Write the equation of the periodic time for an oscillating bar magnet in a uniform magnetic field.

With a standard rectangular bar magnet, the time period of a vibration magnetometer is $4 \,s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

Magnets $A$ and $B$ are geometrically similar,but the magnetic moment of $A$ is twice that of $B$. If $T_1$ and $T_2$ are the time periods of oscillation when their like poles and unlike poles are kept together respectively,then the ratio $\frac{T_1}{T_2}$ will be:

$A$ magnet is placed horizontally on the ground with its north pole pointing towards the geographic north pole of the Earth. Where is the neutral point obtained?

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. What is the total magnetic field (in $G$) on the normal bisector of the magnet at the same distance as the null-point (i.e.,$14 \; cm$) from the centre of the magnet? (At null points,field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)

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