$A$ 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

$A$ 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 $A$)

Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of current and the thumb gives the direction of the magnetic field. In this case,the upper side of the loop may be thought of as

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:

$A$ closely wound solenoid of $800$ turns and area of cross-section $2.5 \times 10^{-4} \; m^2$ carries a current of $3.0 \; A$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

$A$ closely wound solenoid of $1000$ turns and area of cross-section $2 \times 10^{-4} \ m^2$ carrying a current of $5.0 \ A$. The magnetic moment is . . . . . . $A \ m^2$.