$A$ vibration magnetometer works on the principle of

  • A
    Torque acting on the bar magnet
  • B
    Force acting on the bar magnet
  • C
    Both the force and the torque acting on the bar magnet
  • D
    None of these

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In a vibration magnetometer,the time period of a bar magnet oscillating in the horizontal component of the Earth's magnetic field is $2 \ s$. When a magnet is brought near and parallel to it,the time period reduces to $1 \ s$. The ratio $H/F$ of the horizontal component $H$ and the field $F$ due to the magnet will be:

The tangent galvanometer,when connected in series with a standard resistance,can be used as:

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 $; T$)?

$A$ magnet makes $20$ and $15$ oscillations in $1$ minute at two places where the angles of dip are $30^{\circ}$ and $60^{\circ}$ respectively. What is the ratio of the total magnetic fields at these two places?

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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

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