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Finding Moment of Inertia from Magnet Oscillations

Physics · Magnetism And Matter · NEET

A small magnet or needle in a uniform field B does simple harmonic oscillations with time period T = 2π√(I / mB), where I is its moment of inertia and m is its magnetic moment. Rearranging gives I = mBT² / 4π². Memory hook: it is the SAME shape as a pendulum, T = 2π√(inertia / restoring), where here "inertia" = I and "restoring" = mB.
Magnetic needle oscillating in uniform field BBN (m)SθT = 2π√(I / mB)→I = mBT² / 4π²
The needle (moment of inertia I, magnetic moment m) oscillates about the field B. Restoring torque mB sinθ makes it SHM, giving T = 2π√(I/mB); rearrange to find I = mBT²/4π².

Your doubts, answered

How do I get the time period T when the question gives 'N oscillations in t seconds'?

Time period is the time for ONE oscillation. So T = total time / number of oscillations = t / N. Example: 20 oscillations in 5 s gives T = 5/20 = 0.25 s. Never plug the raw 5 s into the formula — always convert to one-oscillation time first. This single step is where most marks are lost.

What is the difference between magnetic moment (m) and moment of inertia (I) here?

They are two different things that both appear in the formula. Magnetic moment m (unit A m²) measures how strong the magnet is — it decides the restoring torque mB that pulls the needle back. Moment of inertia I (unit kg m²) measures how the mass is spread out — it decides how sluggishly the needle turns. In T = 2π√(I/mB), I is on top (more I → slower → bigger T) and mB is on the bottom (stronger magnet or field → faster → smaller T).

How do I rearrange T = 2π√(I/mB) to solve for I?

Square both sides: T² = 4π² (I/mB). Then multiply both sides by mB and divide by 4π²: I = mBT² / 4π². To instead solve for the magnetic moment, rearrange the same equation to m = 4π²I / (BT²). Pick whichever unknown the question asks for.

Does this formula need the field to be uniform?

Yes. The restoring torque is τ = mB sinθ, and for small angles sinθ ≈ θ, giving SHM only when B is uniform over the needle. In the Earth's field experiment, B is the horizontal component B_H of Earth's field, and the same formula holds with B replaced by B_H.

⚠️ The NEET trap
✗ Using the total oscillation time as T, e.g. plugging T = 5 s (for 20 oscillations in 5 s) into I = mBT²/4π².
✓ T is the time for ONE oscillation: T = 5/20 = 0.25 s. Using 5 s instead of 0.25 s makes T² wrong by a factor of 400, so the whole answer is off by 400×.
🧠 'Oscillations per second' is frequency, not period. Period = time ÷ number of oscillations. Convert first, then square.

Real NEET questions

NEET 2024

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:

A · 128π²
B · 50π²
C · 1280π² ✓
D · 5π²
Solution: Step 1 — Time period: T = total time / number of oscillations = 5 / 20 = 0.25 s, so T² = 0.0625 s². Step 2 — Use T = 2π√(I/mB), rearranged for magnetic moment: m = 4π²I / (BT²). Step 3 — Substitute I = 9.8×10⁻⁶, B = 0.049, T² = 0.0625: m = 4π² × 9.8×10⁻⁶ / (0.049 × 0.0625). Step 4 — Denominator = 0.049 × 0.0625 = 3.0625×10⁻³. So 9.8×10⁻⁶ / 3.0625×10⁻³ = 3.2×10⁻³. Step 5 — m = 4π² × 3.2×10⁻³ = 12.8π² × 10⁻³ = 1280π² × 10⁻⁵ A m². Therefore x = 1280π². Answer: C.

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Frequently asked

What is the formula for the time period of an oscillating magnet?

T = 2π√(I / mB), where I is the moment of inertia (kg m²), m is the magnetic moment (A m²), and B is the uniform magnetic field (T). It is the magnetic version of the pendulum formula.

How do you find moment of inertia from oscillations?

Rearrange the period formula: I = mBT² / 4π². Measure the time for one oscillation T, and use the known m and B to compute I in kg m².

Why is the oscillating magnet's motion simple harmonic?

The restoring torque τ = mB sinθ ≈ mBθ for small angles. A torque proportional to the negative of the angular displacement is exactly the condition for SHM, giving the standard T = 2π√(I/mB).

What does B become in the Earth's magnetic field experiment?

B is replaced by the horizontal component of the Earth's field, B_H. A vibration magnetometer uses T = 2π√(I / mB_H) to compare or measure m and B_H.

Related concepts:
Oscillation of a Magnet: Time Period FormulaBent Bar Magnet: New Magnetic Moment CalculationTorque on a Magnetic Dipole in a Uniform Magnetic FieldWhat Is Magnetic Dipole Moment?Stable and Unstable Equilibrium of a Magnetic Dipole
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