The length of a magnet is large compared to its width and breadth. The time period of its oscillatio

Q: Does 'cut along its length' mean cutting it longitudinally?

Linguistically, yes. However, in the context of this specific classic JEE/AIEEE problem, the mathematical solution provided by examiners assumes a transverse cut (cutting the length into three shorter pieces). If cut longitudinally, the time period would actually remain unchanged!

Q: Why does the pole strength remain the same when cut transversely?

Pole strength depends on the cross-sectional area of the magnet. A transverse cut slices across the length, leaving the cross-sectional area intact, hence the pole strength remains unchanged.

Q: Why do we multiply the moment of inertia by 3 for the stacked pieces?

Moment of inertia is a scalar-like additive property for a given axis of rotation. Since the three pieces are stacked and rotate together about the same central vertical axis, their individual moments of inertia simply add up.

Q: Why does the magnetic moment of the stacked combination equal the original magnetic moment?

Each small piece has a magnetic moment of $M/3$. When stacked with like poles together, they act like parallel magnets. Their magnetic moments add up vectorially: $M/3 + M/3 + M/3 = M$.

Q: What would happen to the time period if the pieces were stacked with opposite poles facing each other?

If stacked with opposite poles together, their magnetic moments would cancel each other out. The net magnetic moment would be close to zero, making the time period approach infinity—meaning it wouldn't oscillate at all!

Question

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2	ext{ s}$. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be

Options:

(A) $2	ext{ s}$

(B) $\frac{2}{3}	ext{ s}$

(C) $2\sqrt{3}	ext{ s}$

(D) $\frac{2}{\sqrt{3}}	ext{ s}$

Sigma Percentile · Accepted Answer

### The Anatomy of a Vibration Magnetometer

Imagine a long, slender bar magnet suspended freely in a uniform magnetic field, gently oscillating back and forth. This setup is the heart of a vibration magnetometer. The time period of these oscillations is governed by a beautiful interplay between the magnet's rotational inertia and the magnetic torque acting upon it.

The master equation for this time period is:

$$T = 2\pi \sqrt{\frac{I}{MH}}$$

Here, **$I$** represents the moment of inertia of the magnet, **$M$** is its magnetic dipole moment, and **$H$** is the horizontal component of the Earth's magnetic field. We are given that the initial time period, $T$, is exactly $2	ext{ s}$.

### Dissecting the Original Magnet

Before we start slicing things up, let's establish the baseline properties of our original magnet. For a thin rectangular bar magnet of mass $m$ and length $L$, the moment of inertia about an axis passing through its center and perpendicular to its length is:

$$I = \frac{mL^2}{12}$$

The magnetic dipole moment, $M$, is defined as the product of its pole strength ($m_p$) and its magnetic length ($L$):

$$M = m_p 	imes L$$

### The Art of Cutting a Magnet

The problem states that the magnet is "cut along its length into three equal parts." This phrasing is notoriously ambiguous in physics literature. Does it mean a longitudinal cut (slicing it like a hotdog) or a transverse cut (chopping it into three shorter segments)?

Based on the mathematical framework of the solution, we must interpret this as a **transverse cut**. If we cut it longitudinally, the length would remain $L$, and stacking them back together would simply recreate the exact original magnet, leaving the time period unchanged at $2	ext{ s}$.

Therefore, we chop the magnet into three shorter pieces. For each individual piece:
- The new length is $L' = \frac{L}{3}$.
- The new mass is $m' = \frac{m}{3}$.
- The pole strength remains completely unchanged ($m_p$), because we haven't altered the cross-sectional area of the poles.

### Rebuilding the System: Moment of Inertia

Now, we take these three pieces and stack them on top of each other, ensuring their like poles (North on North, South on South) are aligned. Let's calculate the new moment of inertia, $I'$, of this stacked combination.

The moment of inertia of a single small piece is:

$$I_{	ext{piece}} = \frac{1}{12} \left(\frac{m}{3}ight) \left(\frac{L}{3}ight)^2$$

Since moment of inertia is an additive property for objects rotating about the same axis, and we have three identical pieces stacked together, we multiply this by 3:

$$I' = 3 	imes \left[ \frac{1}{12} \left(\frac{m}{3}ight) \left(\frac{L}{3}ight)^2 ight]$$

Let's simplify the algebra. One of the $3$s in the numerator cancels with the $m/3$, leaving us with:

$$I' = \frac{1}{12} m \left(\frac{L^2}{9}ight) = \frac{1}{9} \left( \frac{mL^2}{12} ight)$$

Notice that the term in the parentheses is our original moment of inertia, $I$. Thus:

$$I' = \frac{I}{9}$$

### Rebuilding the System: Magnetic Moment

Next, we evaluate the new magnetic moment, $M'$. The magnetic moment of a single small piece is its pole strength multiplied by its new length:

$$M_{	ext{piece}} = m_p 	imes \frac{L}{3}$$

Because the three pieces are stacked with their like poles aligned, they act as parallel magnets. Their magnetic moments add up vectorially:

$$M' = 3 	imes \left( m_p 	imes \frac{L}{3} ight) = m_p 	imes L$$

Fascinatingly, this brings us right back to where we started! The new magnetic moment is identical to the original:

$$M' = M$$

### The Final Oscillation

We now have all the pieces of the puzzle. Let's substitute our new $I'$ and $M'$ back into the master equation to find the new time period, $T'$:

$$T' = 2\pi \sqrt{\frac{I'}{M'H}}$$

$$T' = 2\pi \sqrt{\frac{I/9}{MH}}$$

We can pull the $1/9$ out of the square root as a $1/3$:

$$T' = \frac{1}{3} \left( 2\pi \sqrt{\frac{I}{MH}} ight)$$

The term in the parentheses is exactly our original time period, $T$. Therefore:

$$T' = \frac{T}{3}$$

Given that the original time period $T$ was $2	ext{ s}$, our final answer is:

**$$T' = \frac{2}{3}	ext{ s}$$**

By carefully tracking how mass, length, and pole strength scale when a physical object is divided and recombined, we've elegantly solved the problem.

Select Answer:

Visualized Solution

T = 2\pi \sqrt{\frac{I}{MH}}

I = \frac{mL^2}{12}

L' = \frac{L}{3}

I' = 3 \times I_{\text{piece}}

I' = \frac{I}{9}

M' = M

T' = 2\pi \sqrt{\frac{I'}{M'H}}

T' = \frac{2}{3}\text{ s}

The Sigma Insight: Bar Magnet and Magnetic Dipole

The Anatomy of a Vibration Magnetometer

Dissecting the Original Magnet

The Art of Cutting a Magnet

Rebuilding the System

Rebuilding the System

The Final Oscillation

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