Give the stability position of a bar magnet for $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$.

$A$ magnet of magnetic moment $2 \,J \,T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \,T$. What is the net work done to bring the magnet normal to the magnetic field (in $\,J$)?

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

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:

$A$ planet has a magnetic dipole moment of $27 \times 10^{22} \ A \ m^2$. If the radius of the planet is $300 \ km$, what would be the magnetic field at its equator (in $T$)? (Take $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \ m/A$)

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centre (figure). The magnitude of the magnetic field at a distance $d$ from the centre on the bisector of the right angle is