The dipole moment of a short bar magnet is $1.25 \, A-m^2$. The magnetic field on its axis at a distance of $0.5 \, m$ from the centre of the magnet is:

$A$ magnetic dipole of magnetic moment $ 6 \times 10^{-2} \text{ A m}^2 $ and moment of inertia $ 12 \times 10^{-6} \text{ kg m}^2 $ performs oscillations in a magnetic field of $ 2 \times 10^{-2} \text{ T} $. The time taken by the dipole to complete $ 20 $ oscillations is (assume $ \pi \simeq 3 $). (in $\text{ s}$)

$A$ magnet of total magnetic moment $10^{-2} \hat{i} \text{ A} \cdot \text{m}^2$ is placed in a time-varying magnetic field $\vec{B} = B \cos(\omega t) \hat{i}$, where $B = 1 \text{ T}$ and $\omega = 0.125 \text{ rad/s}$. The work done to reverse the direction of the magnetic moment at $t = 1 \text{ s}$ is: (in $\text{ J}$)

$A$ 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 $.........\,J$.

Two short magnets $AB$ and $CD$ are in the $X-Y$ plane and are parallel to the $X$-axis. The coordinates of their centres are $(0,2)$ and $(2,0)$ respectively. The line joining the north-south poles of $CD$ is opposite to that of $AB$ and lies along the positive $X$-axis. The resultant magnetic field induction due to $AB$ and $CD$ at a point $P(2,2)$ is $100 \times 10^{-7} \ T$. When the poles of the magnet $CD$ are reversed,the resultant field induction is $50 \times 10^{-7} \ T$. The values of the magnetic moments of $AB$ and $CD$ (in $Am^2$) are:

$A$ 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 $J$)