In the Biot-Savart law,the direction of the magnetic field is determined by which of the following cross products in the expression $d\vec B = \frac{\mu_0}{4\pi} \frac{I d\vec l \times \vec r}{r^3}$?

The magnetic field due to a current-carrying circular loop of radius $6 \text{ cm}$ at a point on the axis at a distance of $8 \text{ cm}$ from the centre is $27 \mu \text{T}$. The magnetic field at the centre of the current-carrying loop is: (in $\mu \text{T}$)

$A$ circular loop and an infinitely long straight conductor carry equal currents,as shown in the figure. The net magnetic field at the centre of the loop is $B_1$,when the current in the loop is clockwise and $B_2$ when the current in the loop is anti-clockwise. Then $\frac{B_1}{B_2}$ is

$A$ long wire carries a steady current. It is bent into a coil of one turn such that the magnetic induction at the centre is $B$. If the same wire is bent to form a coil of smaller radius with $n$ turns,then the new magnetic induction $B^{\prime}$ at the centre is:

The magnetic field due to a current in a straight wire segment of length $L$ at a point on its perpendicular bisector at a distance $r$ $(r >> L)$ is:

An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current $I$ and the radius of the circular loop is $r$. The magnetic induction at the center $O$ of the circular loop is: