The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20 \ cm$ from its center is $1.5 \times 10^{-5} \ T \cdot m$. The magnetic moment of the dipole is . . . . . . $A \cdot m^2$. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$)

$A$ magnetic dipole of magnetic moment $M$ is freely suspended in a magnetic field of induction $B$. The minimum and maximum values of the potential energy of the dipole,respectively,are

Two short bar magnets of magnetic moments $M$ and $\sqrt{3} M$ are joined like a cross (+). This cross is suspended from its centre with its plane horizontal in the earth's magnetic field. When the cross comes to equilibrium,the angle made by the weaker magnet with the earth's magnetic field $B_H$ is (in $^{\circ}$)

$A$ small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06\, T$ experiences a torque of $0.018\, Nm$. The minimum work required to rotate it from its stable to unstable equilibrium position is

$A$ 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: $B_H = 3 \times 10^{-5} \text{ T}$)

Two identical bar magnets,each of magnetic moment $M$,are kept perpendicular to each other at a certain distance. The magnetic induction at a point that is at the same distance $d$ from the center of both magnets is: (where $\mu_{0}$ is the permeability of free space)