In an ionised sodium atom,an electron is moving in a circular path of radius $r$ with angular velocity $\omega$. The magnetic induction in $Wb/m^2$ produced at the nucleus will be

$A$ wire bent in the shape of a regular $n$-polygonal loop carries a steady current $I$. Let $l$ be the perpendicular distance of a given segment and $R$ be the distance of a vertex from the centre of the loop. The magnitude of the magnetic field at the centre of the loop is given by

Two parallel wires situated at a distance $2a$ are carrying equal currents $i$ in opposite directions as shown in the figure. The value of the magnetic field at a point $P$ situated at equal distances $r$ from both the wires will be:

$A$ long straight wire carrying electric current '$i$' is bent at its point $Q$ to form an angle of $45^{\circ}$ as shown in the figure. The magnetic field at a point $P$ at a distance $d$ from the point $Q$ is:

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

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^\circ$. Another straight thin wire with steady current $I_1$ flowing out of the plane of the paper is kept at the origin. The magnitude of the magnetic field $(B)$ due to the loop $ABCD$ at the origin $(O)$ is: