Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on (axial) position is $9 \ G$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on (equatorial) position (in $G$)?

Two magnetic dipoles $X$ and $Y$ are placed at a separation $d$,with their axes perpendicular to each other. The dipole moment of $Y$ is twice that of $X$. $A$ particle of charge $q$ is passing through their mid-point $P$,at an angle $\theta = 45^\circ$ with the horizontal line as shown in the figure. What would be the magnitude of the force on the particle at that instant? ($d$ is much larger than the dimensions of the dipole)

If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field, then the torque acting on the magnet is:

$A$ short bar magnet of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. Find the value of $n$ (given $\cos 90^{\circ} = 0, \cos 60^{\circ} = 0.5$).

Points $A$ and $B$ are situated perpendicular to the axis of a small 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