Assertion: We cannot think of a magnetic field configuration with three poles.
Reason: $A$ bar magnet does exert a torque on itself due to its own field.

Two identical bar magnets with a length of $10 \, cm$ and a weight of $50 \, g$ are placed freely with their like poles facing each other in an inverted vertical glass tube. The upper magnet hangs in the air above the lower one such that the distance between the nearest poles of the magnets is $3 \, mm$. The pole strength of each magnet is approximately ....... $A \cdot m$.

For a ferromagnetic material,the relative permeability $(\mu_r)$ versus magnetic intensity $(H)$ has the following shape:

$A$ bar magnet is placed with its north pole pointing towards the geographic north. The magnetic field at a point $P$ on its equatorial line is zero. If the magnet is rotated by $90^\circ$,what will be the magnetic field at point $P$? The horizontal component of the Earth's magnetic field is $B_H$.

Two short bar magnets of length $1 \ cm$ each have magnetic moments $1.20 \ Am^2$ and $1.00 \ Am^2$ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0 \ cm$. The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to (Horizontal component of Earth's magnetic induction is $3.6 \times 10^{-5} \ Wb/m^2$)

Demagnetisation of magnets can be done by