$A$ magnetic needle of negligible breadth and thickness compared to its length oscillates in a horizontal plane with a period $T$. What is the period of oscillation of each part obtained after breaking the magnet into $n$ equal parts perpendicular to its length?

The time period of a bar magnet suspended horizontally in the earth's magnetic field and allowed to oscillate:

$A$ magnetic needle has a magnetic moment of $5 \times 10^{-2} \text{ A m}^2$ and a moment of inertia of $8 \times 10^{-6} \text{ kg m}^2$. It has a period of oscillation of $2 \text{ s}$ in a magnetic field $\vec{B}$. The magnitude of the magnetic field is approximately:

The magnetic moments of two bar magnets of same size are in the ratio $1:2$. When they are placed one over the other with their similar poles together,their period of oscillation in a magnetic field is $3 \, s$. If one of the magnets is reversed,then the period of oscillation in the same field will be ......... $s$.

The magnetic needle of a vibration magnetometer makes $12$ oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line,it makes $15$ oscillations per minute. If the poles of the bar magnet are interchanged,the number of oscillations it makes per minute is

$A$ short bar magnet having magnetic moment $4 \text{ Am}^2$,placed in a vibrating magnetometer,vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg m}^2$,the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field.)