The square loop $ABCD$,carrying a current $i$,is placed in a uniform magnetic field $B$,as shown. The loop can rotate about the axis $XX'$. The plane of the loop makes an angle $\theta$ $(\theta < 90^o)$ with the direction of $B$. Through what angle will the loop rotate by itself before the torque on it becomes zero?

The given figure shows the north and south poles of a permanent magnet in which a coil of $n$ turns of cross-sectional area $A$ is placed. When a current $I$ is passed through the coil,the plane of the coil makes an angle $\theta$ with respect to the direction of the magnetic field $B$. If the plane of the magnetic field and the coil are horizontal and vertical respectively,the torque on the coil will be

An electron in the ground state of a hydrogen atom is revolving in an anticlockwise direction in a circular orbit of radius $r$. The atom is placed in a uniform magnetic field $B$ in such a way that the magnetic moment of the orbital electron makes an angle of $30^{\circ}$ with the magnetic field. The torque experienced by the orbital electron is

$A$ circular coil of radius $9 \, cm$ carrying a current of $2 \, A$ is free to rotate about an axis in its plane perpendicular to an external magnetic field of $\pi \times 10^{-2} \, T$. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a time period of $\frac{1}{3} \, s$. If the moment of inertia of the coil about its axis of rotation is $9 \times 10^{-5} \, kg \cdot m^2$, the number of turns of the coil 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?

$A$ 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 . . . . . . $\text{N}\cdot\text{m}$. $(\pi = 3.14)$