$A$ magnetic needle suspended horizontally by an unspun silk fibre,oscillates in the horizontal plane because of the restoring force originating mainly from

Magnets $A$ and $B$ are geometrically similar,but the magnetic moment of $A$ is twice that of $B$. If $T_1$ and $T_2$ are the time periods of oscillation when their like poles and unlike poles are kept together respectively,then the ratio $\frac{T_1}{T_2}$ will be:

The magnetic needle of a vibration magnetometer makes $12$ oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line,it makes $15$ oscillations per minute. If the poles of the bar magnet are interchanged,the number of oscillations it makes per minute is

$A$ 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

At neutral temperature,the thermoelectric power $\left( \frac{dE}{dT} \right)$ has the value:

$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$