$A$ tightly wound $100$ turns coil of radius $10 \,cm$ carries a current of $7 \,A$. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 10^{-7} \,SI$ units):

The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is $64$ times the magnetic field at a distance $x$ on its axis from the centre of the coil. Then,the value of $x$ is

An infinitely long wire carrying $1 \ A$ current in the $+z$ direction is placed at $(1 \ cm, 1 \ cm)$. Another wire carrying $1 \ A$ in the $+x$ direction is placed at $y=1 \ cm$. If the magnetic field due to this configuration at the origin is $B$. Let $B_0$ be the magnitude of the field if only the wire at $(1 \ cm, 1 \ cm)$ was present,then $\frac{B}{B_0}$ is

$A$ straight wire of length $20 \text{ cm}$ carrying a current of $\frac{3}{\pi^2} \text{ A}$ is bent in the form of a circle. The magnetic field at the centre of the circle is

State and explain the Biot-Savart law for the magnetic field produced by a current element. Give the direction of the magnetic field and define its unit.

$A$ 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