If $L$ is the length of a bar magnet,then the separation between the two poles is nearly

$A$ wire of magnetic moment $M$ is bent to form a circular arc which subtends an angle $90^o$ at the center. Then,its new magnetic moment is:

The axial field $(B_{A})$ and the equatorial field $(B_{E})$ due to a short bar magnet at equal distances are related as:

Inside a bar magnet,the magnetic lines of force

The magnetic field at a point $x$ on the axis of a small bar magnet is equal to the field at a point $y$ on the equator of the same magnet. The ratio of the distances of $x$ and $y$ from the centre of the magnet is

$A$ steel wire of length $\ell$ has a magnetic moment $M$. It is then bent into a semicircular arc. The new magnetic moment is