$A$ vibration magnetometer works on the principle of

At neutral temperature,the thermoelectric power $\left( \frac{dE}{dT} \right)$ has the value:

Two different magnets are tied together and allowed to vibrate in a horizontal plane. When their like poles are joined,the time period of oscillation is $5 \, s$,and when their unlike poles are joined,the time period of oscillation is $15 \, s$. The ratio of their magnetic moments is

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

The time period of a thin bar magnet in Earth's magnetic field is $T$. If the magnet is cut into two equal parts perpendicular to its length,the time period of each part in the same field will be

$A$ 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. If the bar magnet is turned around by $180^o$,where will the new null points (in $cm$) be located?