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

$A$ bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along the positive $X$-axis. Likewise, a second bar magnet of magnetic moment $M$ is placed at a distance $2D$ on the positive $Y$-axis and perpendicular to it as shown in the figure. The magnitude of the magnetic field at the origin is $|\vec{B}| = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{D^3} \right]$. The value of $\alpha$ must be (Assume $D \gg l$, where $l$ is the length of the magnets).

If a magnet of length $10 \, cm$ and pole strength $40 \, A-m$ is placed at an angle of $45^\circ$ in a uniform magnetic field of intensity $2 \times 10^{-4} \, T$,the torque acting on it is:

Points $A$ and $B$ are situated perpendicular to the axis of a small bar magnet at large distances $x$ and $3x$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to

$A$ negative charge is given to a nonconducting loop and the loop is rotated in the plane of paper about its centre as shown in the figure. The magnetic field produced by the ring affects a small magnet placed above the ring in the same plane:

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)