The oscillation period of a freely suspended magnetic needle does NOT…

The correct answer is Option (4) → length of suspension fibre

A magnetic needle displaced through a small angle \( \theta \) in the horizontal component \( B_H \) of the earth’s field feels a restoring torque \( MB_H\theta \) and performs simple harmonic oscillations of period

    \( \displaystyle T = 2\pi\sqrt{\frac{I}{MB_H}} \)

Here I is the moment of inertia of the needle, which depends on its mass and dimensions, M is its dipole moment and \( B_H \) is the horizontal component of the earth’s field. All three options (1), (2) and (3) appear in this expression.

The needle is freely suspended, which means the restoring torque comes from the earth’s field alone and not from any twist in the fibre, so the length of the suspension fibre does not enter the period.

Note: A similar question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1).

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
The Dipole in a Uniform Magnetic Field – “… This is done by placing a small compass needle of known magnetic moment m and moment of inertia \( \mathcal{I} \) and allowing it to oscillate in the magnetic field.”

The Dipole in a Uniform Magnetic Field – “… This represents a simple harmonic motion. The square of the angular frequency is \( \omega^2 = mB/\mathcal{I} \) and the time period is,

    \( \displaystyle T = 2\pi\sqrt{\frac{\mathcal{I}}{mB}} \qquad (5.4) \)”