The relation between magnetic moment $(M)$ of a current-carrying circular coil and the length $(L)$ of the wire used is:

$A$ 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 . . . . . . $A \ m^2$.

$A$ current $i$ flows in a horizontal square loop of side $L$. Now,half of the loop is folded to be vertical. If $\overrightarrow {{\mu _1}}$ and $\overrightarrow {{\mu _2}}$ are the magnetic moments before and after folding,respectively,then:

$A$ wire of length $2 \, m$ carrying a current of $1 \, A$ is bent to form a circle. The magnetic moment of the coil is

$A$ 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 field at the centre of a single-turn circular coil of a given length of wire carrying a current $I$ is $B$. If the same wire is bent into a circular coil of two turns and the same current $I$ is passed through it,the new magnetic field at the centre will be: