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

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(C) The wire consists of two semi-infinite segments. Segment $OD$ lies along the $x$-axis. The origin $O$ lies on the axis of this wire segment,so the

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$\frac{{\mu _0 I}}{{8\pi a}}\left( { - \hat i + \hat k} \right)$

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(C) The wire consists of two semi-infinite segments. Segment $OD$ lies along the $x$-axis. The origin $O$ lies on the axis of this wire segment,so the magnetic field due to segment $OD$ at the origin is $B_{OD} = 0$. Segment $AB$ is parallel to the $y$-axis and passes through the point $(a, 0, a)$. The distance of the origin from this wire is $r = \sqrt{a^2 + a^2} = a\sqrt{2}$. The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is $B = \frac{\mu_0 I}{4\pi r}$. Using the right-hand rule,the direction of the magnetic field at the origin due to segment $AB$ is perpendicular to the plane containing the wire and the origin. The vector from the wire to the origin is along $(-\hat{i} - \hat{k})$. The current is along $\hat{j}$. The direction is $\hat{j} 	imes (\hat{i} + \hat{k}) = -\hat{k} + \hat{i}$. Normalizing the direction vector,the field is $B = \frac{\mu_0 I}{4\pi (a\sqrt{2})} \frac{(-\hat{i} + \hat{k})}{\sqrt{2}} = \frac{\mu_0 I}{8\pi a} (-\hat{i} + \hat{k})$.

The magnetic field at the origin due to the current $I$ flowing in the wire as shown in the figure is:

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