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:

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:

$A$ magnetic needle of magnetic moment $6 \times 10^{-2} \text{ A m}^2$ and moment of inertia $9.6 \times 10^{-5} \text{ kg m}^2$ performs simple harmonic motion in a magnetic field of $0.01 \text{ T}$. The time taken to complete $10$ oscillations is (Take $\pi = 3.14$): (in $\text{ s}$)

The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be:

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 $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

Points $A$ and $B$ are situated along the extended axis of a $2 \ cm$ long bar magnet at distances $x$ and $2x$ $cm$ respectively from the nearer pole. The ratio of the magnetic field at $A$ and $B$ will be: