In a tangent galvanometer,a current of $0.1 \, A$ produces a deflection of $30^\circ$. The current required to produce a deflection of $60^\circ$ is ...... $A$.

$A$ magnet makes $40$ oscillations per minute at a place having magnetic field intensity of $0.1 \times 10^{-5} \,T$. At another place,it takes $2.5 \,s$ to complete one vibration. The value of earth's horizontal field at that place is:

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

$A$ short magnet oscillates with a frequency $10 \, Hz$ where the horizontal component of Earth's magnetic field is $12 \, \mu T$. $A$ downward current of $15 \, A$ is established in a vertical wire placed $20 \, cm$ west of the magnet. The new frequency of the magnet is (in $ \, Hz$)

$A$ 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).

$A$ 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?