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

In the figure,the magnetic needle has a magnetic moment $m = 6.7 \times 10^{-2} \; A \cdot m^2$ and a moment of inertia $I = 7.5 \times 10^{-6} \; kg \cdot m^2$. It performs $10$ complete oscillations in $6.70 \; s$. What is the magnitude of the magnetic field $B$ (in $; T$)?

$A$ 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 needle of a deflection galvanometer shows a deflection of $60^\circ$ due to a short bar magnet at a certain distance in $\tan A$ position. If the distance is doubled,the deflection is

With a standard rectangular bar magnet of length $l$, breadth $b$ $(b \ll l)$ and magnetic moment $M$, the time period of the magnet in a vibration magnetometer is $4 \ s$. If the magnet is cut normal to its length into four equal pieces, the time period (in seconds) with one of the pieces is

The error in measuring the current with a tangent galvanometer is minimum when the deflection is about $45^o$. (in $^o$)