$A$ short bar magnet experiences a torque of magnitude $0.64 \ J$ when it is placed in a uniform magnetic field of $0.32 \ T$. The magnetic moment of the magnet is . . . . . .

The intensity of the magnetic field is $H$ and the magnetic moment of the magnet is $M$. The maximum potential energy is $..... MH$.

$A$ bar magnet of length $3 \, cm$ has points $A$ and $B$ along its axis at distances of $24 \, cm$ and $48 \, cm$ on the opposite sides. The ratio of the magnetic fields at these points is

Two short bar magnets, each with a magnetic moment of $9 \text{ Am}^2$, are placed such that one is at $x = -3 \text{ cm}$ and the other is at $y = -3 \text{ cm}$. If their magnetic moments are directed along the positive and negative $X$-directions respectively, then the resultant magnetic field at the origin is: (in $\text{ T}$)

$A$ magnet is parallel to a uniform magnetic field. If it is rotated by $60^\circ$,the work done is $0.8\, J$. How much work is done in moving it $30^\circ$ further?

$A$ magnetic needle of magnetic moment $6 \times 10^{-2} \text{ A m}^2$ and moment of inertia $9.6 \times 10^{-5} \text{ kg m}^2$ performs simple harmonic motion in a magnetic field of $0.01 \text{ T}$. The time taken to complete $10$ oscillations is (Take $\pi = 3.14$): (in $\text{ s}$)