Equal current i is flowing in three infinitel | vedclass

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(A) The point $P(0, 0, -a)$ is at a distance $a$ from the $x, y$ and $z$ axes.$1$. For the wire along the $z$-axis: The point $(0, 0, -a)$ lies on the

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

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(A) The point $P(0, 0, -a)$ is at a distance $a$ from the $x, y$ and $z$ axes.$1$. For the wire along the $z$-axis: The point $(0, 0, -a)$ lies on the $z$-axis itself. Therefore,the magnetic field due to the current in the $z$-direction is $\vec{B}_z = 0$.$2$. For the wire along the $x$-axis: The distance from the point $(0, 0, -a)$ to the $x$-axis is $a$. Using the right-hand rule,the direction of the magnetic field at $(0, 0, -a)$ is in the $+\hat{j}$ direction. Thus,$\vec{B}_x = \frac{\mu_0 i}{2\pi a} \hat{j}$.$3$. For the wire along the $y$-axis: The distance from the point $(0, 0, -a)$ to the $y$-axis is $a$. Using the right-hand rule,the direction of the magnetic field at $(0, 0, -a)$ is in the $-\hat{i}$ direction. Thus,$\vec{B}_y = -\frac{\mu_0 i}{2\pi a} \hat{i}$.Total magnetic field $\vec{B} = \vec{B}_x + \vec{B}_y + \vec{B}_z = \frac{\mu_0 i}{2\pi a} \hat{j} - \frac{\mu_0 i}{2\pi a} \hat{i} = \frac{\mu_0 i}{2\pi a}(\hat{j} - \hat{i})$.

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

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