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 $\frac{F}{H}$.

$A$ vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $2^{5/4} \ s$. One of the magnets is removed and if the other magnet oscillates in the same field,then the time period in seconds is:

$A$ bar-magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

$A$ current-carrying coil is placed with its axis perpendicular to the $N-S$ direction. Let the horizontal component of the Earth's magnetic field be $H_0$ and the magnetic field inside the loop be $H$. If a magnet is suspended inside the loop,it makes an angle $\theta$ with $H$. Then $\theta =$

$A$ very small magnet is placed in the magnetic meridian with its south pole pointing north. The null point is obtained $20 \, cm$ away from the centre of the magnet. If the earth's magnetic field (horizontal component) at this point is $0.3 \, gauss$,the magnetic moment of the magnet is:

$A$ tangent galvanometer has $80$ turns of wire. The internal and external diameters of the coil are $19 \, cm$ and $21 \, cm$ respectively. The reduction factor of the galvanometer at a place where $H = 0.32 \, \text{oersted}$ will be $(1 \, \text{oersted} = 80 \, A/m)$.