$A$ bar magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

$A$ bar magnet is placed on a table along the east-west direction in the Earth's magnetic field. The number of neutral points,where the magnetic field is zero,is

$A$ 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:

$A$ magnet makes $20$ and $15$ oscillations in $1$ minute at two places where the angles of dip are $30^{\circ}$ and $60^{\circ}$ respectively. What is the ratio of the total magnetic fields at these two places?

In a vibration magnetometer,the time period of a bar magnet oscillating in the horizontal component of the Earth's magnetic field is $2 \ s$. When a magnet is brought near and parallel to it,the time period reduces to $1 \ s$. The ratio $H/F$ of the horizontal component $H$ and the field $F$ due to the magnet will be:

The time period of oscillation of a freely suspended bar magnet with usual notations is given by