A straight wire carrying a current of 12 ,A i | vedclass

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(D) According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a current element $i d\vec{l}$ at a point with position vector $\vec{r}$ is

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(D) According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a current element $i d\vec{l}$ at a point with position vector $\vec{r}$ is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{i d\vec{l} 	imes \vec{r}}{r^3}$.For the straight segments of the wire, the current element $d\vec{l}$ and the position vector $\vec{r}$ (pointing from the element to the center $O$) are collinear (either parallel or anti-parallel).Therefore, the cross product $d\vec{l} 	imes \vec{r} = 0$.Consequently, the magnetic field at the center $O$ due to the straight segments is zero.

$A$ straight wire carrying a current of $12 \,A$ is bent into a semi-circular arc of radius $2 \,cm$ as shown in the figure. Then the magnetic field due to the straight segments at the centre of the arc is

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