$A$ bar magnet of $10 \,cm$ length is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \,cm$ from each pole. Given the horizontal component of Earth's magnetic field is $0.4 \,Gauss$, the pole strength of the magnet is: (in $\,A-m$)

$A$ 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 $(1 \, \text{oersted} = 80 \, A/m)$.

$A$ short bar magnet with its north pole facing north forms a neutral point at $P$ in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane,the net magnetic induction at $P$ is (Horizontal component of earth's magnetic field = $B_H$)

$A$ deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling of the deflection is $T$. When the magnet is removed, the period of oscillation of the needle is $T_0$ before settling to $0^{\circ}-0^{\circ}$. If the earth's horizontal magnetic field is $B_H$, the relation between $T$ and $T_0$ is

With a standard rectangular bar magnet, the time period of a vibration magnetometer is $4 \,s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

$A$ short magnet oscillates with a time period $0.1 \, s$ at a place where the horizontal magnetic field is $24 \, \mu T$. $A$ downward current of $18 \, A$ is established in a vertical wire kept at a distance of $20 \, cm$ east of the magnet. The new time period of oscillations of the magnet is (in $s$)