Q.A bar magnet of magnetic moment $m$ and moment of inertia $I$ (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let $T$ be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field $B$. What would be the similar period $T'$ for each piece?

Concept: Magnetic Poles – When a bar magnet is cut perpendicular to its length, each piece becomes a smaller magnet with half the length, half the magnetic moment, and a different moment of inertia.

Reasoning:

1. Original magnet: magnetic moment $m$, length $L$, moment of inertia $I = \frac{1}{12} M L^2$ (about centre, perpendicular to length). Period $T = 2\pi \sqrt{\frac{I}{mB}}$.

2. After cutting into two equal pieces perpendicular to length, each piece has:

   • Mass $M' = M/2$, length $L' = L/2$.
   • Magnetic moment $m' = m/2$ (since pole strength unchanged, length halved). …

The key insight here is that magnetic moment and moment of inertia both change when you cut the magnet, and they change by different factors. The period of a magnetic dipole oscillating in a uniform field depends on the ratio $I/m$, so we need to track how that ratio transforms.

Why this approach works

A bar magnet oscillating in a uniform magnetic field behaves like a torsional pendulum. The restoring torque is $\tau = -mB\sin\theta \approx -mB\theta$ for small angles, giving angular SHM with period:

Since $B$ is the same for both the original and the pieces, the period ratio depends only on how $I$ and $m$ change when we cut the magnet.

Step-by-step reasoning

1. What changes when you cut perpendicular to length?

Cutting perpendicular to length means you slice the magnet into two equal halves along its short axis — like cutting a pencil into two shorter pencils. Each piece has:

• Half the original length: $L' = L/2$
• Same cross-sectional area (the cut doesn't change the width or thickness)
• Same material, so same magnetization per unit volume

2. Magnetic moment halves

Magnetic moment $m$ of a bar magnet equals pole strength $p$ times length $L$: $m = pL$.

When you cut the magnet, each piece still has the same pole strength $p$ at its ends (pole strength depends on the cross-section and material, which haven't changed). But the length is halved, so:

3. Moment of inertia changes by a factor of 8

The original magnet has moment of inertia $I$ about its centre, perpendicular to length. For a uniform rod of mass $M$ and length $L$:

Each piece has: …

Method: Finding How the Oscillation Period of a Magnet Scales When It Is Cut or Resized

Use this whenever a bar magnet (or any oscillating dipole) is physically altered — cut, scaled, or its mass/length changed — and you need the new period in terms of the old one.

Steps

Step 1: Write the general period formula and identify its two "changeable" ingredients

A magnetic dipole oscillating in a uniform field behaves as angular SHM with restoring torque $\tau=-mB\sin\theta\approx-mB\theta$, giving

The field $B$ is fixed by the experiment (unchanged), so only $I$ (moment of inertia) and $m$ (magnetic moment) change when the magnet's geometry changes — track how each one individually scales.

Step 2: Determine how the magnetic moment scales

Magnetic moment is $m=p\cdot L$ (pole strength times length). Pole strength $p$ is an intrinsic property of the end-face material/cross-section and does not change just because the magnet gets shorter — only $L$ changes. So if the length scales by a factor $k$ (e.g. $k=\tfrac12$ for cutting in half), the new moment is

Step 3: Determine how the moment of inertia scales

For a uniform rod about its centre, perpendicular to its length, $I=\tfrac{1}{12}ML^2$. If both the mass scales by $k$ (same material removed proportionally) and the length scales by $k$,

$I' = \frac{1}{12}(kM)(kL)^2 = k^3\left(\frac{1}{12}ML^2\right) = k^3 I.$ …