$A$ tangent galvanometer has a coil of $100$ turns and a radius of $20\, cm$. The horizontal component of the earth's magnetic field is $B_H = 3 \times 10^{-5}\, T$. Find the current which gives a deflection of $45^{\circ}$. (in $, A$)

$A$ 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

The time period in a vibration magnetometer will be infinity at

In a uniform magnetic field of $0.049 \ T$, a magnetic needle performs $20$ complete oscillations in $5 \ s$. The moment of inertia of the needle is $9.8 \times 10^{-5} \ kg \ m^2$. If the magnitude of the magnetic moment of the needle is $x \times 10^{-5} \ A \ m^2$, then the value of '$x$' is: (in $\pi^2$)

$A$ 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

When a freely suspended magnet is heated,its magnetic moment decreases by $19\%$. What is the percentage change in its time period?