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.....$A$

$A$ 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 is placed in the magnetic meridian with its south pole pointing towards the geographic north. The neutral point is at a distance of $20 \, cm$ from the center of the magnet. If the horizontal component of the Earth's magnetic field is $B_H = 0.3 \times 10^{-4} \, Wb/m^2$,what is the magnetic moment of the magnet?

$A$ magnetic needle has a magnetic moment of $5 \times 10^{-2} \text{ A m}^2$ and a moment of inertia of $8 \times 10^{-6} \text{ kg m}^2$. It has a period of oscillation of $2 \text{ s}$ in a magnetic field $\vec{B}$. The magnitude of the magnetic field is approximately:

$A$ tangent galvanometer shows a deflection of $30^{\circ}$ under the influence of the horizontal component of the Earth's magnetic field,which is $0.34 \times 10^{-4} \, T$. What is the magnetic field of the coil?

$A$ short bar magnet having a magnetic moment $4 \text{ Am}^2$,placed in a vibrating magnetometer,vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg-m}^2$,the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field).