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Chapter 28: Problem 32

A thin rectangular magnet suspended freely has a period 38 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 \(T^{\prime}\), the ratio \(T^{\prime} / T\) is : (a) \(1 / 4\) (b) \(1 / 2 \sqrt{2}\) (c) \(1 / 2\) (d) 2

Short Answer

Expert verified
The ratio \( T^{\prime}/T \) is \(1/(2\sqrt{2})\).

Step by step solution

01

Understanding the Relationship

The period of oscillation for a magnet in a magnetic field is given by the formula \( T = 2\pi \sqrt{\frac{I}{mB}} \), where \( I \) is the moment of inertia, \( m \) is the magnetic moment, and \( B \) is the magnetic field strength. We need to consider how \( I \) and \( m \) change when the magnet is cut into two halves.
02

Moment of Inertia with Halved Length

When a rectangular magnet is cut into two equal halves, each half has half the original length. For a rectangular bar magnet, \( I \) is proportional to the square of the length (\( I \propto \frac{L^2}{12} \), for a uniform rod about an axis perpendicular and through the center). Thus, \( I_{new} = \frac{I}{4} \).
03

Magnetic Moment with Halved Length

The magnetic moment \( m \) is proportional to the product of pole strength and the separation between the poles. When the bar is cut into half, the magnetic moment for each piece becomes half, \( m_{new} = \frac{m}{2} \).
04

New Period of Oscillation Formula

Substitute the new values of \( I \) and \( m \) into the period formula for the new piece: \( T' = 2\pi \sqrt{\frac{I_{new}}{m_{new}B}} = 2\pi \sqrt{\frac{\frac{I}{4}}{\frac{m}{2}B}} = 2\pi \sqrt{\frac{I}{2mB}} = \frac{1}{\sqrt{2}} T \).
05

Calculate the Ratio

The ratio of the new period \( T' \) to the original period \( T \) is then \( \frac{T'}{T} = \frac{1}{\sqrt{2}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a fundamental concept in physics, especially in rotational dynamics. It quantifies an object's resistance to changes in its rotational motion. You can think of it as the rotational equivalent of mass in linear motion.

For a rectangular magnet, the moment of inertia depends on its shape and the axis about which it rotates. In the original exercise, we have a thin rectangular magnet, which can be likened to a long uniform rod. The formula for the moment of inertia in this context is given by:
  • \( I = \frac{1}{12} m L^2 \)
where \( L \) is the length of the rod-like magnet, and \( m \) is the mass.

After cutting the magnet in half, the length of each piece is reduced by half. Since moment of inertia is proportional to the square of the length, the new moment of inertia becomes:
  • \( I_{new} = \frac{I}{4} \)
Recognizing this change is crucial for calculating how other physical properties, like the period of oscillation, will be affected.
Magnetic Moment
The magnetic moment is another key term in the study of magnetism and oscillation. A magnetic moment essentially measures the magnet's strength as well as its orientation within a magnetic field.

The magnetic moment \( m \) of a rectangular magnet is affected by its pole strength and the distance between the poles. In simpler terms, it relates to how strong the magnet is and how far apart the magnetic forces spread.

When the magnet is split into two, the magnetic moment of each piece is also halved because the pole strength remains the same, but the separation between the poles is reduced by half. Thus, the new magnetic moment for each half is:
  • \( m_{new} = \frac{m}{2} \)
Understanding this decrease helps in determining the new period of oscillation for the halved magnet. A reduced magnetic moment essentially means a weaker interaction with the surrounding magnetic field.
Rectangular Magnet
When dealing with a rectangular magnet, understanding its geometric and magnetic properties help in solving related physics problems. A rectangular magnet is essentially a magnet shaped like a rectangle, where its length plays a key role in determining both its moment of inertia and magnetic moment.

In problems involving magnetic oscillation, it is often suspended freely, and its physical attributes directly influence its behavior in a magnetic field. The length of the magnet affects how these properties鈥攎oment of inertia and magnetic moment鈥攁re calculated.
For example, when the original magnet mentioned in the exercise is cut into two pieces, its rectangular shape determines how the moment of inertia and magnetic moment are recalculated. The length diminishes, leading to significant changes in its oscillatory behavior.

These concepts of geometry and magnetic properties are deeply intertwined and crucial for holistically understanding the motions and forces acting on the magnetic piece. By considering its rectangular shape, you can draw valuable insights into how oscillations and other physical phenomena will play out.

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Most popular questions from this chapter

A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of original magnet is \(T_{0}\) in a uniform magnetic field \(B\). Then the time period of each part in the same magnetic field is: (a) \(T_{0} / 2\) (b) \(T_{0} / 3\) (c) \(T_{0} / 4\) (d) none of these

A uniform magnetic needle of strength of each pole is \(98.1\) amp. \(\mathrm{cm}\) is suspended from its centre by a thread. When a mass of \(50 \mathrm{mg}\) is loaded to its upper end, the needle become horizontal, then the vertical component of earth's magnetic induction is : \(\left(g=981 \mathrm{~cm} / \mathrm{sec}^{2}\right)\) (a) \(0.50\) gauss (b) \(0.25\) gauss (c) \(0.05\) gauss (d) \(0.005\) gauss

An iron rod is subjected to cycles of magnetisation at the rate of \(50 \mathrm{~Hz}\). Given the density of the rod is \(8 \times 10^{3}\). \(\mathrm{kg} / \mathrm{m}^{3}\) and specific heat is \(0.11 \times 10^{-3} \mathrm{cal} / \mathrm{kg}^{\circ} \mathrm{C}\). The rise in temperature per minute, if the area inclosed by the B-H loop corresponds to energy of \(10^{-2} \mathrm{~J}\), is : [Assume there is no radiation losses] (a) \(78^{\circ} \mathrm{C}\) (b) \(88^{\circ} \mathrm{C}\) (c) \(8.1^{\circ} \mathrm{C}\) (d) none of these

A magnetic needle lying parallel to a magnetic field requires \(W\) units of work to turn it through \(60^{\circ}\). The torque needed to maintain the needle in this position will be: (a) \(2 \mathrm{~W}\) (b) \(\sqrt{3} \mathrm{~W}\) (c) \(\mathrm{W}\) (d) \(\frac{\sqrt{3}}{2} \mathrm{~W}\)

Inside a long solenoid wounded with 300 turns/metre, an iron rod is placed. An iron rod is \(0.2 \mathrm{~m}\) long, \(10 \mathrm{~mm}\) in diameter and of permeability \(10^{3}\). The magnetic moment of the rod, if \(0.5\) amp of current is passed through the rod, is : (a) \(2.356\) SI unit (b) \(1.335\) SI unit (c) \(3.664\) SI unit (d) \(1.664\) SI unit
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