At two places $A$ and $B$,using a vibration magnetometer,a magnet vibrates in a horizontal plane. Its respective periodic times are $2 \ s$ and $3 \ s$. At these places,the Earth's horizontal components are $H_A$ and $H_B$ respectively. Then the ratio between $H_A$ and $H_B$ will be:

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

$A$ short magnet oscillates with a frequency $10 \, Hz$ where the horizontal component of Earth's magnetic field is $12 \, \mu T$. $A$ downward current of $15 \, A$ is established in a vertical wire placed $20 \, cm$ west of the magnet. The new frequency of the magnet is (in $ \, Hz$)

$A$ magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of Earth's magnetic field is $20 \mu T$. If the neutral point is found at a distance of $20 \ cm$ in the plane bisecting the dipole, then the magnetic moment of the dipole is (Assume $\mu_0 = 4 \pi \times 10^{-7} \text{ S.I. units}$) (in $\text{ A m}^2$)

$A$ bar magnet is placed in the magnetic meridian with its south pole pointing towards the geographic north. The neutral point is at a distance of $20 \, cm$ from the center of the magnet. If the horizontal component of the Earth's magnetic field is $B_H = 0.3 \times 10^{-4} \, Wb/m^2$,what is the magnetic moment of the magnet?

$A$ thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T'$,then the ratio $\frac{T'}{T}$ is