Current $i$ is carried in a wire of length $L$. If the wire is turned into a circular coil,the maximum magnitude of torque in a given magnetic field $B$ will be

$A$ circular coil having $200$ turns,$2.5 \times 10^{-4} \text{ m}^2$ area and carrying $100 \mu\text{A}$ current is placed in a uniform magnetic field of $1 \text{ T}$. Initially,the magnetic dipole moment $(\vec{M})$ was directed along $\vec{B}$. The amount of work required to rotate the coil through $90^{\circ}$ from its initial orientation such that $\vec{M}$ becomes perpendicular to $\vec{B}$ is . . . . $\mu\text{J}$.

$A$ circular coil having $N$ turns and radius $r$ carries a current $I$. It is held in the $XZ$ plane in a magnetic field $\vec{B} = B\hat{i}$. The torque on the coil due to the magnetic field is

$A$ circular current loop of magnetic moment $M$ is in an arbitrary orientation in an external uniform magnetic field $\vec{B}$. The work done to rotate the loop by $30^{\circ}$ about an axis perpendicular to its plane is

$(a)$ $A$ current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e.,turns about the vertical axis)?
$(b)$ $A$ current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn,what is its orientation of stable equilibrium? Show that in this orientation,the flux of the total field (external field $+$ field produced by the loop) is maximum.
$(c)$ $A$ loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible,why does it change to a circular shape?

Write an expression for the torque acting on a current-carrying loop suspended in a uniform magnetic field.