$A$ bar magnet is placed on a table along the east-west direction in the Earth's magnetic field. The number of neutral points,where the magnetic field is zero,is

The time period of a bar magnet suspended horizontally in the earth's magnetic field and allowed to oscillate:

$A$ magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of Earth's magnetic field is $20 \mu T$. If the neutral point is found at a distance of $20 \ cm$ in the plane bisecting the dipole, then the magnetic moment of the dipole is (Assume $\mu_0 = 4 \pi \times 10^{-7} \text{ S.I. units}$) (in $\text{ A m}^2$)

$A$ 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 $20 \, cm$ east of the magnet. Find the new time period of the oscillator. (in $, s$)

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

$A$ 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?