The effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows:

In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?

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 $\cos 90^{\circ} = 0, \cos 60^{\circ} = 0.5$).

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 short bar magnets have their magnetic moments $1.2 \text{ A m}^2$ and $1.0 \text{ A m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards geographic south. They have a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. The resultant horizontal magnetic induction at the midpoint of the line joining their centers is $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

The magnitude of the axial field due to a short bar magnet at a distance of $50 \,cm$ from its mid-point is (The magnetic moment of the bar magnet is $0.4 \,Am^2$)