$A$ tangent galvanometer shows a deflection of $30^{\circ}$ under the influence of the horizontal component of the Earth's magnetic field,which is $0.34 \times 10^{-4} \, T$. What is the magnetic field of the coil?

In a uniform magnetic field,a magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, A \cdot m^{2}$ and a moment of inertia $5 \times 10^{-6} \, kg \cdot m^{2}$. If it performs $10$ complete oscillations in $5 \, s$,then the magnitude of the magnetic field is $.... \, mT$. [ Take $\pi^{2} = 9.85$ ]

The periodic time of a magnet in a vibration magnetometer is $T_0$. If the magnet is replaced by another magnet whose moment of inertia is three times and magnetic moment is one-third of the original magnet, then the time period of the new magnet will be:

$A$ bar magnet is placed in the north-south direction with its north pole pointing towards the geographic north. In which direction from the center of the magnet will the points of zero magnetic field (neutral points) be located?

$A$ 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?

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: