$A$ square coil of side $10 \; cm$ consists of $20$ turns and carries a current of $12 \; A$. The coil is suspended vertically and the normal to the plane of the coil makes an angle of $30^{\circ}$ with the direction of a uniform horizontal magnetic field of magnitude $0.80 \; T$. What is the magnitude of torque (in $N \; m$) experienced by the coil?

$A$ square loop of side $l$ is kept in a uniform magnetic field $B$ such that its plane makes an angle $\alpha$ with $\vec{B}$. The loop carries a current $i$. The torque experienced by the loop in this position is

$A$ $100$ turns coil shown in the figure carries a current of $2\, A$ in a magnetic field $B = 0.2\, Wb/m^2$. The torque acting on the coil is

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

$A$ current loop in a magnetic field:

$A$ coil in the shape of an equilateral triangle of side $10 \, cm$ lies in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of $20 \, mT$. The torque acting on the coil when a current of $0.2 \, A$ is passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{x} \times 10^{-5} \, Nm$. The value of $x$ is ..... .