The moment of inertia of a magnetic needle is $40 \, g \cdot cm^2$ and it has a time period of $3 \, s$ in the Earth's horizontal magnetic field of $3.6 \times 10^{-5} \, Wb/m^2$. Its magnetic moment will be (in $A \cdot m^2$):

$A$ bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

The magnetic moments of two bar magnets of same size are in the ratio $1:2$. When they are placed one over the other with their similar poles together,their period of oscillation in a magnetic field is $3 \, s$. If one of the magnets is reversed,then the period of oscillation in the same field will be ......... $s$.

If magnetic lines of force are drawn by keeping a magnet vertical,then the number of neutral points will be

$A$ 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 kept at a distance of $20 \, cm$ east of the magnet. The new time period of oscillations of the magnet is (in $s$)

$A$ tangent galvanometer is used to measure: