$A$ 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 $\,A-m^2$)

$A$ magnet freely suspended in a vibration magnetometer makes $40$ oscillations per minute at place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$, then its value at $B$ is

How many neutral points will be obtained when a bar magnet is kept with its magnetic moment parallel to the Earth's magnetic field?

If a brass bar is placed on a vibrating magnet, then its time period

$A$ short bar magnet is placed in the magnetic meridian of the earth with its north pole pointing towards the geographic north. Neutral points are found at a distance of $30 \, cm$ from the magnet on the East-West line, drawn through the center of the magnet. The magnetic moment of the magnet in $A \cdot m^2$ is close to: (Given $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$ and $B_H = 3.6 \times 10^{-5} \, T$)

At a certain place,a magnet makes $30$ oscillations per minute. At another place where the magnetic field is double,its time period will be: