$A$ tangent galvanometer has a coil of $100$ turns and a radius of $20\, cm$. The horizontal component of the earth's magnetic field is $B_H = 3 \times 10^{-5}\, T$. Find the current which gives a deflection of $45^{\circ}$. (in $, A$)

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \, s$. The magnet is cut along its length into three equal parts and these three parts are then placed on each other with their like poles together. The time period of this combination will be

$A$ bar magnet is in simple harmonic motion with time period $T$ in an Earth's magnetic field. If its mass is increased by $9$ times,the time period becomes:

$A$ tangent galvanometer has $80$ turns of wire. The internal and external diameters of the coil are $19 \, cm$ and $21 \, cm$ respectively. The reduction factor of the galvanometer at a place where $H = 0.32 \, \text{oersted}$ will be $(1 \, \text{oersted} = 80 \, A/m)$.

Two magnets are placed with their like poles together,and they perform $12$ oscillations in $1$ minute. When they are placed with their unlike poles together,they perform $4$ oscillations in $1$ minute. What is the ratio of the magnetic moments of the magnets?

$A$ short bar magnet having a magnetic moment $4 \text{ Am}^2$,placed in a vibrating magnetometer,vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg-m}^2$,the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field).