$A$ bar magnet of length $10 \, cm$ is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \, cm$ from each pole. Given the horizontal component of the earth's magnetic field is $0.4 \, Gauss$, the pole strength of the magnet is: (in $ \, A-m$)

In two separate experiments,the neutral points due to two small magnets are at a distance of $r$ and $2r$ in the broadside-on position. The ratio of their magnetic moments will be:

$A$ short magnet of magnetic moment $6.75 \, Am^2$ produces a neutral point on its axis. If the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, Wb/m^2$, then the distance of the neutral point from the center of the magnet is.....$cm$.

$A$ 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 $\text{ A m}^2$)

When a freely suspended magnet is heated,its magnetic moment decreases by $19\%$. What is the percentage change in its time period?

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 $A \cdot m^2$):