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 $60^{\circ}$. The value of $n$ is:

Two identical small bar magnets, each of dipole moment $3\sqrt{5} \text{ J/T}$, are placed at a center-to-center separation of $10 \text{ cm}$, with their axes perpendicular to each other as shown in the figure. The value of the magnetic field at the point $P$ midway between the magnets is $\alpha \times 10^{-3} \text{ T}$. The value of $\alpha$ is . . . . . . .

$A$ short bar magnet produces a magnetic field of $6.4 \times 10^{-5} \,T$ at a distance of $20 \,cm$ from the centre of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 \,cm$ from the centre of the magnet on the axis,is

$A$ magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^{\circ}$. One of the fields has a magnitude of $15 \, mT$. The dipole attains stable equilibrium at an angle of $30^{\circ}$ with this field. The magnitude of the other field (in $mT$) is close to:

Two equal bar magnets are kept as shown in the figure. The direction of the resultant magnetic field,indicated by the arrowhead at point $P$,is (approximately):

$A$ bar magnet of length $3 \, cm$ has points $A$ and $B$ along its axis at distances of $24 \, cm$ and $48 \, cm$ on the opposite sides. The ratio of the magnetic fields at these points is