Two bar magnets of the same mass,length,and breadth,but with magnetic moments $M$ and $2M$ respectively,have a time period of $3 \, s$ when placed in the same position (sum position). What will be the time period when they are placed in the different position (difference position)?

$A$ 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 $20 \, cm$ east of the magnet. Find the new time period of the oscillator. (in $, s$)

Twists of suspension fibre should be removed in a vibration magnetometer so that:

$A$ magnetic needle oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^{\circ}$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian,its period will be

$A$ 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

$A$ 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 $\left(1 \, G = 10^{-4} \, T\right)$.