The correct Biot-Savart law in vector form is

Two infinitely long thin wires are placed at $(1 \text{ cm}, 0 \text{ cm})$ and $(2 \text{ cm}, 0 \text{ cm})$ as shown in the figure. The same current $i$ flows in both the wires in the same direction,into the page. Let the magnetic field at the origin due to these wires be $\vec{B}$. If $B_0$ is the magnitude of the magnetic field if only the wire at $(1 \text{ cm}, 0 \text{ cm})$ was present,then the value of $B / B_0$ is

$A$ straight wire carrying current $I$ is bent into a semi-circular arc of radius $r$,as shown. The magnitude of the magnetic field at point $O$ due to the semi-circular arc is ($\mu_{0} =$ permeability of free space).

$A$ direct current is passing through a wire. It is bent to form a coil of one turn. Now,it is further bent to form a coil of two turns but at a smaller radius. Find the ratio of the magnetic induction at the centre of this two-turn coil to that at the centre of the one-turn coil.

The current of $5 \ A$ flows in a square loop of side $1 \ m$ placed in air. The magnetic field at the centre of the loop is $X \sqrt{2} \times 10^{-7} \ T$. The value of $X$ is . . . . . . .

$A$ coil of one turn is made of a wire of a certain length,and then from the same length of wire,a coil of two turns is made. If the same current is passed in both cases,what is the ratio of the magnetic inductions at their centers?