The time period of oscillation of a magnet in a vibration magnetometer is $1.5 \ s$. The time period of oscillation of another magnet similar in size,shape,and mass but having one-fourth magnetic moment than that of the first magnet,oscillating at the same place,will be.....$s$.

$A$ tangent galvanometer shows a deflection of $45^{\circ}$ when a $10 \, mA$ current passes through it. If the horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \, T$ and the radius of the coil is $10 \, cm$,find the number of turns in the coil.

$A$ 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

$A$ 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 of a thin bar magnet in Earth's magnetic field is $T$. If the magnet is cut into two equal parts perpendicular to its length,the time period of each part in the same field will be

$A$ bar magnet is in simple harmonic motion with time period $T$ in an Earth's magnetic field. If its mass is increased by $9$ times,the time period becomes: