A compass needle of oscillation magnetometer oscillates $20$ times per minute at a place $P$ of dip $30^{\circ}$. The number of oscillations per minute become $10$ at another place $Q$ of $60^{\circ} dip$. The ratio of the total magnetic field at the two places $\left(B_{Q}: B_{p}\right)$ is.

  • [JEE MAIN 2022]
  • A

    $\sqrt{3}: 4$

  • B

    $4: \sqrt{3}$

  • C

    $\sqrt{3}: 2$

  • D

    $2: \sqrt{3}$

Similar Questions

At a certain place the angle of dip is $30^{\circ}$ and the horizontal component of earth's magnetic field is $0.5 G$. The earth's total magnetic field (in $G$ ), at that certain place, is ................

  • [JEE MAIN 2022]

The earth's magnetic field was flipped by $180^{\circ}$ a million years ago. This flip was relatively rapid and took $10^5 \,yrs$. Then, the average change in orientation per year during the flip was closest to ............ $s$

  • [KVPY 2020]

The vertical component of the earth's magnetic field is zero at a place where the angle of dip is.....$^o$

Which is incorrect

A long straight horizontal cable carries a current of $2.5\;A$ in the direction $10^{\circ}$ south of west to $10^{\circ}$ north of east. The magnetic meridian of the place happens to be $10^{\circ}$ west of the geographic meridian. The earth's magnetic field at the location is $0.33\; G ,$ and the angle of $dip$ is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)