Question

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

• A. $\frac{{\mu }_{0}I}{8\pi a}(\hat{i}+\hat{k})$
• B. $\frac{{\mu }_{0}I}{2\pi a}(\hat{i}+\hat{k})$
• C. $\frac{{\mu }_{0}I}{8\pi a}(-\hat{i}+\hat{k})$
• D. $\frac{{\mu }_{0}I}{4\pi a\sqrt{2}}(\hat{i}-\hat{k})$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{8\pi a}(-\hat{i}+\hat{k})$

One part of the wire is lying along positive x-axis, one part along the line $x=z$ and one part along positive y-axis.


There is no magnetic field at the origin due to the segments 1 and 2 (since they both pass through the origin). Hence, net magnetic field at origin is only due to 3.

The magnetic field due to 3 at origin is given by the formula for the magnetic field due to a semi infinite wire at a point.



$B=\frac{{\mu }_{0}i}{4\pi d}(\sin \alpha +\sin \beta )$ where $\alpha$ and $\beta$ are the angles made by the line joining the point O with the extremes of the wire.
Here, $\alpha =0^{\circ },\beta ={90}^{\circ }$
$\therefore B=\frac{{\mu }_{0}i}{4\pi d}$

To find the direction of field, plot the magnetic field lines around the wire passing through point O.


Then, we can say that the magnetic field at O must be perpendicular to the radius of the circle passing through O shown in the diagram.
Since the radius is along the line $x=z$, the line perpendicular to it is $x=-z$ i.e the direction of magnetic field at O is along ($-\hat{i}+\hat{k}$)
Thus, magnetic field at O can be written as
$\overrightarrow{B}=\frac{{\mu }_{0}i}{4\pi d}\cos {45}^{\circ }(-\hat{i})+\frac{{\mu }_{0}i}{4\pi d}\sin {45}^{\circ }\left(\hat{k}\right)$
$=\frac{{\mu }_{0}i}{4\pi \sqrt{2}a}\cos {45}^{\circ }(-\hat{i}+\hat{k})$
$=\frac{{\mu }_{0}i}{8\pi a}(-\hat{i}+\hat{k})$