The magnetic field due to a small magnetic dipole of magnetic moment $M$,at a distance $r$ from the centre on the equatorial line is given by (in $M.K.S.$ system):

$A$ bar magnet with a magnetic moment $5.0\,Am^2$ is placed in a parallel position relative to a magnetic field of $0.4\,T$. The amount of work done in turning the magnet from a parallel to an antiparallel position relative to the field direction is $.........\,J$.

$A$ magnetic dipole is under the influence of two orthogonal magnetic fields, $B_1 = 0.5 \times 10^{-3} \ T$ and $B_2 = 0.866 \times 10^{-3} \ T$. If the dipole comes to stable equilibrium at an angle $\theta$ with respect to the $B_2$ field, then the value of $\theta$ is (in $^{\circ}$)

Two short magnetic dipoles $m_{1}$ and $m_{2}$, each having a magnetic moment of $1\, \text{Am}^{2}$, are placed at points $O$ and $P$ respectively. The distance between $O$ and $P$ is $1\, \text{m}$. The torque experienced by the magnetic dipole $m_{2}$ due to the presence of $m_{1}$ is ...... $\times 10^{-7}\, \text{Nm}$.

Two identical short bar magnets,each having a magnetic moment of $10 \, Am^2$,are arranged such that their axial lines are perpendicular to each other and their centres lie along the same straight line in a horizontal plane. If the distance between their centres is $0.2 \, m$,find the resultant magnetic induction at a point midway between them. (Given: $\mu_0 = 4\pi \times 10^{-7} \, Hm^{-1}$)

The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field $B$ is .........