$A$ 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:

Two mutually perpendicular uniform magnetic fields $F$ and $H$ act on a magnetic needle. The needle makes an angle of $60^\circ$ with the field $F$. Find the ratio $\frac{F}{H}$.

$A$ 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

The deflection produced in a tangent galvanometer,whose coil has a resistance of $9 \Omega$,is $30^{\circ}$. The potential difference across the coil is $4.5 \ V$. If the number of turns in the coil is $10$,the radius of the coil is (Given,$B_{H} = 3.14 \times 10^{-5} \ T$)

$A$ 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.)

$A$ 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?