Physics · Magnetism And Matter · NEET
m is the MAGNETIC MOMENT (unit A·m²), not the mass. This confuses many students because the pendulum formula uses mass. Here the restoring effort comes from the magnetic torque τ = mB sinθ, so the magnetic moment m sits in the denominator. The mass of the magnet is already hidden inside I, the moment of inertia (kg·m²). Never plug the mass in kg into the m slot.
When a magnet is turned by a small angle θ from the field direction, the field applies a restoring torque τ = -mB sinθ. For small θ, sinθ ≈ θ, so τ ≈ -(mB)θ. This is the exact SHM condition: torque is proportional to angular displacement and points back to equilibrium. Comparing with I(d²θ/dt²) = -(mB)θ gives angular frequency ω = √(mB/I), so T = 2π√(I/mB).
For SMALL swings, no. The time period is independent of amplitude, exactly like a simple pendulum with small angle. This only holds while the swing is small enough that sinθ ≈ θ. For large swings the motion is still periodic but no longer perfect SHM, and the simple T = 2π√(I/mB) starts to fail. In NEET numericals you always assume small oscillations.
First get the time period: T = (total time)/(number of oscillations). Then rearrange the formula. Squaring T = 2π√(I/mB) gives T² = 4π²I/(mB), so m = 4π²I/(B T²). Plug in I, B, and T. This is exactly what NEET 2024 asked — the needle did 20 oscillations in 5 s, so T = 0.25 s, and m came out as 1280π² × 10⁻⁵ A·m².
They are the same physics. A vibration magnetometer is just the apparatus that lets a small magnet oscillate horizontally in the Earth's field (or an applied field) and measures T. The working formula is the same T = 2π√(I/mB). If the field is the Earth's horizontal component, you write B_H instead of B.
In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is 9.8 × 10⁻⁶ kg·m². If the magnetic moment of the needle is x × 10⁻⁵ A·m², the value of x is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2π√(I/mB), where I is the moment of inertia (kg·m²), m is the magnetic moment (A·m²), and B is the magnetic field (T). It describes small-angle SHM of a magnet free to rotate in the field.
I is in kg·m², m (magnetic moment) is in A·m², and B is in tesla (T). These combine so that I/mB has units of s², and the square root gives seconds — the correct unit for T.
A stronger field B gives a larger restoring torque, so the magnet swings faster and the time period DECREASES. Since T ∝ 1/√B, doubling B reduces T by a factor of √2.
Cutting along the length halves both the magnetic moment and the moment of inertia proportionally, so T can stay similar; but cutting across the length changes both differently. Always recompute I and m from the new geometry rather than guessing — this is a common exam twist covered under moment of inertia of an oscillating needle.
