The length $l$ of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \ s$. The magnet is cut into three equal parts of length $l/3$ each. If these parts are placed on each other with their like poles together,then the time period of this combination 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

$A$ 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:

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. What is the total magnetic field (in $G$) on the normal bisector of the magnet at the same distance as the null-point (i.e.,$14 \; cm$) from the centre of the magnet? (At null points,field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)

In a uniform magnetic field,a magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, A \cdot m^{2}$ and a moment of inertia $5 \times 10^{-6} \, kg \cdot m^{2}$. If it performs $10$ complete oscillations in $5 \, s$,then the magnitude of the magnetic field is $.... \, mT$. [ Take $\pi^{2} = 9.85$ ]

$A$ certain amount of current when flowing in a properly set tangent galvanometer,produces a deflection of $45^{\circ}$. If the current is reduced by a factor of $\sqrt{3}$,the deflection would: