In a hydrogen atom,an electron revolves around the nucleus at a frequency of $6.6 \times 10^{15} \, \text{rev/sec}$ in an orbit of radius $0.528 \, \mathring{A}$. The magnetic moment in $\text{A} \cdot \text{m}^2$ is:

$A$ small bar magnet experiences a torque of $0.016 \ Nm$ when placed with its axis at $30^{\circ}$ with an external magnetic field of $0.04 \ T$. If the bar magnet is replaced by a solenoid of cross-sectional area $1 \ cm^2$ and $1000$ turns, having the same magnetic moment as that of the bar magnet, then the current flowing through the solenoid is: (in $A$)

$A$ thin circular wire carrying current $I$ has magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have magnetic moment $M_1$. The ratio $M$ to $M_1$ 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.

$A$ 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 . . . . . . .

Write the equation of the magnetic dipole moment in terms of electric current.