$A$ tangent galvanometer has $80$ turns of wire. The internal and external diameters of the coil are $19 \, cm$ and $21 \, cm$ respectively. The reduction factor of the galvanometer at a place where $H = 0.32 \, \text{oersted}$ will be $(1 \, \text{oersted} = 80 \, A/m)$.

$A$ 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 $\,A-m^2$)

$A$ bar magnet of length $14 \, cm$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. $A$ neutral point is obtained at a distance of $18 \, cm$ from the center of the magnet. If $B_{H} = 0.4 \, G$,the magnetic moment of the magnet is $\left(1 \, G = 10^{-4} \, T\right)$.

$A$ 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?

The period of oscillation of a vibration magnetometer depends on which of the following factors? (where $I$ is the moment of inertia of the magnet about the axis of suspension,$M$ is the magnetic moment of the magnet,and $H$ is the external magnetic field.)

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 $\pi^{2} = 9.85$ ]