In the figure,the magnetic needle has a magnetic moment $m = 6.7 \times 10^{-2} \; A \cdot m^2$ and a moment of inertia $I = 7.5 \times 10^{-6} \; kg \cdot m^2$. It performs $10$ complete oscillations in $6.70 \; s$. What is the magnitude of the magnetic field $B$ (in $; T$)?

$A$ 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:

$A$ tangent galvanometer is used to measure

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \, s$. The magnet is cut along its length into three equal parts and these three parts are then placed on each other with their like poles together. The time period of this combination will be

$A$ bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

$A$ 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?