$A$ galvanometer coil has $500$ turns and each turn has an average area of $3 \times 10^{-4} \ m^{2}$. If a torque of $1.5 \ Nm$ is required to keep this coil parallel to the magnetic field when a current of $0.5 \ A$ is flowing through it,the strength of the magnetic field (in $T$) is:

$A$ triangular loop of side $l$ carries a current $I$. It is placed in a magnetic field $B$ such that the plane of the loop is in the direction of $B$. The torque on the loop is

$A$ single circular loop of radius $1.00 \, m$ carries a current of $10.0 \, mA$. It is placed in a uniform magnetic field of magnitude $0.500 \, T$ that is directed parallel to the plane of the loop as shown in the figure. The magnitude of the torque exerted on the loop by the magnetic field 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:

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 figure represents four positions of a current-carrying coil in a magnetic field directed towards the right. $\hat{n}$ represents the direction of the area vector of the coil. The correct order of potential energy is: