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(C) The magnetic field due to an infinite wire at a distance $r$ is $B = \frac{\mu_0 i}{2\pi r}$.$1$. Wire $(1)$ along the $x$-axis carries current $i

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$\frac{\mu_0 i}{4\pi r}\hat{i} + \frac{\mu_0 i}{4\pi r}\hat{j} - \frac{\mu_0 i}{\pi r}\hat{k}$

Vedclass · Answer

(C) The magnetic field due to an infinite wire at a distance $r$ is $B = \frac{\mu_0 i}{2\pi r}$.$1$. Wire $(1)$ along the $x$-axis carries current $i$ in the $+x$ direction. At point $A(r, -r, 0)$,the position vector is $r\hat{i} - r\hat{j}$. The magnetic field is given by $\vec{B}_1 = \frac{\mu_0 i}{2\pi r} (\hat{i} 	imes \hat{n})$,where $\hat{n}$ is the unit vector from the wire to $A$. The direction is $-\hat{k}$. Thus,$\vec{B}_1 = \frac{\mu_0 i}{2\pi r} (-\hat{k})$.$2$. Wire $(2)$ along the $y$-axis carries current $i$ in the $+y$ direction. At point $A(r, -r, 0)$,the magnetic field is $\vec{B}_2 = \frac{\mu_0 i}{2\pi r} (\hat{k})$.$3$. Wire $(3)$ along the $z$-axis carries current $i$ out of the plane ($+\hat{k}$ direction). The distance from the $z$-axis to $A$ is $d = \sqrt{r^2 + (-r)^2} = r\sqrt{2}$. The magnetic field magnitude is $B_3 = \frac{\mu_0 i}{2\pi (r\sqrt{2})}$. The direction is perpendicular to the position vector $\vec{r} = r\hat{i} - r\hat{j}$,which is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$. Thus,$\vec{B}_3 = \frac{\mu_0 i}{2\pi r\sqrt{2}} \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} ight) = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j}$.Summing the fields: $\vec{B}_{net} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 = 0 + \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j} = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j}$.Wait,re-evaluating the directions: Wire $(1)$ is at $y=0$,$z=0$. Point $A$ is at $(r, -r, 0)$. The vector from wire to $A$ is $-r\hat{j}$. $\vec{B}_1 = \frac{\mu_0 i}{2\pi r} (\hat{i} 	imes (-\hat{j})) = -\frac{\mu_0 i}{2\pi r} \hat{k}$.Wire $(2)$ is at $x=0$,$z=0$. Point $A$ is at $(r, -r, 0)$. The vector from wire to $A$ is $r\hat{i}$. $\vec{B}_2 = \frac{\mu_0 i}{2\pi r} (\hat{j} 	imes \hat{i}) = -\frac{\mu_0 i}{2\pi r} \hat{k}$.Wire $(3)$ is at $x=0$,$y=0$. $\vec{B}_3 = \frac{\mu_0 i}{2\pi (r\sqrt{2})} \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j}$.Total $\vec{B} = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j} - \frac{\mu_0 i}{\pi r} \hat{k}$.

Three infinite wires are arranged in space in three dimensions (along $x$,$y$,and $z$ axes) as shown. Each wire carries current $i$. Find the magnetic field at point $A$ located at $(r, -r, 0)$.

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