$A$ 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 $, A$)

The period of oscillation of a magnet is $2 \, s$. When it is remagnetised so that the pole strength is $4$ times its original value,its new period will be......$s$.

With a standard rectangular bar magnet, the time period of a vibration magnetometer is $4 \,s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

$A$ magnet freely suspended in a vibration magnetometer makes $40$ oscillations per minute at 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 time period of a vibration magnetometer is $T_0$. Its magnet is replaced by another magnet whose moment of inertia is $3$ times and magnetic moment is $1/3$ of the initial magnet. The time period now will be

$A$ 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: