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(A) The magnetic field due to an infinitely long wire carrying current $I$ at a perpendicular distance $a$ is given by $B = \frac{\mu_0 I}{2\pi a}$. T

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$\frac{\mu_0 I}{2\pi a}(\hat{j} - \hat{i})$

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(A) The magnetic field due to an infinitely long wire carrying current $I$ at a perpendicular distance $a$ is given by $B = \frac{\mu_0 I}{2\pi a}$. The direction is given by the right-hand rule.$1$. For the wire along the positive $x$-axis: The position vector from the wire to the point $(0, 0, -a)$ is in the $-z$ direction. Using the right-hand rule,the magnetic field at $(0, 0, -a)$ is $\vec{B}_1 = \frac{\mu_0 I}{2\pi a} \hat{j}$.$2$. For the wire along the positive $y$-axis: The position vector from the wire to the point $(0, 0, -a)$ is in the $-z$ direction. Using the right-hand rule,the magnetic field at $(0, 0, -a)$ is $\vec{B}_2 = \frac{\mu_0 I}{2\pi a} (-\hat{i})$.$3$. For the wire along the positive $z$-axis: The point $(0, 0, -a)$ lies on the line of the wire itself. Therefore,the magnetic field due to this wire is $\vec{B}_3 = 0$.$4$. The net magnetic field is $\vec{B}_{net} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 = \frac{\mu_0 I}{2\pi a} (\hat{j} - \hat{i})$.

Equal currents $I$ are flowing in three infinitely long wires along the positive $x$,$y$,and $z$ directions. The magnetic field at a point $(0, 0, -a)$ would be:

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