$A$ 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 of magnetic moments $m$ each are arranged at the opposite corners of a square of side $d$ such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

$A$ magnetic dipole experiences a torque of $80 \sqrt{3} \ N \ m$ when placed in a uniform magnetic field in such a way that the dipole moment makes an angle of $60^{\circ}$ with the magnetic field. The potential energy of the dipole is

$A$ bar magnet of length $10 \text{ cm}$ and having the pole strength equal to $10^{-3} \text{ A-m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \text{ T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is

$A$ magnetic dipole is under the influence of two orthogonal magnetic fields, $B_1 = 0.5 \times 10^{-3} \ T$ and $B_2 = 0.866 \times 10^{-3} \ T$. If the dipole comes to stable equilibrium at an angle $\theta$ with respect to the $B_2$ field, then the value of $\theta$ is (in $^{\circ}$)

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: