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(C) 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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perpendicular to the plane of the paper,directed into the paper

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(C) 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 every current element $I d\vec{l}$ on the quarter-circular arc $AB$,the direction of the magnetic field at the center $C$ is determined by the right-hand rule.Since the current flows along the arc in the plane of the paper,the cross product $d\vec{l} 	imes \vec{r}$ for every element on the arc points perpendicularly into the plane of the paper.For the straight segments connected to $A$ and $B$,the current is directed along the lines passing through $C$ (or collinear with the radius vector),so the magnetic field contribution from these straight segments at point $C$ is zero.Therefore,the net magnetic field at $C$ is solely due to the arc $AB$ and is directed perpendicular to the plane of the paper,into the paper.

$A$ wire carrying current $I$ is shaped as shown in the figure. Section $AB$ is a quarter circle of radius $r$. The magnetic field at point $C$ is directed:

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