$A$ magnet is suspended horizontally in the earth's magnetic field. When it is displaced and then released,it oscillates in a horizontal plane with a period $T$. If a piece of wood of the same moment of inertia (about the axis of rotation) as the magnet is attached to the magnet,what would the new period of oscillation of the system become?

$A$ 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 has a coil of $100$ turns and a radius of $20\, cm$. The horizontal component of the earth's magnetic field is $B_H = 3 \times 10^{-5}\, T$. Find the current which gives a deflection of $45^{\circ}$. (in $, A$)

If a brass bar is placed on a vibrating magnet, then its time period

$A$ short bar magnet is placed in the magnetic meridian of the earth with its north pole pointing towards the geographic north. Neutral points are found at a distance of $30 \, cm$ from the magnet on the East-West line, drawn through the center of the magnet. The magnetic moment of the magnet in $A \cdot m^2$ is close to: (Given $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$ and $B_H = 3.6 \times 10^{-5} \, T$)

$A$ neutral point is obtained at the centre of a vertical circular coil carrying current. The angle between the plane of the coil and the magnetic meridian is.......$^o$