Two bar magnets oscillate in a horizontal plane in Earth's magnetic field with time periods of $3\,s$ and $4\,s$ respectively. If their moments of inertia are in the ratio of $3: 2$,then the ratio of their magnetic moments will be.

$A$ vibration magnetometer is placed at the south pole; then its time period will be

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 $\pi^{2} = 9.85$ ]

$A$ magnetic needle of negligible breadth and thickness compared to its length oscillates in a horizontal plane with a period $T$. What is the period of oscillation of each part obtained after breaking the magnet into $n$ equal parts perpendicular to its length?

$A$ dip circle lies initially in the magnetic meridian,it shows an angle of dip $\delta$ at a place. The dip circle is rotated through an angle $\alpha$ in the horizontal plane and then it shows an angle of dip $\delta^{\prime}$. Hence $\frac{\tan \delta^{\prime}}{\tan \delta}$ is

$A$ long vertical wire in which a current is flowing produces a neutral point with the Earth's magnetic field at a distance of $5 \ cm$ from the wire. If the horizontal component of the Earth's magnetic induction is $0.18 \ G$,then the current in the wire is: (in $A$)