$A$ 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....$s$

Two magnets $A$ and $B$ are identical in mass,length,and breadth but have different magnetic moments. In a vibration magnetometer,if the time period of $B$ is twice the time period of $A$,then the ratio of the magnetic moments $M_A/M_B$ of the magnets will be:

Two tangent galvanometers $A$ and $B$ have coils of radii $8 \text{ cm}$ and $16 \text{ cm}$ respectively and have a resistance of $8 \Omega$ each. They are connected in parallel with a cell of emf $4 \text{ V}$ and negligible internal resistance. The deflections produced in the tangent galvanometers $A$ and $B$ are $30^{\circ}$ and $60^{\circ}$, respectively. If $A$ has $2$ turns, then $B$ must have: (in $turns$)

The time period of oscillation of a magnet is $2 \ s$. When it is remagnetised so that its pole strength is $4$ times,its period will be ..... $s$.

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

The radius of the coil of a Tangent galvanometer,which has $10$ turns,is $0.1\,m$. The current required to produce a deflection of $60^{\circ}$ $(B_H = 4 \times 10^{-5}\,T)$ is.....$A$