Two tangent galvanometers having coils of the same radius are connected in series. $A$ current flowing in them produces deflections of $60^{\circ}$ and $45^{\circ}$ respectively. The ratio of the number of turns in the coils is

$A$ short bar magnet with its north pole facing north forms a neutral point at $P$ in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane,the net magnetic induction at $P$ is (Horizontal component of earth's magnetic field = $B_H$)

When two magnets of equal pole strength and length are placed with their dissimilar poles on the same side,what will be their time period?

$A$ deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling of the deflection is $T$. When the magnet is removed, the period of oscillation of the needle is $T_0$ before settling to $0^{\circ}-0^{\circ}$. If the earth's horizontal magnetic field is $B_H$, the relation between $T$ and $T_0$ is

$A$ short magnet oscillates with a time period $0.1 \, s$ at a place where the horizontal magnetic field is $24 \, \mu T$. $A$ downward current of $18 \, A$ is established in a vertical wire $20 \, cm$ east of the magnet. Find the new time period of the oscillator. (in $, s$)

The period of oscillation of a magnet in a vibration magnetometer is $2 \, s$. The period of oscillation of a magnet whose magnetic moment is four times that of the first magnet is ... $s$.