$A$ vibration magnetometer works on the principle of

The magnetic needle of a tangent galvanometer is deflected at an angle $30^\circ$ due to a magnet. The horizontal component of Earth's magnetic field $0.34 \times 10^{-4} \, T$ is along the plane of the coil. The magnetic intensity is

$A$ bar magnet is placed in the north-south direction with its north pole pointing towards the geographic north. In which direction from the center of the magnet will the points of zero magnetic field (neutral points) be located?

The time period of a magnet is $T$. If it is divided into four equal parts along its axis and perpendicular to its axis as shown,then the time period for each part will be

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

$A$ tangent galvanometer has a coil with $50$ turns and a radius equal to $4$ $cm$. $A$ current of $0.1$ $A$ is passing through it. The plane of the coil is set parallel to the earth's magnetic meridian. If the value of the earth's horizontal component of the magnetic field is $7 \times 10^{-5}$ $T$ and $\mu_0 = 4\pi \times 10^{-7}$ $T \cdot m/A$,then the deflection in the galvanometer needle will be.....$^o$