \hat{i}$.Adding the two fields, the $\hat{i}$ components cancel out: $\vec{B}_{net} = \vec{B}_L + \vec{B}_M = 2 	imes \frac{\mu_0 I}{2 \pi r} 	imes \sin(	heta) \hat{j}$, where $\sin(	heta) = \frac{1}{\sqrt{2}}$.$\vec{B}_{net} = 2 	imes \frac{\mu_0 I}{2 \pi \sqrt{2}} 	imes \frac{1}{\sqrt{2}} \hat{j} = \frac{\mu_0 I}{2 \pi} \hat{j}$.

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