A straight section PQ of a circuit lies along | 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 position vector $\vec{r}$ is given by $d

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Zero

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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 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 any point lying on the axis of the current-carrying wire,the current element $d\vec{l}$ and the position vector $\vec{r}$ are collinear (i.e.,the angle between them is $0^\circ$ or $180^\circ$).Since the cross product of two collinear vectors is zero,the magnetic field at any point on the axis of the wire is zero.In this problem,the point $x = a$ lies on the $X$-axis,which is the same axis along which the wire $PQ$ is placed.Therefore,the magnetic field at $x = a$ is zero.

$A$ straight section $PQ$ of a circuit lies along the $X$-axis from $x = -\frac{a}{2}$ to $x = \frac{a}{2}$ and carries a steady current $i$. The magnetic field due to the section $PQ$ at a point $x = a$ on the $X$-axis will be:

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