$A$ coil carrying electric current is placed in a uniform magnetic field,then

$(a)$ $A$ 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?

$A$ current-carrying circular loop of radius $2 \text{ cm}$ with unit normal $\hat{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$ is placed in a magnetic field $\vec{B} = B_0(3\hat{i} + 2\hat{j})$. If $B_0 = 4 \times 10^{-3} \text{ T}$ and current $I = 100\sqrt{2} \text{ A}$, the torque experienced by the loop is . . . . . . $\text{N}\cdot\text{m}$. $(\pi = 3.14)$

The torque required to hold a small circular coil of $10$ turns, area $1 \,mm^{2}$ and carrying a current of $\left(\frac{21}{44}\right) \,A$ in the middle of a long solenoid of $10^{3} \,turns/m$ carrying a current of $2.5 \,A$, with its axis perpendicular to the axis of the solenoid is

$(a)$ $A$ circular coil of $30$ turns and radius $8.0 \; cm$ carrying a current of $6.0 \; A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \; T$. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
$(b)$ Would your answer change,if the circular coil in $(a)$ were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

$A$ coil is placed in $y-z$ plane making an angle of $30^{\circ}$ with the $x$-axis. The current through the coil is $I$,and the number of turns is $N$. If a magnetic field of strength $B$ is applied in the positive $x$-direction,find the torque experienced by the coil (Radius of the coil is $R$).
$(N=100, I=1 \text{ A}, R=2 \text{ m}, B=\frac{1}{\pi} \text{ T})$