A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places (B_Q: B_P) is :

Solution

Physics sir · 4 steps

The needle swings only under the horizontal field, and frequency squared tracks it, not frequency itself.

1. What the needle feels. An oscillation magnetometer needle swings in a horizontal plane, so only the horizontal component B_H = B cos δ provides the restoring torque. Here δ is the angle of dip.

   f = 1/2π√(mB_H/I) ⇒ f² ∝ B cos δ

2. Take the ratio. The same needle is used at both places, so m and I cancel: (B_Q cos 60°)/(B_P cos 30°) = (10/20)² = 1/4.
3. Solve for the ratio. B_Q/B_P = 1/4×(cos 30°)/(cos 60°) = 1/4×(√3/2)/(1/2) = √3/4.
4. So. B_Q : B_P = √3 : 4, which is option (a).

The idea to keep. For any oscillation magnetometer question, write f² ∝ B_H = B cos δ. Square the frequency ratio first, then divide out the cosines of dip.

Answer from the NTA final answer key. Solution by PrepWiser.