$A$ circular loop having radius $r$,carrying current $I$,produces a magnetic field at the centre of the loop equal to $B$. What will be the magnetic dipole moment of this loop?

$A$ charged particle going around in a circle can be considered to be a current loop. $A$ particle of mass $m$ carrying charge $q$ is moving in a plane with speed $v$ under the influence of magnetic field $\overrightarrow{ B }$. The magnetic moment of this moving particle is:

An electron in a circular orbit of radius $0.05 \ nm$ performs $10^{16}$ revolutions per second. The magnetic moment due to this rotation of the electron is (in $A \ m^{2}$):

Two particles each of mass $m$ and charge $q$ are attached to the two ends of a light rigid rod of length $2R$. The rod is rotated at a constant angular speed $\omega$ about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is:

$A$ wire of length $314 \, cm$ carrying a current of $14 \, A$ is bent to form a circle. The magnetic moment of the coil is $........ \, A \cdot m^2$. [Given $\pi = 3.14$]

$A$ square loop of side $l$ is placed in the neighbourhood of an infinitely long straight wire carrying a current $I_1$. The loop carries a current $I_2$ as shown in the figure.