$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

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Similar Questions

The time period of a magnet is $2 \, s$. The horizontal component of the Earth's magnetic field is $H$. When an external magnetic field $F$ is applied in the same direction,the new time period becomes $1 \, s$. What is the ratio $H/F$?

$A$ magnet makes $40$ oscillations per minute at a place having magnetic field intensity of $0.1 \times 10^{-5} \,T$. At another place,it takes $2.5 \,s$ to complete one vibration. The value of earth's horizontal field at that place is:

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

$A$ vibration magnetometer is used at two different places $A$ and $B$ on the earth. The time period of a magnet suspended freely in the magnetometer at $A$ is twice that at $B$. If the horizontal component of the earth's magnetic field at $B$ is $32 \times 10^{-6} \,T$, then its value at $A$ is

At a certain place, a magnet makes $30$ oscillations per minute. At another place where the magnetic field is double, its time period will be:

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