The period of oscillation of a magnet in a vibration magnetometer is $2 \, s$. The period of oscillation of a magnet whose magnetic moment is four times that of the first magnet is ... $s$.

To compare magnetic moments of two magnets by vibration magnetometer,the 'sum and difference method' is better because:

$A$ tangent galvanometer shows a deflection of $45^{\circ}$ when a $10 \, mA$ current passes through it. If the horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \, T$ and the radius of the coil is $10 \, cm$,find the number of turns in the coil.

$A$ magnet freely suspended in a vibration magnetometer makes $40$ oscillations per minute at place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$,then its value at $B$ is

Two small magnets have their masses and lengths in the ratio $1:2$. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations,the ratio of their time periods 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