The ratio of magnetic intensities for an axial point and a point on the broad side-on (equatorial) position at an equal distance $d$ from the centre of a short bar magnet is:

There are two current-carrying planar coils made from identical wires of length $L$. $C_1$ is circular (radius $R$) and $C_2$ is square (side $a$). They are constructed such that they have the same frequency of oscillation when placed in the same uniform magnetic field $B$ and carry the same current $I$. Find $a$ in terms of $R$.

The magnetic moment of a magnet is $20 \, C.G.S$ units. How much work is required to rotate it by $30^{\circ}$ from its equilibrium position in a magnetic field of $0.3 \, C.G.S$ units?

Two identical bar magnets of magnetic moment $M$ each,are placed along $X$ and $Y$-axes,respectively at a distance $d$ from the origin (as shown in the figure). The origin lies on the perpendicular bisector of the magnet placed on the $X$-axis and on the magnetic axis of the magnet placed on the $Y$-axis. If the magnitude of the total magnetic field at the origin is $B = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{d^3} \right]$,then the value of the constant $\alpha$ will be (given $d >> l$,where $l$ is the length of the bar magnets and the direction of $N$ to $S$ in the magnets is opposite with respect to each other).

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centre (figure). The magnitude of the magnetic field at a distance $d$ from the centre on the bisector of the right angle is

The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field $B$ is .........