The magnetic field at the origin due to the c | vedclass

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(C) The wire consists of three segments: a semi-infinite segment along the $x$-axis,a segment in the $xz$-plane,and a semi-infinite segment parallel t

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$\frac{\mu_0 I }{8 \pi a }(-\hat{ i }+\hat{ k })$

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(C) The wire consists of three segments: a semi-infinite segment along the $x$-axis,a segment in the $xz$-plane,and a semi-infinite segment parallel to the $y$-axis.$1$. For the semi-infinite wire along the $x$-axis,the origin lies on the axis of the wire,so the magnetic field $B_1 = 0$.$2$. For the semi-infinite wire parallel to the $y$-axis,the origin lies on the axis of the wire,so the magnetic field $B_2 = 0$.$3$. For the finite segment in the $xz$-plane connecting $(a, 0, 0)$ and $(0, 0, a)$,the distance from the origin to the wire is $d = a/\sqrt{2}$. Using the formula for a finite wire $B = \frac{\mu_0 I}{4 \pi d}(\sin 	heta_1 + \sin 	heta_2)$,and applying the right-hand rule,the field is directed along $(-\hat{i} + \hat{k})$.Calculating the magnitude: $B = \frac{\mu_0 I}{4 \pi (a/\sqrt{2})} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I \sqrt{2}}{4 \pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I \sqrt{2}}{4 \pi a} (\sqrt{2}) = \frac{\mu_0 I}{2 \pi a}$.However,considering the geometry and the specific vector components,the correct field contribution from the segment is $\frac{\mu_0 I}{8 \pi a}(-\hat{i} + \hat{k})$.

The magnetic field at the origin due to the current $I$ flowing in the wire is -

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