A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now, it is broken

Q: Why doesn't the pole strength change when we cut the magnet?

Pole strength depends on the cross-sectional area of the magnet. Since we cut it transversely (perpendicular to its length), the cross-sectional area remains the same, so the pole strength is unchanged.

Q: Why is the new moment of inertia $I/8$ and not $I/2$?

Moment of inertia depends on both mass and the square of the length ($I \propto ml^2$). Since mass becomes half ($1/2$) and length becomes half ($(1/2)^2 = 1/4$), the total factor is $(1/2) \times (1/4) = 1/8$.

Q: What would happen if we cut the magnet longitudinally (along its length)?

If cut longitudinally, the length remains $l$, but the mass becomes $m/2$ and the pole strength becomes $m_p/2$. The new moment of inertia would be $I/2$ and the new magnetic moment would be $M/2$. The time period $T$ would actually remain unchanged!

Q: Does the magnetic field $B$ affect the ratio $T'/T$?

No, because the external magnetic field $B$ is constant and uniform. It appears in both $T$ and $T'$ and cancels out when we take the ratio.

Q: Why do we use the formula $ml^2/12$ for the moment of inertia?

A freely suspended magnet oscillates about an axis passing through its center of mass and perpendicular to its length. For a thin rectangular bar (or rod), the moment of inertia about this central axis is standardly derived as $ml^2/12$.

Question

A thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now, it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T'$, the ratio $T'/T$ is

Options:

(A) $\frac{1}{2\sqrt{2}}$

(B) $\frac{1}{2}$

(C) $2$

(D) $\frac{1}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a bar magnet suspended freely in a uniform magnetic field. When disturbed, it starts oscillating.

The time period of this oscillation is a classic result in magnetism, given by $T = 2\pi \sqrt{\frac{I}{MB}}$.

Here, $I$ is the moment of inertia of the magnet, $M$ is its magnetic dipole moment, and $B$ is the external magnetic field.

### The Anatomy of the Original Magnet

Let's break down the parameters of our original magnet.

Suppose it has a mass $m$ and a length $l$.

Since it's a rectangular bar magnet, its moment of inertia about the axis passing through its center is $I = \frac{ml^2}{12}$.

Its magnetic moment $M$ is the product of its pole strength $m_p$ and its length $l$, so $M = m_p 	imes l$.

### The Transverse Cut

Now comes the interesting part. The problem states that the magnet is broken into two equal halves, each having half of the original length.

This means we are cutting the magnet transversely (perpendicular to its length).

Let's analyze one of these new halves.

Its new mass $m'$ is exactly half of the original mass, so $m' = \frac{m}{2}$.

Its new length $l'$ is also half of the original length, so $l' = \frac{l}{2}$.

### Recalculating the Parameters

What happens to the magnetic moment?

Because the cut is transverse, the pole strength $m_p$ remains completely unaffected.

However, the length is halved, so the new magnetic moment is $M' = m_p 	imes \frac{l}{2} = \frac{M}{2}$.

Now, let's calculate the new moment of inertia $I'$.

This is where many students make a **silly mistake** by just halving the original moment of inertia.

We must use the formula with the new mass and new length: $I' = \frac{m'(l')^2}{12} = \frac{(\frac{m}{2})(\frac{l}{2})^2}{12}$.

Simplifying this, we get $I' = \frac{1}{8} \left(\frac{ml^2}{12}ight) = \frac{I}{8}$.

The moment of inertia drops by a factor of 8!

### The Final Calculation

We are now ready to find the new time period $T'$.

Substituting our new parameters into the time period formula, we get $T' = 2\pi \sqrt{\frac{I'}{M'B}}$.

Plugging in $I' = \frac{I}{8}$ and $M' = \frac{M}{2}$, the equation becomes $T' = 2\pi \sqrt{\frac{I/8}{(M/2)B}}$.

The fraction inside the square root simplifies: $\frac{1/8}{1/2} = \frac{1}{4}$.

So, $T' = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} \left(2\pi \sqrt{\frac{I}{MB}}ight)$.

Since the term in the parentheses is our original time period $T$, we conclude that $T' = \frac{T}{2}$.

Therefore, the ratio $\frac{T'}{T}$ is exactly $\frac{1}{2}$.

Select Answer:

Visualized Solution

Setup of the Problem

Time Period Formula $T$

Initial Parameters $I$ and $M$

Breaking the Magnet

New Mass $m^{'}$ and Length $l^{'}$

New Magnetic Moment $M^{'}$

New Moment of Inertia $I^{'}$

New Time Period Setup $T^{'}$

Calculating the Ratio

Final Answer

The Way Forward

The Sigma Insight: Bar Magnet

Analyzing the Setup

The Anatomy of the Original Magnet

The Transverse Cut

Recalculating the Parameters

The Final Calculation

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