The magnitude of the axial field due to a short bar magnet at a distance of $50 \,cm$ from its mid-point is (The magnetic moment of the bar magnet is $0.4 \,Am^2$)

$A$ short magnetic needle is placed in a magnetic field $B \hat{i}$. The needle is oriented in the direction $(\sqrt{3} \hat{i} + \hat{j})$. The needle experiences a torque of $0.06 \ N-m$. If the same magnetic needle is placed in a magnetic field $2B \hat{j}$ and is oriented in the direction $(\hat{i} + \sqrt{3} \hat{j})$,the torque experienced by it is: (in $N-m$)

$A$ bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

$A$ bar magnet of magnetic moment $1.5 \, J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment perpendicular to the field direction? (in $J$)

Two short bar magnets have magnetic moments $1.2 \text{ Am}^2$ and $1.0 \text{ Am}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres, such that their north poles point towards the geographic south. They share a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. Calculate the resultant horizontal magnetic induction at the midpoint of the line joining their centres. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2$)

$A$ bar magnet of magnetic moment $\vec{M}$ is placed in a magnetic field of induction $\vec{B}$. The torque exerted on it is . . . . . . .