A non-conducting rod of length $L$ with linear charge density $\lambda ={\lambda }_{0}x$ where $x$ is the distance from end $A$ is rotating with constant angular speed $\omega$ about the same end. If the angular velocity of the rod $\left(\omega \right)$ is large, then the magnetic dipole moment of the system is $\frac{\omega {\lambda }_0{L}^4}{n}$. What is the value of $n$?

The ratio of the magnetic field at the centre of a current-carrying circular coil to magnetic moment is $x$. If the current and the radius both are doubled. The new ratio will become

A uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current-carrying coil in the shape of,

(i) an equilateral triangle of side $a$.
(ii) a square of side $a$.
The magnetic dipole moments of the coil in each case respectively are:

A particle of mass $M$ and charge $q$ is moving in a circle of radius $R$ with speed $v$, where $v\ll$ speed of light. The ratio of the magnetic moment of the particle to its angular momentum is

A wire carrying current $\mathrm{I}$ is bent in the shape$\mathrm{ABCDEFA}$ as shown, where rectangle $\mathrm{ABCDA}$ and $\mathrm{ADEFA}$ are perpendicular to each other. If the sides of the rectangles are of lengths $\mathrm{a}$ and $\mathrm{b}$, then the magnitude and direction of magnetic moment of the loop $\mathrm{ABCDEFA}$ is :