$A$ circular coil of $20$ turns and radius $10\, cm$ is placed in a uniform magnetic field of $0.10\, T$ normal to the plane of the coil. If the current in the coil is $5\, A$,then the torque acting on the coil will be...... $Nm$.

$A$ wire of length $L$ carries a current $I$. If the wire is turned into a square coil of single turn,the maximum magnitude of the torque in a given magnetic field $\overrightarrow{B}$ is

$A$ circular coil of radius $R$ and $N$ turns has negligible resistance. As shown in the schematic figure,its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge $Q$ through a switch. The coil is in a horizontal uniform magnetic field $B_0$ parallel to the plane of the coil. When the switch is closed,the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully,the magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time):

$(a)$ $A$ 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 rectangular coil is placed in a uniform magnetic field. In which orientation will the coil not tend to rotate?

$A$ 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 $\vec{B} = B_0(3\hat{i} + 2\hat{k})$. If $B_0 = 4 \times 10^{-3} \text{ T}$ and current $I = 100\sqrt{2} \text{ A}$, the torque experienced by the loop is . . . . . . $\text{N}\cdot\text{m}$. $(\pi = 3.14)$