$A$ square coil of side $10 \ cm$ having $200$ turns is placed in a uniform magnetic field of $2 \ T$ such that the plane of the coil is in the direction of the magnetic field. If the current through the coil is $3 \ mA$, then the torque acting on the coil is

$A$ circular coil having $N$ turns and radius $R$ carrying current $I$ is held in the $z-x$ plane in a magnetic field $B\hat{k}$. The torque on the coil due to the magnetic field in $N-m$ is:

$A$ 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 ..... .

$A$ rectangular coil of $400$ turns and $10^{-2} \ m^2$ area, carrying a current of $0.5 \ A$ is placed in a uniform magnetic field of $1 \ T$ such that the plane of the coil makes an angle of $60^{\circ}$ with the direction of the magnetic field. The initial moment of force acting on the coil in $Nm$ is

$A$ circular coil has a moment of inertia of $0.8 \, kg \cdot m^2$ around any diameter and is carrying a current to produce a magnetic moment of $20 \, A \cdot m^2$. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 \, T$ is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60^{\circ}$ will be:

$A$ 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...... $Nm$.