$A$ bar magnet is placed in the magnetic meridian with its south pole pointing towards the geographic north. The neutral point is at a distance of $20 \, cm$ from the center of the magnet. If the horizontal component of the Earth's magnetic field is $B_H = 0.3 \times 10^{-4} \, Wb/m^2$,what is the magnetic moment of the magnet?

The period of oscillation of a magnet in a vibration magnetometer is $2 \, s$. The period of oscillation of a magnet whose magnetic moment is four times that of the first magnet is ... $s$.

$A$ magnet makes $40$ oscillations per minute at a place having magnetic field intensity of $0.1 \times 10^{-5} \,T$. At another place,it takes $2.5 \,s$ to complete one vibration. The value of earth's horizontal field at that place is:

$A$ 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.)

Before use,a vibration magnetometer should be set:

$A$ vibration magnetometer is used at two different places $A$ and $B$ on the earth. The time period of a magnet suspended freely in the magnetometer at $A$ is twice that at $B$. If the horizontal component of the earth's magnetic field at $B$ is $32 \times 10^{-6} \,T$, then its value at $A$ is