$A$ magnet freely suspended in a vibration magnetometer makes $10$ oscillations per minute at a place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$,then its value at $B$ 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.....$^o$

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. What is the total magnetic field (in $G$) on the normal bisector of the magnet at the same distance as the null-point (i.e.,$14 \; cm$) from the centre of the magnet? (At null points,field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)

$A$ short magnetic needle is pivoted in a uniform magnetic field of induction $1 \text{ T}$. Now, simultaneously another magnetic field of induction $\sqrt{3} \text{ T}$ is applied at right angles to the first field; the needle deflects through an angle $\theta$ whose value is (in $^{\circ}$)

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

At a certain place, a magnet makes $30$ oscillations per minute. At another place where the magnetic field is double, its time period will be: