The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field $B$ is .........

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:

$A$ bar magnet of magnetic moment $\overrightarrow{M}$ is placed in a magnetic field of induction $\overrightarrow{B}$. The torque exerted on it is

The magnetic moment of a current-carrying loop is $2.1 \times 10^{-25} \text{ A m}^2$. The magnetic field at a point on its axis at a distance of $1 \text{ Å}$ is:

$A$ 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 magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.