$A$ 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 $, s$)

$A$ bar magnet is oscillating in the earth's magnetic field with time period $T.$ If its mass is increased four times,then its time period will be

$A$ bar magnet of length $10 \, cm$ is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \, cm$ from each pole. Given the horizontal component of the earth's magnetic field is $0.4 \, Gauss$, the pole strength of the magnet is: (in $ \, A-m$)

In a uniform magnetic field,a magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, A \cdot m^{2}$ and a moment of inertia $5 \times 10^{-6} \, kg \cdot m^{2}$. If it performs $10$ complete oscillations in $5 \, s$,then the magnitude of the magnetic field is $.... \, mT$. [ Take $\pi^{2} = 9.85$ ]

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 $turns$)

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