The magnetic moment of a circular coil carrying current is

$A$ current-carrying small loop behaves like a small magnet. If $A$ is its area and $M$ is its magnetic moment,the current in the loop will be

$A$ circular coil and a square coil are prepared from two identical metal wires,and the same current is passed through them. The ratio of the magnetic dipole moment associated with the circular coil to that of the square coil is:

The magnetic moment vectors $\mu_{s}$ and $\mu_{l}$ associated with the intrinsic spin angular momentum $S$ and orbital angular momentum $L$ respectively,of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by $\mu_{s} = -(e/m)S$ and $\mu_{l} = -(e/2m)L$. Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Due to the flow of current in a circular loop of radius $R$,the magnetic induction produced at the centre of the loop is $B$. The magnetic moment of the loop is (where ${\mu _0}$ is the permeability constant).

$A$ conducting ring of radius $R$ and one turn is formed from a conducting wire of length $L$ and on passing current $I$ the obtained magnetic dipole moment is $m$. If this wire is then converted to a ring of two turns and on passing electric current $I$,the new magnetic dipole moment obtained is . . . . . . .