The magnetic dipole moment of a rectangular current-carrying loop is:

An insulated wire is wound so that it forms a flat coil with $N = 200$ turns. The radius of the innermost turn is $r_1 = 3\text{ cm}$, and of the outermost turn $r_2 = 6\text{ cm}$. If $20\text{ mA}$ current flows in it, then the magnetic moment will be $\alpha \times 10^{-2}\text{ A.m}^2$. The value of $\alpha$ 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

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).

$A$ current of $3 \, A$ is flowing in a plane circular coil of radius $4 \, cm$ and number of turns $20$. The coil is placed in a uniform magnetic field of magnetic induction $0.5 \, T$. The magnetic dipole moment of the coil is $... \, A \cdot m^2$.

$A$ current $i$ is flowing in a conductor of length $l$. When it is bent in the form of a circular loop,then its magnetic moment will be-