$A$ tangent galvanometer is used to measure

$A$ 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 $s$)

$A$ bar magnet of length $10 \, cm$ 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 the earth's magnetic field is $0.4 \, Gauss$, the pole strength of the magnet is: (in $ \, A-m$)

$A$ thin magnetic iron rod of length $30 \ cm$ is suspended in a uniform magnetic field. Its time period of oscillation is $4 \ s$. It is broken into three equal parts. The time period in seconds of oscillation of one part when suspended in the same magnetic field is

The length $l$ 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 into three equal parts of length $l/3$ each. If these parts are placed on each other with their like poles together,then the time period of this combination is

$A$ magnet makes $40$ oscillations per minute at a place having a magnetic intensity of $0.1 \times 10^{-5} \, T$. At another place, it takes $2.5 \, s$ to complete one oscillation. The value of the Earth's horizontal magnetic field at that place is: