$A$ circular coil of radius $4\, cm$ and of $20$ turns carries a current of $3$ amperes. It is placed in a magnetic field of intensity $0.5\, Wb/m^2$. The magnetic dipole moment of the coil is.......$A \cdot m^2$.

$A$ steady current $i$ flows in a small square loop of wire of side $L$ in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let $\overrightarrow {{\mu _1}} $ and $\overrightarrow {{\mu _2}} $ respectively denote the magnetic moments due to the current loop before and after folding. Then

$A$ 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 $A$)

$2 \ A$ current is flowing through two different small circular copper coils having radii in the ratio $1:2$. The ratio of their respective magnetic moments 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:

Magnetic induction at the centre of a circular loop of area $\pi \ m^2$ is $0.1 \ T$. The magnetic moment of the loop is ( $\mu_0 = \text{permeability of air}$ ).