$A$ bar magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. If the bar magnet is turned around by $180^o$,where will the new null points (in $cm$) be located?

$A$ bar-magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

Two tangent galvanometer coils of the same radius are connected in series. The current flowing produces deflections of $60^o$ and $45^o$. The ratio of the number of turns in the coils is:

Two bar magnets $A$ and $B$ are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make $20$ oscillations per minute when the similar poles of $A$ and $B$ are on the same side, while they make $15$ oscillations per minute when their opposite poles lie on the same side. If $M_A$ and $M_B$ are the magnetic moments of $A$ and $B$ and if $M_A > M_B$, the ratio of $M_A$ and $M_B$ is

$A$ vibration magnetometer is used at two different places $A$ and $B$ on the earth. The time period of a magnet suspended freely in the magnetometer at $A$ is twice that at $B$. If the horizontal component of the earth's magnetic field at $B$ is $32 \times 10^{-6} \,T$, then its value at $A$ is