Points $A$ and $B$ are situated perpendicular to the axis of a $2 \ cm$ long 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:

$A$ 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).

$A$ bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it,then the angle by which it is to be rotated is....$^o$

The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?

$A$ magnetic dipole of moment $2.5 \text{ A m}^2$ is free to rotate about a vertical axis passing through its centre. It is released from the East-West direction. What is its kinetic energy at the moment it takes the North-South position (in $\mu\text{J}$)? (Given: $B_H = 3 \times 10^{-5} \text{ T}$)

$A$ bar magnet of magnetic moment $1.5 \, J \, T^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$.
$(a)$ What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: $(i)$ normal to the field direction,$(ii)$ opposite to the field direction?
$(b)$ What is the torque on the magnet in cases $(i)$ and $(ii)?$