If a brass bar is placed on a vibrating magnet, then its time period

$A$ 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 needle is made to vibrate in a uniform magnetic field $H$. Its time period is $T$. If it vibrates in a field of intensity $4H$,its time period will be:

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

At a certain place, a magnet makes $30$ oscillations per minute. At another place where the magnetic field is double, its time period will be:

$A$ bar magnet of length $14 \, cm$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. $A$ neutral point is obtained at a distance of $18 \, cm$ from the center of the magnet. If $B_{H} = 0.4 \, G$,the magnetic moment of the magnet is $\left(1 \, G = 10^{-4} \, T\right)$.