The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $36\%$. By doing this,the periodic time of the magnetometer will

$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$

In two separate experiments,the neutral points due to two small magnets are at a distance of $r$ and $2r$ in the broadside-on position. The ratio of their magnetic moments will be:

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

The magnetic needle of a tangent galvanometer is deflected at an angle $30^{\circ}$ due to a magnet. The horizontal component of the earth's magnetic field is $0.34 \times 10^{-4} \text{ T}$,which is along the plane of the coil. The magnetic field due to the magnet is:

$A$ 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